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Theorem 7.2.6. Let \( A \in {M}_{n} \) be Hermitian and positive semidefinite, let \( r = \operatorname{rank}A \) , and let \( k \in \{ 2,3,\ldots \} \).\n\n(a) There is a unique Hermitian positive semidefinite matrix \( B \) such that \( {B}^{k} = A \).\n\n(b) There is a polynomial \( p \) with real coefficients such ... | Proof. Represent \( A = {U\Lambda }{U}^{ * } \), in which \( U = \left\lbrack \begin{array}{ll} {U}_{1} & {U}_{2} \end{array}\right\rbrack \) is unitary, \( {U}_{1} \in {M}_{n, r},\Lambda = \) \( \operatorname{diag}\left( {{\lambda }_{1},\ldots ,{\lambda }_{r}}\right) \oplus {0}_{n - r} \), and \( {\lambda }_{1},\ldots... | Yes |
Theorem 7.2.7. Let \( A \in {M}_{n} \) be Hermitian.\n\n(a) \( A \) is positive semidefinite if and only if there is a \( B \in {M}_{m, n} \) such that \( A = {B}^{ * }B \) . | Proof. (a) If \( A = {B}^{ * }B \) for some \( B \in {M}_{m, n} \), then \( {x}^{ * }{Ax} = {x}^{ * }{B}^{ * }{Bx} = \parallel {Bx}{\parallel }_{2}^{2} \geq 0 \) . The asserted factorization can be achieved, for example, with \( B = {A}^{1/2} \) and \( m = n \) . | Yes |
Corollary 7.2.8. A Hermitian matrix \( A \) is positive definite if and only if it is \( {}^{ * } \) congruent to the identity. | Proof. This is simply a restatement of (7.2.7). | No |
Corollary 7.2.9 (Cholesky factorization). Let \( A \in {M}_{n} \) be Hermitian. Then \( A \) is positive semidefinite (respectively, positive definite) if and only if there is a lower triangular matrix \( L \in {M}_{n} \) with nonnegative (respectively, positive) diagonal entries such that \( A = L{L}^{ * } \) . If \( ... | Proof. Let \( {A}^{1/2} = {QR} \) be a \( {QR} \) factorization and let \( L = {R}^{ * } \) . Then \( A = {A}^{1/2}{A}^{1/2} = \) \( {R}^{ * }{Q}^{ * }{QR} = {R}^{ * }R = L{L}^{ * } \) . The asserted properties of \( L \) follow from the properties of \( R \) stated in (2.1.14). | No |
Theorem 7.2.10. Let \( {v}_{1},\ldots ,{v}_{m} \) be vectors in an inner product space \( V \) with inner product \( \langle \cdot , \cdot \rangle \), and let \( G = {\left\lbrack \left\langle {v}_{j},{v}_{i}\right\rangle \right\rbrack }_{i, j = 1}^{m} \in {M}_{m} \). Then\n\n(a) \( G \) is Hermitian and positive semid... | Proof. (a) Let \( \parallel \cdot \parallel \) be the norm derived from the given inner product and let \( x = \left\lbrack {x}_{i}\right\rbrack \in \) \( {\mathbf{C}}^{m} \). The properties listed in (5.1.3) ensure that \( G \) is Hermitian and\n\n\[ \n{x}^{ * }{Gx} = \mathop{\sum }\limits_{{i, j = 1}}^{m}\left\langle... | Yes |
Theorem 7.3.1 (Polar decomposition). Let \( A \in {M}_{n, m} \). (a) If \( n < m \), then \( A = {PU} \), in which \( P \in {M}_{n} \) is positive semidefinite and \( U \in {M}_{n, m} \) has orthonormal rows. The factor \( P = {\left( A{A}^{ * }\right) }^{1/2} \) is uniquely determined; it is a polynomial in \( A{A}^{ ... | Proof. We adopt the notation of (2.6.3), which ensures that there are unitary matrices \( V \in {M}_{n} \) and \( W \in {M}_{m} \), and a nonnegative diagonal matrix \( \sum \in {M}_{n, m} \) with a special structure, such that \( A = {V\sum }{W}^{ * } \). Let \( q = \min \{ n, m\} \) and let \( {\sum }_{q} \in {M}_{q}... | Yes |
Theorem 7.3.2. Let \( A \in {M}_{n, m} \), let \( q = \min \{ n, m\} \), and let \( r = \operatorname{rank}A \) . Suppose that \( {A}^{ * }A = {W\Lambda }{W}^{ * } \), in which \( W \in {M}_{m} \) is unitary, \( \Lambda = \operatorname{diag}\left( {{\sigma }_{1}^{2},\ldots ,{\sigma }_{r}^{2}}\right) \oplus {0}_{m - r} ... | Proof. (a) Let \( D = {\sum }_{r} \oplus {I}_{m - r} \in {M}_{m} \) and partition \( X = {AW}{D}^{-1} = \left\lbrack {{V}_{1}Z}\right\rbrack \in {M}_{n, m} \), in which \( {V}_{1} \in {M}_{n, r} \) . Then \( {X}^{ * }X = {D}^{-1}{W}^{ * }{A}^{ * }{AW}{D}^{-1} = {D}^{-1}\Lambda {D}^{-1} = {I}_{r} \oplus {0}_{m - r} \), ... | Yes |
Theorem 7.3.3. Let \( A \in {M}_{n, m} \), let \( q = \min \{ n, m\} \), let \( {\sigma }_{1} \geq \cdots \geq {\sigma }_{q} \) be the ordered singular values of \( A \), and define the Hermitian matrix\n\n\[ \mathcal{A} = \left\lbrack \begin{matrix} 0 & A \\ {A}^{ * } & 0 \end{matrix}\right\rbrack \]\n\nThe ordered ei... | Proof. Suppose that \( n \geq m \) and let \( A = {V\sum }{W}^{ * } \) be a singular value decomposition, in which \( \sum = {\left\lbrack {\sum }_{m}0\right\rbrack }^{T} \in {M}_{n, m} \) . Write the left unitary factor as \( V = \left\lbrack \begin{array}{ll} {V}_{1} & {V}_{2} \end{array}\right\rbrack \in {M}_{n} \),... | Yes |
Corollary 7.3.5. Let \( A, B \in {M}_{n, m} \) and let \( q = \min \{ m, n\} \) . Let \( {\sigma }_{1}\left( A\right) \geq \cdots \geq {\sigma }_{q}\left( A\right) \) and \( {\sigma }_{1}\left( B\right) \geq \cdots \geq {\sigma }_{q}\left( B\right) \) be the nonincreasingly ordered singular values of \( A \) and \( B \... | Proof. (a) Let \( E = A - B \) and apply (6.3.4.1) to \( \mathcal{A} = \left\lbrack \begin{matrix} 0 & A \\ {A}^{ * } & 0 \end{matrix}\right\rbrack \) and \( \mathcal{E} = \left\lbrack \begin{matrix} 0 & E \\ {E}^{ * } & 0 \end{matrix}\right\rbrack \) .\n\n(b) Apply (6.3.9) to \( \mathcal{A} \) and \( \mathcal{E} \) ; ... | No |
Corollary 7.3.6. Let \( A \in {M}_{n, m} \), let \( q = \min \{ m, n\} \), and let \( \widehat{A} \) be a matrix obtained from A by deleting any one of its columns or rows. Let \( {\sigma }_{1} \geq \cdots \geq {\sigma }_{q} \) and \( {\widehat{\sigma }}_{1} \geq \cdots \geq {\widehat{\sigma }}_{q} \) denote the respec... | Proof. Let \( \mathcal{A} = \left\lbrack \begin{matrix} 0 & A \\ {A}^{ * } & 0 \end{matrix}\right\rbrack \) . Deleting row \( i \) from \( A \) corresponds to deleting row \( i \) and column \( i \) from \( \widetilde{\mathcal{A}} \) ; deleting column \( j \) from \( A \) corresponds to deleting row \( n + j \) and col... | Yes |
Theorem 7.3.8. Let \( A \in {M}_{n, m} \), let \( q = \min \{ n, m\} \), let \( {\sigma }_{1}\left( A\right) \geq \cdots \geq {\sigma }_{q}\left( A\right) \) be the ordered singular values of \( A \), and let \( k \in \{ 1,\ldots, q\} \) . Then\n\n\[ \n{\sigma }_{k}\left( A\right) = \mathop{\min }\limits_{{\{ S : \dim ... | Proof. These characterizations follow from (4.2.7) and (4.2.8). If \( {\lambda }_{1} \leq {\lambda }_{2} \leq \cdots \leq {\lambda }_{m} \) are the ordered eigenvalues of the positive semidefinite Hermitian matrix \( {A}^{ * }A \), then \( {\sigma }_{k}^{2}\left( A\right) = {\lambda }_{m - k + 1}\left( {{A}^{ * }A}\rig... | Yes |
Theorem 7.3.11. Let \( n, p \), and \( q \) be positive integers with \( p \leq q \) . Let \( A \in {M}_{p, n} \) and \( B \in {M}_{q, n} \) . Then \( {A}^{ * }A = {B}^{ * }B \) if and only if there is a \( V \in {M}_{q, p} \) with orthonormal columns such that \( B = {VA} \) . If \( A \) and \( B \) are real, then \( ... | Proof. If \( B = {VA} \), then \( {B}^{ * }B = {A}^{ * }{V}^{ * }{VA} = {A}^{ * }A \) . Conversely, if \( {A}^{ * }A = {B}^{ * }B \), then use (7.3.2) and its notation to write \( A = {V}_{1}{\sum }_{r}{W}_{1}^{ * } \) and \( B = {V}_{2}{\sum }_{r}{W}_{1}^{ * } \), in which \( {V}_{1} \in {M}_{p, r} \) and \( {V}_{2} \... | Yes |
Lemma 7.5.2. Let \( A, B \in {M}_{n} \) and \( x, y \in {\mathbf{C}}^{n} \) be given. Let \( \operatorname{diag}x \) and \( \operatorname{diag}y \) be the \( n \) -by- \( n \) diagonal matrices whose respective main diagonal entries are the respective entries of \( x \) and \( y \) ; see (0.9.1). Then\n\n\[ \n{x}^{ * }... | Proof. Let \( A = \left\lbrack {a}_{ij}\right\rbrack, B = \left\lbrack {b}_{ij}\right\rbrack, x = \left\lbrack {x}_{i}\right\rbrack \), and \( y = \left\lbrack {y}_{i}\right\rbrack \) . Then (diag \( \bar{x} \) ) \( A = \left\lbrack {{\bar{x}}_{i}{a}_{ij}}\right\rbrack \) and \( B \) diag \( y = \left\lbrack {{b}_{ij}{... | Yes |
Theorem 7.5.3. Let \( A, B \in {M}_{n} \) be positive semidefinite.\n\n(a) \( A \circ B \) is positive semidefinite.\n\n(b) If \( A \) is positive definite and every main diagonal entry of \( B \) is positive, then \( A \circ B \) is positive definite.\n\n(c) If both \( A \) and \( B \) are positive definite, then \( A... | Proof. Let \( A = \left\lbrack {a}_{ij}\right\rbrack, B = \left\lbrack {b}_{ij}\right\rbrack \), and \( x = \left\lbrack {x}_{i}\right\rbrack \) .\n\n(a) Let \( C = \left( {\operatorname{diag}x}\right) {\bar{B}}^{1/2} \) and use the preceding lemma to compute\n\n\[{x}^{ * }\left( {A \circ B}\right) x = \operatorname{tr... | Yes |
Theorem 7.5.4 (Moutard). Let \( A = \left\lbrack {a}_{ij}\right\rbrack \in {M}_{n} \) . Then \( A \) is positive semidefinite if and only if \( \operatorname{tr}\left( {A{B}^{T}}\right) = \mathop{\sum }\limits_{{i, j = 1}}^{n}{a}_{ij}{b}_{ij} \geq 0 \) for every positive semidefinite \( B = \left\lbrack {b}_{ij}\right\... | Proof. Suppose that \( A \) and \( B \) are positive semidefinite, and let \( e \in {\mathbf{C}}^{n} \) be the all-ones vector. Then \( \operatorname{diag}\left( e\right) = I \) and \( \operatorname{tr}\left( {A{B}^{T}}\right) = \operatorname{tr}\left( {\left( {\operatorname{diag}e}\right) A\left( {\operatorname{diag}e... | Yes |
Theorem 7.5.9. Let \( A = \left\lbrack {a}_{ij}\right\rbrack \in {M}_{n} \) be positive semidefinite.\n\n(a) The Hadamard powers \( {A}^{\left( k\right) } = \left\lbrack {a}_{ij}^{k}\right\rbrack \) are positive semidefinite for all \( k = 1,2,\ldots \) ; they are positive definite if \( A \) is positive definite.\n\n(... | Proof. Only the assertions about positive definiteness require justification. The assertion in (a) follows from (7.5.3) and an induction. The assertion in (b) follows from (a). See (7.5.P18 to P21) for a proof of the assertion in (c). | No |
Theorem 7.6.1. Let \( A, B \in {M}_{n} \) be Hermitian.\n\n(a) If \( A \) is positive definite, then there is a nonsingular \( S \in {M}_{n} \) such that \( A = {SI}{S}^{ * } \) and \( B = {S}^{- * }\Lambda {S}^{-1} \), in which \( \Lambda \) is real diagonal. The inertias of \( B \) and \( \Lambda \) are the same, so ... | Proof. (a) Theorem 4.5.7 ensures that there is a nonsingular \( T \in {M}_{n} \) such that \( {T}^{-1}A{T}^{- * } = I \) . The matrix \( {T}^{ * }{BT} \) is Hermitian, so there is a unitary \( U \in {M}_{n} \) such that \( {U}^{ * }\left( {{T}^{ * }{BT}}\right) U = \Lambda \) is diagonal. Let \( S = {TU} \) . Then \( {... | Yes |
Corollary 7.6.2. Let \( A, B \in {M}_{n} \) be Hermitian.\n\n(a) If \( A \) is positive definite, then \( {AB} \) is diagonalizable and has real eigenvalues. If, in addition, \( B \) is positive definite or positive semidefinite, then \( \Lambda \) has positive or nonnegative eigenvalues, respectively.\n\n(b) If \( A \... | Proof. (a) Use part (a) of the preceding theorem to represent \( A = S{S}^{ * } \) and \( B = \) \( {S}^{- * }\Lambda {S}^{-1} \) . Then \( {AB} = S{S}^{ * }{S}^{- * }\Lambda {S}^{-1} = {S\Lambda }{S}^{-1} \).\n\n(b) Use part (b) of the preceding theorem to represent \( A = S\left( {{I}_{r} \oplus {0}_{n - r}}\right) {... | Yes |
Theorem 7.6.3. Let \( A, B \in {M}_{n} \) be Hermitian, and assume that \( A \) is positive semidefinite and singular. Then \( {AB} \) is similar to \( \Lambda \oplus N \), in which \( \Lambda \) is real diagonal and \( N = {J}_{2}\left( 0\right) \oplus \cdots \oplus {J}_{2}\left( 0\right) \) is a direct sum of 2-by-2 ... | Proof. Choose a nonsingular \( S \) such that \( {S}^{-1}A{S}^{- * } = {I}_{r} \oplus {0}_{n - r} \) and partition \( {S}^{ * }{BS} = \) \( \left\lbrack {B}_{ij}\right\rbrack \) conformally to \( {I}_{r} \oplus {0}_{n - r} \) . Then \( {S}^{-1}{ABS} = \left( {{S}^{-1}A{S}^{- * }}\right) \left( {{S}^{ * }{BS}}\right) = ... | Yes |
Theorem 7.6.4. Let \( A, B \in {M}_{n} \) be Hermitian.\n\n(a) If \( A \) is positive definite, then there is a nonsingular \( S \in {M}_{n} \) such that \( A = {SI}{S}^{ * } \) and \( B = {S\Lambda }{S}^{ * } \), in which \( \Lambda \) is real diagonal. The inertias of \( B \) and \( \Lambda \) are the same, so \( \La... | Proof. (a) Choose a nonsingular \( T \in {M}_{n} \) such that \( {T}^{-1}A{T}^{- * } = I \) . Choose a unitary \( U \in {M}_{n} \) such that \( {U}^{ * }\left( {{T}^{-1}B{T}^{- * }}\right) U = \Lambda \) is diagonal. Let \( S = {TU} \) . Then \( {S}^{-1}A{S}^{- * } = \) \( {U}^{ * }{T}^{-1}A{T}^{- * }U = {U}^{ * }{IU} ... | Yes |
Theorem 7.6.5. Let \( A, B \in {M}_{n} \) . If \( A \) is positive definite and \( B \) is complex symmetric, then there is a nonsingular \( S \in {M}_{n} \) such that \( A = {SI}{S}^{ * } \) and \( B = {S\Lambda }{S}^{T} \), in which \( \Lambda \) is nonnegative diagonal. The main diagonal entries of \( {\Lambda }^{2}... | Proof. Choose a nonsingular matrix \( R \in {M}_{n} \) such that \( {R}^{-1}A{R}^{- * } = I \) . Use (4.4.4(c)) to choose a unitary \( U \in {M}_{n} \) such that \( {U}^{ * }{R}^{-1}B{R}^{-T}\bar{U} = \Lambda \) is nonnegative diagonal. Let \( S = {RU} \) . Then \( {S}^{-1}A{S}^{- * } = {U}^{ * }{R}^{-1}A{R}^{- * }U = ... | Yes |
Theorem 7.6.6. The function \( f\left( A\right) = \log \det A \) is a strictly concave function on the convex set of positive definite Hermitian matrices in \( {M}_{n} \) . | Proof. Let \( A, B \in {M}_{n} \) be positive definite. We must show that\n\n\[ \log \det \left( {{\alpha A} + \left( {1 - \alpha }\right) B}\right) \geq \alpha \log \det A + \left( {1 - \alpha }\right) \log \det B \]\n\n(7.6.7)\n\nfor all \( \alpha \in \left( {0,1}\right) \), with equality if and only if \( A = B \) .... | Yes |
Theorem 7.6.10. The function \( f\left( A\right) = \operatorname{tr}{A}^{-1} \) is a strictly convex function on the convex set of positive definite Hermitian matrices in \( {M}_{n} \) . | Proof. Let \( A, B \in {M}_{n} \) be positive definite. We must show that\n\n\[ \operatorname{tr}{\left( \alpha A + \left( 1 - \alpha \right) B\right) }^{-1} \leq \alpha \operatorname{tr}{A}^{-1} + \left( {1 - \alpha }\right) \operatorname{tr}{B}^{-1} \]\n\nfor all \( \alpha \in \left( {0,1}\right) \), with equality if... | Yes |
Theorem 7.7.2. Let \( A, B \in {M}_{n} \) be Hermitian and let \( S \in {M}_{n, m} \) . Then\n\n(a) if \( A \succcurlyeq B \), then \( {S}^{ * }{AS} \succcurlyeq {S}^{ * }{BS} \)\n\n(b) if \( \operatorname{rank}S = m \), then \( A \succ B \) implies \( {S}^{ * }{AS} \succ {S}^{ * }{BS} \)\n\n(c) if \( m = n \) and \( S... | Proof. (a) If \( \left( {A - B}\right) \succcurlyeq 0 \), then (7.1.8(a)) ensures that \( {S}^{ * }\left( {A - B}\right) S = {S}^{ * }{AS} - {S}^{ * }{BS} \) \( \succcurlyeq 0 \).\n\n(b) This assertion follows in the same way from (7.1.8(b)).\n\n(c) If \( {S}^{ * }{AS} \succ {S}^{ * }{BS} \), then \( {S}^{- * }\left( {... | Yes |
Theorem 7.7.3. Let \( A, B \in {M}_{n} \) be Hermitian and suppose that \( A \) is positive definite.\n\n(a) If \( B \) is positive semidefinite, then \( A \succcurlyeq B \) (respectively, \( A \succ B \) ) if and only if \( \rho \left( {{A}^{-1}B}\right) \leq 1 \) (respectively, \( \rho \left( {{A}^{-1}B}\right) < 1 \... | Proof. (a) It follows from (a) and (d) of the preceding theorem that \( A \succcurlyeq B \) if and only if \( I = {A}^{-1/2}A{A}^{-1/2} \succcurlyeq {A}^{-1/2}B{A}^{-1/2} \) if and only if \( 1 \geq {\sigma }_{1}\left( {{A}^{-1/2}B{A}^{-1/2}}\right) \) . But \( {A}^{-1/2}B{A}^{-1/2} \) is positive semidefinite, so\n\n\... | Yes |
Corollary 7.7.4. Let \( A, B \in {M}_{n} \) be Hermitian. Let \( {\lambda }_{1}\left( A\right) \leq \cdots \leq {\lambda }_{n}\left( A\right) \) and \( {\lambda }_{1}\left( B\right) \leq \cdots \leq {\lambda }_{n}\left( B\right) \) be the ordered eigenvalues of \( A \) and \( B \), respectively.\n\n(a) If \( A \succ 0 ... | Proof. (a) The preceding theorem ensures that \( A \succcurlyeq B \) if and only if \( \rho \left( {{A}^{-1}B}\right) = \rho \left( {B{A}^{-1}}\right) \leq 1 \) if and only if \( {B}^{-1} \succcurlyeq {A}^{-1} \) . | Yes |
Lemma 7.7.6. Let \( X \in {M}_{p, q} \) and let \( K = \left\lbrack \begin{matrix} {I}_{p} & X \\ {X}^{ * } & {I}_{q} \end{matrix}\right\rbrack \in {M}_{p + q} \) . Then\n\n(a) \( K \) is positive definite if and only if \( X \) is a strict contraction\n\n(b) \( K \) is positive semidefinite if and only if \( X \) is a... | Proof. The identity (7.7.5) and (7.7.2(d)) ensure that \( \left\lbrack \begin{matrix} {I}_{p} & X \\ {X}^{ * } & {I}_{q} \end{matrix}\right\rbrack \succ 0 \) if and only if \( {I}_{q} - \) \( {X}^{ * }X \succ 0 \) if and only if \( {\sigma }_{1}\left( X\right) < 1 \) . The assertion in (b) follows in a similar fashion.... | Yes |
Theorem 7.7.7. Let \( H = \left\lbrack \begin{matrix} A & B \\ {B}^{ * } & C \end{matrix}\right\rbrack \in {M}_{p + q} \) be Hermitian with \( A \in {M}_{p} \) and \( C \in {M}_{q} \) . The following are equivalent:\n\n(a) \( H \) is positive definite.\n\n(b) \( A \) is positive definite and \( C - {B}^{ * }{A}^{-1}B \... | Proof. (a) \( \Leftrightarrow \) (b): Demonstrated in our discussion of (7.7.5).\n\n(b) \( \Rightarrow \) (c): Follows from (7.7.3(a)).\n\n(c) \( \Leftrightarrow \) (d): Let \( X = {A}^{-1/2}B{C}^{-1/2} \) . Then \( 1 > \rho \left( {{B}^{ * }{A}^{-1}B{C}^{-1}}\right) = \rho \left( {{C}^{-1/2}{B}^{ * }{A}^{-1}}\right. \... | Yes |
Lemma 7.7.8. Let \( A \in {M}_{n} \) be positive semidefinite and singular, and let \( {A}_{k} = A + \) \( {k}^{-1}{I}_{n} \) for each \( k = 1,2,\ldots \) . Let \( {X}_{k} \in {M}_{m, n} \) be a contraction for each \( k = 1,2,\ldots \) . Then\n\n(a) each \( {A}_{k} \) is positive definite and \( \mathop{\lim }\limits... | Proof. (a) Let \( r = \operatorname{rank}A \), let \( {\lambda }_{1},\ldots ,{\lambda }_{r} \) be the positive eigenvalues of \( A \), and let \( A = \) \( U\left( {\operatorname{diag}\left( {{\lambda }_{1},\ldots ,{\lambda }_{r}}\right) \oplus {0}_{n - r}}\right) {U}^{ * } \) be a spectral decomposition. Then continui... | Yes |
Theorem 7.7.9. Let \( H = \left\lbrack \begin{matrix} A & B \\ {B}^{ * } & C \end{matrix}\right\rbrack \in {M}_{p + q} \) be Hermitian, with \( A \in {M}_{p} \) and \( C \in {M}_{q} \) . The following two statements are equivalent:\n\n(a) \( H \) is positive semidefinite.\n\n(b) A and \( C \) are positive semidefinite,... | Proof. (a) \( \Rightarrow \) (b): Consider \( {H}_{k} = H + {k}^{-1}{I}_{n} \) for each \( k = 1,2,\ldots \) Then \( {H}_{k} \) , \( {A}_{k} = A + {k}^{-1}{I}_{p} \), and \( {C}_{k} = C + {k}^{-1}{I}_{q} \) are positive definite for each \( k = 1,2,\ldots \) , so (7.7.7(e)) ensures that there is a contraction \( {X}_{k... | Yes |
Corollary 7.7.10. Let \( A, C \in {M}_{p} \) be Hermitian.\n\n(a) If \( \left\lbrack \begin{array}{ll} A & {I}_{p} \\ {I}_{p} & C \end{array}\right\rbrack \succcurlyeq 0 \), then \( A \succ 0, C \succ 0, A \succcurlyeq {C}^{-1} \), and \( C \succcurlyeq {A}^{-1} \) . | Proof. (a) The hypothesis ensures that \( A \succcurlyeq 0 \) and \( C \succcurlyeq 0 \) . Theorem 7.7.9(b) ensures that there is a contraction \( X \) such that \( I = {A}^{1/2}X{C}^{1/2} \), so both \( {A}^{1/2} \) and \( {C}^{1/2} \) (and hence both \( A \) and \( C \) ) are nonsingular. Then \( A \succcurlyeq {C}^{... | Yes |
Theorem 7.7.11. Let \( A \in {M}_{p} \) and \( C \in {M}_{q} \) be positive semidefinite and let \( B \in {M}_{p, q} \) . The following four statements are equivalent:\n\n(a) \( \left( {{x}^{ * }{Ax}}\right) \left( {{y}^{ * }{Cy}}\right) \geq {\left| {x}^{ * }By\right| }^{2} \) for all \( x \in {\mathbf{C}}^{p} \) and ... | Proof. (a) \( \Rightarrow \) (b): This implication follows from the arithmetic-geometric mean inequality: \( \frac{1}{2}\left( {{x}^{ * }{Ax} + {y}^{ * }{Cy}}\right) \geq {\left( {x}^{ * }Ax\right) }^{1/2}{\left( {y}^{ * }Cy\right) }^{1/2} \geq \left| {{x}^{ * }{By}}\right| \) . (b) \( \Rightarrow \) (c): Let \( z = {\... | Yes |
Corollary 7.7.12. Let \( A \in {M}_{n} \) be positive semidefinite and let \( B \in {M}_{n} \) be Hermitian. The following four statements are equivalent:\n\n(a) \( {x}^{ * }{Ax} \geq \left| {{x}^{ * }{Bx}}\right| \) for all \( x \in {\mathbf{C}}^{n} \) .\n\n(b) \( {x}^{ * }{Ax} + {y}^{ * }{Ay} \geq 2\left| {{x}^{ * }{... | Proof. (a) \( \Rightarrow \) (b): Let \( x, y \in {\mathbf{C}}^{n} \) and use the inequality in (a), the triangle inequality, and Hermicity of \( B \) to compute\n\n\[ 2\left( {{x}^{ * }{Ax} + {y}^{ * }{Ay}}\right) = {\left( x + y\right) }^{ * }A\left( {x + y}\right) + {\left( x - y\right) }^{ * }A\left( {x - y}\right)... | Yes |
Corollary 7.7.13. Let \( A, B \in {M}_{n} \) be positive semidefinite. The following statements are equivalent:\n\n(a) \( A \succcurlyeq B \) .\n\n(b) \( \left\lbrack \begin{array}{ll} A & B \\ B & A \end{array}\right\rbrack \succcurlyeq 0 \) .\n\n(c) There is a positive semidefinite contraction \( X \in {M}_{n} \) suc... | Proof. Since \( A \succcurlyeq B \) if and only if \( {x}^{ * }{Ax} \geq {x}^{ * }{Bx} \) for all \( x \in {\mathbf{C}}^{n} \), the asserted equivalences follow from (7.7.12) in the special case in which the Hermitian matrix \( B \) is positive semidefinite, provided we can choose \( X \) to be positive semidefinite in... | Yes |
Corollary 7.7.14. Let \( A, B, C, D \in {M}_{n} \) be Hermitian and suppose that \( A \) and \( C \) are positive semidefinite. If \( {x}^{ * }{Ax} \geq \left| {{x}^{ * }{Bx}}\right| \) and \( {x}^{ * }{Cx} \geq \left| {{x}^{ * }{Dx}}\right| \) for all \( x \in {\mathbf{C}}^{n} \), then \( {x}^{ * }\left( {A \circ C}\r... | Proof. The hypotheses and the implication (a) \( \Rightarrow \) (c) in (7.7.12) ensure that \( \left\lbrack \begin{array}{ll} A & B \\ B & A \end{array}\right\rbrack \succcurlyeq \) 0 and \( \left\lbrack \begin{array}{ll} C & D \\ D & C \end{array}\right\rbrack \succcurlyeq 0 \), so (7.5.3) tells us that \( \left\lbrac... | Yes |
Theorem 7.7.15. Let \( H \in {M}_{n} \) be positive definite and let \( \alpha \subset \{ 1,\ldots, n\} \) . Then \( {H}^{-1}\left\lbrack \alpha \right\rbrack \succcurlyeq {\left( H\left\lbrack \alpha \right\rbrack \right) }^{-1}. \) | Proof. Since a permutation congruence of a positive definite matrix is positive definite, we may assume that \( H = \left\lbrack \begin{matrix} A & B \\ {B}^{ * } & C \end{matrix}\right\rbrack ,\alpha = \{ 1,\ldots, k\} \), and \( H\left\lbrack \alpha \right\rbrack = A \) . The identity (0.7.3.1) ensures that \( {H}^{-... | Yes |
Theorem 7.7.16. Let \( A \in {M}_{n} \) be positive definite. The following matrices are positive semidefinite and singular:\n\n(a) \( \left\lbrack \begin{matrix} A & X \\ {X}^{ * } & {X}^{ * }{A}^{-1}X \end{matrix}\right\rbrack \) for any \( X \in {M}_{n, m} \).\n\n(b) \( \left\lbrack \begin{matrix} A & {I}_{n} \\ {I}... | Proof. Use (7.7.9). In each case, we verify that for \( \left\lbrack \begin{matrix} A & B \\ {B}^{ * } & C \end{matrix}\right\rbrack \) we have \( A \) nonsingular and \( C = {B}^{ * }{A}^{-1}B \).\n\n(a) \( {X}^{ * }{A}^{-1}X - {X}^{ * }{A}^{-1}X = 0 \).\n\n(b) Take \( X = {I}_{n} \) in (a).\n\n(c) Take \( X = A \) in... | Yes |
Theorem 7.7.17. Let \( A, B \in {M}_{n} \) be positive definite. Then\n\n(a) \( {A}^{-1} \circ {B}^{-1} \succcurlyeq {\left( A \circ B\right) }^{-1} \) | Proof. (a) The preceding theorem and the Schur product theorem ensure that\n\n\[ \left\lbrack \begin{matrix} A & {I}_{n} \\ {I}_{n} & {A}^{-1} \end{matrix}\right\rbrack \circ \left\lbrack \begin{matrix} B & {I}_{n} \\ {I}_{n} & {B}^{-1} \end{matrix}\right\rbrack = \left\lbrack \begin{matrix} A \circ B & {I}_{n} \\ {I}_... | Yes |
Theorem 7.7.18. Let \( A, B \in {M}_{n} \) be positive definite. Then \( {\lambda }_{\min }\left( {A \circ B}\right) \geq \) \( \max \left\{ {{\lambda }_{\min }\left( {AB}\right) ,{\lambda }_{\min }\left( {A{B}^{T}}\right) }\right\} \) . | Proof. Since \( {\lambda }_{\min }\left( {{B}^{1/2}A{B}^{1/2}}\right) = {\lambda }_{\min }\left( {AB}\right) \), we have \( {B}^{1/2}A{B}^{1/2} \succcurlyeq {\lambda }_{\min }\left( {AB}\right) I \), which is equivalent to \( A \succcurlyeq {\lambda }_{\min }\left( {AB}\right) {B}^{-1} \) . It follows from (7.7.17(c)) ... | Yes |
Theorem 7.8.1 (Hadamard’s inequality). Let \( A = \left\lbrack {a}_{ij}\right\rbrack \in {M}_{n} \) be positive definite. Then\n\n\[ \det A \leq {a}_{11}\cdots {a}_{nn} \]\n\nwith equality if and only if \( A \) is diagonal. | Proof. Since \( A \) is positive definite, it has positive main diagonal entries and is diagonally *congruent to a correlation matrix. Let \( D = \operatorname{diag}\left( {{a}_{11}^{1/2},\ldots ,{a}_{nn}^{1/2}}\right) \) and define \( C = {D}^{-1}A{D}^{-1} \), which is also positive definite; it has unit diagonal entr... | Yes |
Corollary 7.8.3 (Hadamard’s inequality). Let \( B \in {M}_{n} \) be nonsingular and partition \( B = \left\lbrack \begin{array}{lll} {b}_{1} & \ldots & {b}_{n} \end{array}\right\rbrack \) and \( {B}^{ * } = \left\lbrack \begin{array}{lll} {\beta }_{1} & \ldots & {\beta }_{n} \end{array}\right\rbrack \) according to the... | Proof. Apply (7.8.2) to the positive definite matrix \( A = {B}^{ * }B : \det A = {\left| \det B\right| }^{2} \), and the main diagonal entries of \( A \) are \( {\begin{Vmatrix}{b}_{1}\end{Vmatrix}}_{2}^{2},\ldots ,{\begin{Vmatrix}{b}_{n}\end{Vmatrix}}_{2}^{2} \) . The columns of \( B \) are orthogonal if and only if ... | Yes |
Theorem 7.8.5 (Fischer's inequality). Suppose that the partitioned Hermitian matrix \[ H = \left\lbrack \begin{matrix} A & B \\ {B}^{ * } & C \end{matrix}\right\rbrack \in {M}_{p + q},\;A \in {M}_{p}\text{ and }C \in {M}_{q} \] is positive definite. Then \[ \det H \leq \left( {\det A}\right) \left( {\det C}\right) \] | Proof. Let \( A = {U\Lambda }{U}^{ * } \) and \( C = {V\Gamma }{V}^{ * } \) be spectral decompositions, in which \( U \) and \( V \) are unitary and \( \Lambda = \operatorname{diag}\left( {{\lambda }_{1},\ldots ,{\lambda }_{p}}\right) \) and \( \Gamma = \operatorname{diag}\left( {{\gamma }_{1},\ldots ,{\gamma }_{q}}\ri... | Yes |
Lemma 7.8.8. Let \( B \in {M}_{m} \) be positive definite. Let \( \alpha ,\beta \subset \{ 1,\ldots, m\} \) . Suppose that \( {\alpha }^{c} \) and \( {\beta }^{c} \) are nonempty and disjoint, and \( \alpha \cup \beta = \{ 1,\ldots, m\} \) . Then \( \det B\left\lbrack {{\alpha }^{c} \cup {\beta }^{c}}\right\rbrack \leq... | Proof. There is no loss of generality to assume that \( {\beta }^{c} = \{ 1,\ldots, k\} ,{\alpha }^{c} = \{ j,\ldots, m\} \) , and \( 1 < k < j < m \) . Then \( A\left\lbrack {\alpha }^{c}\right\rbrack \) and \( A\left\lbrack {\beta }^{c}\right\rbrack \) are complementary principal submatrices of \( A\left\lbrack {{\al... | Yes |
Theorem 7.8.9 (Koteljanskii’s inequality). Let \( A \in {M}_{n} \) be positive definite and let \( \alpha ,\beta \subset \{ 1,\ldots, n\} \) . Then\n\n\[ \left( {\det A\left\lbrack {\alpha \cup \beta }\right\rbrack }\right) \left( {\det A\left\lbrack {\alpha \cap \beta }\right\rbrack }\right) \leq \left( {\det A\left\l... | Proof. There is no loss of generality to assume that \( \alpha \cup \beta = \{ 1,\ldots, n\} \) (if not, work within the principal submatrix \( A\left\lbrack {\alpha \cup \beta }\right\rbrack \) )). We may also assume that \( \alpha \cap \beta \) is nonempty (if it is empty, then \( \beta = {\alpha }^{c} \) and (7.8.10... | Yes |
Theorem 7.8.11 (Szász’s inequality). Let \( A \in {M}_{n} \) be positive definite. Then\n\n\[ \n{P}_{k + 1}{\left( A\right) }^{{\left( \begin{matrix} n - 1 \\ k \end{matrix}\right) }^{-1}} \leq {P}_{k}{\left( A\right) }^{{\left( \begin{matrix} n - 1 \\ k - 1 \end{matrix}\right) }^{-1}}\;\text{ for each }\;k = 1,\ldots,... | Proof. The identity \( {A}^{-1} = {\left( \det A\right) }^{-1} \) adj \( A \) reminds us that each diagonal entry of \( {A}^{-1} \) is the ratio of a principal minor of \( A \) of size \( n - 1 \) and det \( A \) . Thus, an application of (7.8.2) to the positive definite matrix \( {A}^{-1} \) gives the inequality\n\n\[... | Yes |
Lemma 7.8.15. Let \( A = \left\lbrack {a}_{ij}\right\rbrack \in {M}_{n} \) be positive semidefinite and partitioned as \( A = \) \( \left\lbrack \begin{matrix} {a}_{11} & {x}^{ * } \\ x & {A}_{22} \end{matrix}\right\rbrack \), in which \( {A}_{22} \in {M}_{n - 1} \) . Define\n\n\[ \alpha \left( A\right) = \left\{ \begi... | Proof. There is nothing to prove if \( A \) is singular, so assume that \( A \) is positive definite. Apply Sylvester's criterion (7.2.5) to the trailing principal minors. The trailing minors \( \det \left( {\widetilde{A}\left\lbrack {\{ k,\ldots, n\} }\right\rbrack }\right) = \det A\left\lbrack {\{ k,\ldots, n\} }\rig... | Yes |
Theorem 7.8.19 (Ostrowski-Taussky inequality). Let \( H, K \in {M}_{n} \) be Hermitian and let \( A = H + {iK} \) . If \( H \) is positive definite, then\n\n\[ \det H \leq \left| {\det \left( {H + {iK}}\right) }\right| = \left| {\det A}\right| \] | Proof. We have \( A = H\left( {I + i{H}^{-1}K}\right) \), so (7.8.20) is equivalent to the inequality \( \left| {\det \left( {I + i{H}^{-1}K}\right) }\right| \geq 1 \) . Corollary 7.6.2(a) ensures that \( {H}^{-1}K \) is diagonalizable and has real eigenvalues \( {\lambda }_{1},\ldots ,{\lambda }_{n} \) . Then\n\n\[ \l... | Yes |
Theorem 7.8.21 (Minkowski’s determinant inequality). Let \( A, B \in {M}_{n} \) be positive definite. Then\n\n\[ \n{\left( \det A\right) }^{1/n} + {\left( \det B\right) }^{1/n} \leq {\left( \det \left( A + B\right) \right) }^{1/n} \n\]\n\n(7.8.22)\n\nwith equality if and only if \( A = {cB} \) for some \( c > 0 \) . | Proof. Theorem 7.6.4 ensures that there is a nonsingular \( S \in {M}_{n} \) such that \( A = {SI}{S}^{ * } \) and \( B = {S\Lambda }{S}^{ * } \), in which \( \Lambda = \operatorname{diag}\left( {{\lambda }_{1},\ldots ,{\lambda }_{n}}\right) \) is positive diagonal. The assertion (7.8.22) is that\n\n\[ \n{\left( \det S... | Yes |
Theorem 7.8.27 (Fan’s determinant inequality). Let \( H, K \in {M}_{n} \) be Hermitian and let \( A = H + {iK} \) . If \( H \) is positive definite, then\n\n\[ \n{\left( \det H\right) }^{2/n} + {\left| \det K\right| }^{2/n} \leq {\left| \det \left( H + iK\right) \right| }^{2/n} = {\left| \det A\right| }^{2/n} \n\]\n\n(... | Proof. Theorem 7.6.4 ensures that there is a nonsingular \( S \in {M}_{n} \) such that \( H = {SI}{S}^{ * } \) and \( K = {S\Lambda }{S}^{ * } \), in which \( \Lambda = \operatorname{diag}\left( {{\lambda }_{1},\ldots ,{\lambda }_{n}}\right) \) is real diagonal; its diagonal entries are the eigenvalues of the diagonali... | Yes |
Proposition 8.1.8. Let \( A = \left\lbrack {a}_{ij}\right\rbrack \in {M}_{n} \) and \( x = \left\lbrack {x}_{i}\right\rbrack \in {\mathbf{C}}^{n} \) be given.\n\n(a) \( \left| {Ax}\right| \leq \left| A\right| \left| x\right| \) . | Proof. (a) The assertion follows from the triangle inequality:\n\n\[ \n{\left| Ax\right| }_{k} = \left| {\mathop{\sum }\limits_{j}{a}_{kj}{x}_{j}}\right| \leq \mathop{\sum }\limits_{j}\left| {{a}_{kj}{x}_{j}}\right| = \mathop{\sum }\limits_{j}\left| {a}_{kj}\right| \left| {x}_{j}\right| = {\left( \left| A\right| \left|... | Yes |
Theorem 8.1.18. Let \( A, B \in {M}_{n} \) and suppose that \( B \) is nonnegative. If \( \left| A\right| \leq \) \( B \), then \( \rho \left( A\right) \leq \rho \left( \left| A\right| \right) \leq \rho \left( B\right) \) . | Proof. Invoking (8.1.10) and (8.1.12), we have \( \left| {A}^{m}\right| \leq {\left| A\right| }^{m} \leq {B}^{m} \) for each \( m = \) \( 1,2,\ldots \) . Thus,(8.1.16) and (8.1.17) ensure that\n\n\[ \n{\begin{Vmatrix}{A}^{m}\end{Vmatrix}}_{2} \leq {\begin{Vmatrix}{\left| A\right| }^{m}\end{Vmatrix}}_{2} \leq {\begin{Vm... | Yes |
Corollary 8.1.20. Let \( A = \left\lbrack {a}_{ij}\right\rbrack \in {M}_{n} \) be nonnegative.\n\n(a) If \( \widetilde{A} \) is principal submatrix of \( A \), then \( \rho \left( \widetilde{A}\right) \leq \rho \left( A\right) \) . | Proof. (a) If \( r = n \), there is nothing to prove. Suppose that \( 1 \leq r < n \), let \( \widetilde{A} \) be an \( r \) -by- \( r \) principal square submatrix of \( A \), and let \( P \) be a permutation matrix such that \( {PA}{P}^{T} = \left\lbrack \begin{array}{ll} \widehat{A} & B \\ C & D \end{array}\right\rb... | Yes |
Lemma 8.1.21. Let \( A = \left\lbrack {a}_{ij}\right\rbrack \in {M}_{n} \) be nonnegative. Then \( \rho \left( A\right) \leq \parallel A{\parallel }_{\infty } = \) \( \mathop{\max }\limits_{{1 \leq i \leq n}}\mathop{\sum }\limits_{{j = 1}}^{n}{a}_{ij} \) and \( \rho \left( A\right) \leq \parallel A{\parallel }_{1} = \m... | Proof. We know that \( \left| \lambda \right| \leq \rho \left( A\right) \leq \parallel A\parallel \) for any eigenvalue \( \lambda \) of \( A \) and any matrix norm \( \parallel \mid \cdot \parallel \mid \) . If all the row sums of \( A \) are equal, then \( e = {\left\lbrack \begin{array}{lll} 1 & \ldots & 1 \end{arra... | Yes |
Theorem 8.1.22. Let \( A = \left\lbrack {a}_{ij}\right\rbrack \in {M}_{n} \) be nonnegative. Then\n\n\[ \mathop{\min }\limits_{{1 \leq i \leq n}}\mathop{\sum }\limits_{{j = 1}}^{n}{a}_{ij} \leq \rho \left( A\right) \leq \mathop{\max }\limits_{{1 \leq i \leq n}}\mathop{\sum }\limits_{{j = 1}}^{n}{a}_{ij} \]\n\n(8.1.23)\... | Proof. Let \( \alpha = \mathop{\min }\limits_{{1 \leq i \leq n}}\mathop{\sum }\limits_{{j = 1}}^{n}{a}_{ij} \) . If \( \alpha = 0 \), let \( B = 0 \) . If \( \alpha > 0 \), define \( B = \left\lbrack {b}_{ij}\right\rbrack \) by letting each \( {b}_{ij} = \alpha {a}_{ij}{\left( \mathop{\sum }\limits_{{k = 1}}^{n}{a}_{ik... | Yes |
Corollary 8.1.29. Let \( A = \left\lbrack {a}_{ij}\right\rbrack \in {M}_{n} \) be nonnegative and let \( x = \left\lbrack {x}_{i}\right\rbrack \in {\mathbf{R}}^{n} \) be a positive vector. If \( \alpha ,\beta \geq 0 \) are such that \( {\alpha x} \leq {Ax} \leq {\beta x} \), then \( \alpha \leq \rho \left( A\right) \le... | Proof. If \( {\alpha x} \leq {Ax} \), then \( \alpha {x}_{i} \leq {\left( Ax\right) }_{i} \) and \( \alpha \leq \mathop{\min }\limits_{{1 \leq i \leq n}}{x}_{i}^{-1}\mathop{\sum }\limits_{{j = 1}}^{n}{a}_{ij}{x}_{j} \), so the preceding theorem ensures that \( \alpha \leq \rho \left( A\right) \) . If \( {\alpha x} < {A... | Yes |
Corollary 8.1.30. Let \( A \in {M}_{n} \) be nonnegative. If \( x \) is a positive eigenvector of \( A \), then \( \rho \left( A\right), x \) is an eigenpair for \( A \) ; that is, if \( A \geq 0, x > 0 \), and \( {Ax} = {\lambda x} \), then \( \lambda = \rho \left( A\right) \) . | Proof. If \( x > 0 \) and \( {Ax} = {\lambda x} \), then \( \lambda \geq 0 \) and \( {\lambda x} \leq {Ax} \leq {\lambda x} \) . But then (8.1.29) ensures that \( \lambda \leq \rho \left( A\right) \leq \lambda \) . | Yes |
Corollary 8.1.31. Let \( A = \left\lbrack {a}_{ij}\right\rbrack \in {M}_{n} \) be nonnegative. If \( A \) has a positive eigenvector, then\n\n\[ \rho \left( A\right) = \mathop{\max }\limits_{{x > 0}}\mathop{\min }\limits_{{1 \leq i \leq n}}\frac{1}{{x}_{i}}\mathop{\sum }\limits_{{j = 1}}^{n}{a}_{ij}{x}_{j} = \mathop{\m... | Exercise. Prove the preceding corollary. Hint: Use the positive eigenvector \( x \) in (8.1.27). | No |
Corollary 8.1.33. Let \( A = \left\lbrack {a}_{ij}\right\rbrack \in {M}_{n} \) be nonnegative and write \( {A}^{m} = \left\lbrack {a}_{ij}^{\left( m\right) }\right\rbrack \) . If \( A \) has a positive eigenvector \( x = \left\lbrack {x}_{i}\right\rbrack \), then for all \( m = 1,2,\ldots \) and for all \( i = 1,\ldots... | Proof. Let \( x = \left\lbrack {x}_{i}\right\rbrack \) be a positive eigenvector of \( A \) . Then (8.1.30) ensures that \( {Ax} = \) \( \rho \left( A\right) x \), so \( {A}^{m}x = \rho {\left( A\right) }^{m}x \) for each \( m = 1,2,\ldots \) . Since \( {A}^{m} \geq 0 \), for any \( i = 1,\ldots, n \) we have\n\n\[ \rh... | Yes |
Lemma 8.2.1. Let \( A \in {M}_{n} \) be positive. If \( \lambda, x \) is an eigenpair of \( A \) and \( \left| \lambda \right| = \rho \left( A\right) \) , then \( \left| x\right| > 0 \) and \( A\left| x\right| = \rho \left( A\right) \left| x\right| \) . | Proof. The hypotheses ensure that \( z = A\left| x\right| > 0 \) (8.1.14). We have \( z = A\left| x\right| \geq \left| {Ax}\right| = \) \( \left| {\lambda x}\right| = \left| \lambda \right| \left| x\right| = \rho \left( A\right) \left| x\right| \), so \( y = z - \rho \left( A\right) \left| x\right| \geq 0 \) . If \( y ... | Yes |
Theorem 8.2.2. If \( A \in {\mathbf{M}}_{n} \) is positive, there are positive vectors \( x \) and \( y \) such that \( {Ax} = \rho \left( A\right) x \) and \( {y}^{T}A = \rho \left( A\right) {y}^{T}. \) | Proof. There is an eigenpair \( \lambda, x \) of \( A \) with \( \left| \lambda \right| = \rho \left( A\right) \) . The preceding lemma ensures that \( \rho \left( A\right) ,\left| x\right| \) is also an eigenpair of \( A \) and \( \left| x\right| > 0 \) . The assertion about \( y \) follows from considering \( {A}^{T}... | Yes |
Lemma 8.2.3. Let \( A \in {M}_{n} \) be positive. If \( \lambda, x \) is an eigenpair of \( A \) and \( \left| \lambda \right| = \rho \left( A\right) \) , then there is a \( \theta \in \mathbf{R} \) such that \( {e}^{-{i\theta }}x = \left| x\right| > 0 \) . | Proof. The hypothesis is that \( x \in {\mathbf{C}}^{n} \) is nonzero and \( \left| {Ax}\right| = \left| {\lambda x}\right| = \rho \left( A\right) \left| x\right| \) ; (8.2.1) ensures that \( A\left| x\right| = \rho \left( A\right) \left| x\right| \) and \( \left| x\right| > 0 \) . Since \( \left| {Ax}\right| = \rho \l... | Yes |
Theorem 8.2.4. Let \( A \in {M}_{n} \) be positive. If \( \lambda \) is an eigenvalue of \( A \) and \( \lambda \neq \rho \left( A\right) \), then \( \left| \lambda \right| < \rho \left( A\right) \) | Proof. Let \( \lambda, x \) be an eigenpair of \( A \), so \( \left| \lambda \right| \leq \rho \left( A\right) \) . If \( \left| \lambda \right| = \rho \left( A\right) \) ,(8.2.3) ensures that \( w = {e}^{-{i\theta }}x > 0 \) for some \( \theta \in \mathbf{R} \) . Since \( {Aw} = {\lambda w} \) and \( w > 0 \), it foll... | Yes |
Theorem 8.2.5. If \( A \in {M}_{n} \) is positive, then the geometric multiplicity of \( \rho \left( A\right) \) as an eigenvalue of \( A \) is 1 . | Proof. Suppose that \( w, z \in {\mathbf{C}}^{n} \) are nonzero vectors such that \( {Aw} = \rho \left( A\right) w \) and \( {Az} = \rho \left( A\right) z \) . Then \( w = {\alpha z} \) for some \( \alpha \in \mathbf{C} \) . Lemma 8.2.3 ensures that there are real numbers \( {\theta }_{1} \) and \( {\theta }_{2} \) suc... | Yes |
Theorem 8.2.7. Let \( A \in {M}_{n} \) be positive. The algebraic multiplicity of \( \rho \left( A\right) \) as an eigenvalue of \( A \) is \( 1 \) . If \( x \) and \( y \) are the right and left Perron vectors of \( A \), then \( \mathop{\lim }\limits_{{m \rightarrow \infty }}{\left( \rho {\left( A\right) }^{-1}A\righ... | Proof. We know that \( \rho \left( A\right) > 0 \), and that \( x \) and \( y \) are positive vectors such that \( {Ax} = \) \( \rho \left( A\right) x,{y}^{T}A = \rho \left( A\right) {y}^{T} \), and \( {y}^{ * }x = {y}^{T}x = 1 \) . Theorem 1.4.12b ensures that \( \rho \left( A\right) \) has algebraic multiplicity 1, a... | Yes |
Theorem 8.2.9 (Fan). Let \( A = \left\lbrack {a}_{ij}\right\rbrack \in {M}_{n} \) . Suppose that \( B = \left\lbrack {b}_{ij}\right\rbrack \in {M}_{n} \) is nonnegative and \( {b}_{ij} \geq \left| {a}_{ij}\right| \) for all \( i \neq j \) . Then every eigenvalue of \( A \) is in the union of \( n \) discs\n\n\[ \mathop... | Proof. First, assume that \( B > 0 \) . Theorem 8.2.8 ensures that there is a positive vector \( x \) such that \( {Bx} = \rho \left( B\right) x \), and hence\n\n\[ \mathop{\sum }\limits_{{j \neq i}}\left| {a}_{ij}\right| {x}_{j} \leq \mathop{\sum }\limits_{{j \neq i}}{b}_{ij}{x}_{j} = \rho \left( B\right) {x}_{i} - {b... | Yes |
Theorem 8.3.1. If \( A \in {M}_{n} \) is nonnegative, then \( \rho \left( A\right) \) is an eigenvalue of \( A \) and there is a nonnegative nonzero vector \( x \) such that \( {Ax} = \rho \left( A\right) x \) . | Proof. For any \( \epsilon > 0 \), define \( A\left( \epsilon \right) = A + \epsilon {J}_{n} \) . Let \( x\left( \epsilon \right) = \left\lbrack {x{\left( \epsilon \right) }_{i}}\right\rbrack \) be the Perron vector of \( A\left( \epsilon \right) \), so \( x\left( \epsilon \right) > 0 \) and \( \mathop{\sum }\limits_{{... | Yes |
Theorem 8.3.2. Let \( A \in {M}_{n} \) be nonnegative, and let \( x \in {\mathbf{R}}^{n} \) be nonnegative and nonzero. If \( \alpha \in \mathbf{R} \) and \( {Ax} \geq {\alpha x} \), then \( \rho \left( A\right) \geq \alpha \) . | Proof. Let \( A = \left\lbrack {a}_{ij}\right\rbrack \), let \( \epsilon > 0 \), and define \( A\left( \epsilon \right) = A + \epsilon {J}_{n} > 0 \) . Then \( A\left( \epsilon \right) \) has a positive left Perron vector \( y\left( \epsilon \right) : y{\left( \epsilon \right) }^{T}A\left( \epsilon \right) = \rho \left... | Yes |
Corollary 8.3.3. If \( A \in {M}_{n} \) is nonnegative, then\n\n\[ \rho \left( A\right) = \mathop{\max }\limits_{\substack{{x \geq 0} \\ {x \neq 0} }}\mathop{\min }\limits_{\substack{{1 \leq i \leq n} \\ {{x}_{i} \neq 0} }}\frac{1}{{x}_{i}}\mathop{\sum }\limits_{{j = 1}}^{n}{a}_{ij}{x}_{j} \]\n\n(8.3.3a) | Proof. Let \( x \) be any nonzero nonnegative vector and let \( \alpha = \mathop{\min }\limits_{{{x}_{i} \neq 0}}\mathop{\sum }\limits_{j}{a}_{ij}{x}_{j}/{x}_{i} \) . Then \( {Ax} \geq {\alpha x} \), so the preceding theorem ensures that \( \rho \left( A\right) \geq a \), and hence\n\n\[ \rho \left( A\right) \geq \math... | Yes |
Theorem 8.3.4. Let \( A \in {M}_{n} \) be nonnegative. Suppose that there is a positive vector \( x \) and a nonnegative real number \( \lambda \) such that either \( {Ax} = {\lambda x} \) or \( {x}^{T}A = \lambda {x}^{T} \) . Then \( \lambda = \rho \left( A\right) \) . | Proof. Suppose that \( x = \left\lbrack {x}_{i}\right\rbrack \in {\mathbf{R}}^{n} \) and \( {Ax} = {\lambda x} \) . Let \( D = \operatorname{diag}\left( {{x}_{1},\ldots ,{x}_{n}}\right) \) and define \( B = {D}^{-1}{AD} \), which has the same eigenvalues as \( A \) . Then \( {Be} = {D}^{-1}{ADe} = \) \( {D}^{-1}{Ax} = ... | Yes |
Theorem 8.3.5. Suppose that \( A \in {M}_{n} \) is nonnegative and has a positive left eigenvector.\n\n(a) If \( x \in {\mathbf{R}}^{n} \) is nonzero and \( {Ax} \geq \rho \left( A\right) x \), then \( x \) is an eigenvector of \( A \) corresponding to the eigenvalue \( \rho \left( A\right) \) . | Proof. Let \( y \) be a positive left eigenvector of \( A \) . The preceding theorem ensures that \( {A}^{T}y = \rho \left( A\right) y. \) (a) We know that \( x \neq 0 \) and \( {Ax} - \rho \left( A\right) x \geq 0 \) . We need to show that \( {Ax} - \rho \left( A\right) x = \) 0 . If \( {Ax} - \rho \left( A\right) x \... | Yes |
Lemma 8.4.2. Let \( {\lambda }_{1},\ldots ,{\lambda }_{n} \) be the eigenvalues of \( A \in {M}_{n} \) . Then \( {\lambda }_{1} + 1,\ldots ,{\lambda }_{n} + 1 \) are the eigenvalues of \( I + A \) and \( \rho \left( {I + A}\right) \leq \rho \left( A\right) + 1 \) . If \( A \) is nonnegative, then \( \rho \left( {I + A}... | Proof. The first assertion is a consequence of (2.4.2). We have \( \rho \left( {I + A}\right) = \) \( \mathop{\max }\limits_{{1 \leq i \leq n}}\left| {{\lambda }_{i} + 1}\right| \leq \mathop{\max }\limits_{{1 \leq i \leq n}}\left| {\lambda }_{i}\right| + 1 = \rho \left( A\right) + 1 \) . However,(8.3.1) ensures that \(... | Yes |
Lemma 8.4.3. If \( A \in {M}_{n} \) is nonnegative and \( {A}^{m} \) is positive for some \( m \geq 1 \), then \( \rho \left( A\right) \) is the only maximum-modulus eigenvalue of \( A \) ; it is positive and algebraically simple. | Proof. Let \( {\lambda }_{1},\ldots ,{\lambda }_{n} \) be the eigenvalues of \( A \) . Then \( {\lambda }_{1}^{m},\ldots ,{\lambda }_{n}^{m} \) are the eigenvalues of \( {A}^{m} \) . Theorem 8.2.8 ensures that exactly one of \( {\lambda }_{1}^{m},\ldots ,{\lambda }_{n}^{m} \) is equal to \( \rho \left( {A}^{m}\right) =... | Yes |
Theorem 8.4.4 (Perron-Frobenius). Let \( A \in {M}_{n} \) be irreducible and nonnegative, and suppose that \( n \geq 2 \) . Then\n\n(a) \( \rho \left( A\right) > 0 \)\n\n(b) \( \rho \left( A\right) \) is an algebraically simple eigenvalue of \( A \)\n\n(c) there is a unique real vector \( x = \left\lbrack {x}_{i}\right... | Proof. (a) Corollary 8.1.25 shows that \( \rho \left( A\right) > 0 \) under conditions even weaker than irreducibility.\n\n(b) If \( \rho \left( A\right) \) is a multiple eigenvalue of \( A \), then (8.4.2) ensures that \( \rho \left( A\right) + 1 = \rho (I + \) \( A) \) is a multiple eigenvalue of \( I + A \) and henc... | Yes |
Theorem 8.4.5. Let \( A, B \in {M}_{n} \) . Suppose that \( A \) is nonnegative and irreducible, and \( A \geq \left| B\right| \) . Let \( \lambda = {e}^{i\varphi }\rho \left( B\right) \) be a given maximum-modulus eigenvalue of \( B \) . If \( \rho \left( A\right) = \) \( \rho \left( B\right) \), then there is a diago... | Proof. Let \( x \) be a nonzero vector such that \( {Bx} = {\lambda x} \), and let \( \rho = \rho \left( A\right) = \rho \left( B\right) \) . Then\n\n\[ \rho \left| x\right| = \left| {\lambda x}\right| = \left| {Bx}\right| \leq \left| B\right| \left| x\right| \overset{\left( \alpha \right) }{ \leq }A\left| x\right| \]\... | Yes |
Corollary 8.4.7. Suppose that \( A \in {M}_{n} \) is irreducible and nonnegative. If \( A \) has \( k > 1 \) eigenvalues of maximum modulus, then every main diagonal entry of \( A \) is zero. Moreover, every main diagonal entry of \( {A}^{m} \) is zero for each positive integer \( m \) that is not divisible by \( k \) ... | Proof. Let \( \varphi = {2\pi }/k \) . Corollary 8.4.6a ensures that \( A \) is similar to \( {e}^{i\varphi }A \), so \( {A}^{m} \) is similar to \( {e}^{im\varphi }A \) for each \( m = 1,2,3,\ldots \) and \( \operatorname{tr}{A}^{m} = {e}^{im\varphi }\operatorname{tr}{A}^{m} \) . Since \( {e}^{im\varphi } \) is real a... | Yes |
Theorem 8.5.1. If \( A \in {M}_{n} \) is nonnegative and primitive, and if \( x \) and \( y \) are, respectively, the right and left Perron vectors of \( A \), then \( \mathop{\lim }\limits_{{m \rightarrow \infty }}{\left( \rho {\left( A\right) }^{-1}A\right) }^{m} = x{y}^{T} \), which is a positive rank-one matrix. | Proof. We have in hand all the ingredients required in the proof of (8.2.7): \( \rho \left( A\right) \) is a simple eigenvalue with positive associated right and left eigenvectors \( x \) and \( y \) such that \( {x}^{T}y = 1 \) . We can perform the factorization (8.2.7a), in which every eigenvalue of \( B \) has modul... | Yes |
Theorem 8.5.2. If \( A \in {M}_{n} \) is nonnegative, then \( A \) is primitive if and only if \( {A}^{m} > 0 \) for some \( m \geq 1 \) . | Proof. If \( {A}^{m} \) is positive, there is a directed path of length \( m \) between every pair of nodes of the directed graph \( \Gamma \left( A\right) \) of \( A \), so \( \Gamma \left( A\right) \) is strongly connected and \( A \) is irreducible. In addition,(8.4.3) ensures that there are no maximum-modulus eigen... | Yes |
Theorem 8.5.3. Let \( A \in {M}_{n} \) be irreducible and nonnegative, and let \( {P}_{1},\ldots ,{P}_{n} \) be the nodes of the directed graph \( \Gamma \left( A\right) \) . Let \( {L}_{i} = \left\{ {{k}_{1}^{\left( i\right) },{k}_{2}^{\left( i\right) },\ldots }\right\} \) be the set of lengths of all directed paths i... | Proof. Irreducibility of \( A \) implies that no set \( {L}_{i} \) is empty: For each \( i \) and for any \( j \neq i \) , there is a path in \( \Gamma \left( A\right) \) that joins \( {P}_{i} \) to \( {P}_{j} \) ; there is also a path in \( \Gamma \left( A\right) \) that joins \( {P}_{j} \) to \( {P}_{i} \) . If \( A ... | Yes |
Lemma 8.5.4. If \( A \in {M}_{n} \) is irreducible and nonnegative, and if all its main diagonal entries are positive, then \( {A}^{n - 1} > 0 \), so \( A \) is primitive. | Proof. If every main diagonal entry of \( A \) is positive, let \( \alpha = \min \left\{ {{a}_{11},\ldots ,{a}_{nn}}\right\} > 0 \) and define \( B = A - \operatorname{diag}\left( {{a}_{11},\ldots ,{a}_{nn}}\right) \) . Then \( B \) is nonnegative and irreducible (because \( A \) is irreducible), and \( A \geq {\alpha ... | Yes |
Lemma 8.5.5. Let \( A \in {M}_{n} \) be nonnegative and primitive. Then \( {A}^{m} \) is nonnegative and primitive for every integer \( m \geq 1 \) . | Proof. Since all sufficiently large powers of \( A \) are positive, the same is true for \( {A}^{m} \) for any \( m \) . If \( {A}^{m} \) were reducible, then \( {A}^{mp} \) would be reducible for all \( p = 2,3,\ldots \), and hence these matrices cannot be positive. This contradiction shows that no power of \( A \) ca... | No |
Theorem 8.5.6. Let \( A \in {M}_{n} \) be nonnegative. If \( A \) is primitive, then \( {A}^{k} > 0 \) for some positive integer \( k \leq \left( {n - 1}\right) {n}^{n} \) . | Proof. Because \( A \) is irreducible, there is a directed path from the node \( {P}_{1} \) in \( \Gamma \left( A\right) \) back to itself; let \( {k}_{1} \) be the shortest such path, so that \( {k}_{1} \leq n \) . The matrix \( {A}^{{k}_{1}} \) has a positive entry in its 1,1 position, and any power of \( {A}^{{k}_{1... | Yes |
Corollary 8.5.8 (Wielandt). Let \( A \in {M}_{n} \) be nonnegative. Then \( A \) is primitive if and only if \( {A}^{{n}^{2} - {2n} + 2} > 0 \) . | Proof. If some power of \( A \) is positive, then \( A \) is primitive, so only the converse implication is of interest. If \( n = 1 \), the result is trivial, so assume that \( n > 1 \) . If \( A \) is primitive, then it is irreducible and there are cycles in \( \Gamma \left( A\right) \) . If the shortest cycle in \( ... | Yes |
Theorem 8.5.9. Let \( A \in {M}_{n} \) be irreducible and nonnegative, and suppose that \( A \) has \( d \) positive main diagonal entries, \( 1 \leq d \leq n \) . Then \( {A}^{{2n} - d - 1} > 0 \) ; that is, \( \gamma \left( A\right) \leq \) \( {2n} - d - 1 \) . | Proof. Under the stated hypotheses, \( A \) must be primitive, and \( \Gamma \left( A\right) \) has \( d \) cycles with (minimum) length one. We may assume that \( {P}_{1},\ldots ,{P}_{d} \) are the nodes in \( \Gamma \left( A\right) \) that have loops. Consider \( {A}^{{2n} - d - 1} = {A}^{n - d}{\left( {A}^{1}\right)... | Yes |
Theorem 8.6.1. Let \( A \in {M}_{n} \) be irreducible and nonnegative, let \( n \geq 2 \), and let \( x \) and \( y \), respectively, be the right and left Perron vectors of \( A \) . Then \[ \mathop{\lim }\limits_{{N \rightarrow \infty }}\frac{1}{N}\mathop{\sum }\limits_{{m = 1}}^{N}{\left( \rho {\left( A\right) }^{-1... | Proof. If \( A \) is primitive, \( \rho {\left( A\right) }^{-1}A \) can be factored as in (8.2.7a), in which \( x \) is the first column of \( S \) and \( y \) is the first column of \( {S}^{-1} \) . We have \[ \frac{1}{N}\mathop{\sum }\limits_{{m = 1}}^{N}{\left( \rho {\left( A\right) }^{-1}A\right) }^{m} = S\left\lbr... | Yes |
Lemma 8.7.1. Let \( A = \left\lbrack {a}_{ij}\right\rbrack \in {M}_{n} \) be a doubly stochastic matrix that is not the identity matrix. There is a permutation \( \sigma \) of \( \{ 1,\ldots, n\} \) that is not the identity permutation and is such that \( {a}_{{1\sigma }\left( 1\right) }\cdots {a}_{{n\sigma }\left( n\r... | Proof. Suppose that every permutation \( \sigma \) of \( \{ 1,\ldots, n\} \) that is not the identity permutation \( {\sigma }_{0} \) has the property that \( {a}_{{1\sigma }\left( 1\right) }\cdots {a}_{{n\sigma }\left( n\right) } = 0 \) . This assumption and (0.3.2.1) permit us to compute the characteristic polynomial... | Yes |
Theorem 8.7.2 (Birkhoff). A matrix \( A \in {M}_{n} \) is doubly stochastic if and only if there are permutation matrices \( {P}_{1},\ldots ,{P}_{N} \in {M}_{n} \) and positive scalars \( {t}_{1},\ldots ,{t}_{N} \in \mathbf{R} \) such that \( {t}_{1} + \cdots + {t}_{N} = 1 \) and\n\n\[ A = {t}_{1}{P}_{1} + \cdots + {t}... | Proof. The sufficiency of the representation (8.7.3) is clear; we prove its necessity by exhibiting an algorithm that constructs it in finitely many steps.\n\nIf \( A \) is a permutation matrix, there is nothing to prove. If not, the preceding lemma ensures that there is a nonidentity permutation \( \sigma \) of \( \{ ... | Yes |
Corollary 8.7.4. The maximum (respectively, minimum) of a convex (respectively, concave) real-valued function on the set of doubly stochastic \( n \) -by- \( n \) matrices is attained at a permutation matrix. | Proof. Let \( f \) be a convex real-valued function on the set of \( n \) -by- \( n \) doubly stochastic matrices, let \( A \) be a doubly stochastic matrix at which \( f \) attains its maximum value, represent \( A = {t}_{1}{P}_{1} + \cdots + {t}_{N}{P}_{N} \) as a convex combination of permutation matrices, and let \... | Yes |
Lemma 8.7.5. Let \( A \in {M}_{n} \) be doubly substochastic. There is a doubly stochastic matrix \( S \in {M}_{n} \) such that \( A \leq S \) . | Proof. For any doubly substochastic matrix \( S \in {M}_{n} \), let \( N\left( S\right) \) denote the number of row sums and column sums of \( S \) that are less than one, that is, the number of entries of the vectors \( {Ae} \) and \( {A}^{T}e \) whose entries are less than one.\n\nLet \( A \in {M}_{n} \) be doubly su... | Yes |
Theorem 1.1 If \( \left( {P, \leq }\right) \) is a partially ordered set, then the relation\n\n\[ a < b\;\text{ if }\;a \leq b, a \neq b \]\n\nis a strict order on \( P \) . Conversely, if \( < \) is a strict order on \( P \), then the relation\n\n\[ a \leq b\;\text{ if }\;a < b\text{ or }a = b \]\n\nis a partial order... | \[ ▱ \] | No |
Theorem 1.4 Let \( P \) be a poset.\n\n1) P has the ACC if and only if it has the maximal condition.\n\n2) P has the DCC if and only if it has the minimal condition. | Proof. Suppose \( P \) has the ACC and let \( S \subseteq P \) be nonempty. Let \( {s}_{1} \in S \) . If \( {s}_{1} \) is maximal we are done. If not, then we can pick \( {s}_{2} \in S \) such that \( {s}_{2} > {s}_{1} \) . Continuing in this way, we either arrive at a maximal element in \( S \) or we get a strictly in... | Yes |
Theorem 1.5 Let \( P \) be a poset.\n\n1) The following are equivalent:\n\na) P has no infinite chains.\n\nb) \( P \) has both chain conditions.\n\nIf these conditions hold, then for any \( a < b \) in \( P \), there is a maximal finite chain from a to \( b \) . | Proof. It is clear that 1a) implies 1b). For the converse, suppose that \( P \) has BCC and let \( \mathcal{C} \) be an infinite chain. The ACC implies that \( \mathcal{C} \) has a maximal element \( {x}_{1} \), which must be maximum in \( \mathcal{C} \) since \( \mathcal{C} \) is totally ordered. Then \( \mathcal{C} \... | Yes |
Theorem 1.7 Suppose that \( \left( {P, \leq }\right) \) is a partially ordered set for which every subset of \( P \) has a meet. Then \( \left( {P, \leq }\right) \) is a complete lattice, where the join of a subset \( S \) of \( P \) is the meet of all upper bounds for \( S \) . | Proof. First, note that the meet of the empty set is the maximum element of \( P \) and the meet of \( P \) is the minimum element of \( P \) and so \( P \) is bounded. In particular, the join of \( \varnothing \) exists.\n\nLet \( S \) be a nonempty subset of \( P \) . The family \( U \) of upper bounds for \( S \) is... | Yes |
Theorem 1.18 Let \( p \) be a prime and let \( d \geq 1 \) . Let \( {o}_{p}\left( n\right) \) denote the largest exponent \( e \) for which \( {p}^{e} \) divides \( n \) .\n\n1) For \( 1 \leq k \leq {p}^{d} \) ,\n\n\[ \n{o}_{p}\left\lbrack \left( \begin{matrix} {p}^{d} \\ k \end{matrix}\right) \right\rbrack = d - {o}_{... | Proof. For part 1), write\n\n\[ \n\left( \begin{matrix} {p}^{d} \\ k \end{matrix}\right) = \frac{{p}^{d}}{k}\frac{\left( {p}^{d} - 1\right) }{1}\cdots \frac{\left( {p}^{d} - u\right) }{u}\cdots \frac{{p}^{d} - \left( {k - 1}\right) }{k - 1} \n\]\n\nwhere \( 1 \leq u \leq k - 1 \) . Now, if \( 1 \leq i \leq d \), then \... | Yes |
Theorem 2.1 Let \( G \) be a group. If \( a \in G \) has finite order \( o\left( a\right) \), then the exponents of a are precisely the integral multiples of \( o\left( a\right) \) . | Proof. Let \( n = o\left( a\right) \) . Any integral multiple of \( n \) is clearly an exponent of \( a \) . Conversely, if \( {a}^{m} = 1 \), then \( m = {qn} + r \) where \( 0 \leq r < n \) . Hence,\n\n\[ \n1 = {a}^{m} = {a}^{{qn} + r} = {a}^{qn}{a}^{r} = {a}^{r} \n\] \n\nand so the minimality of \( n \) implies that... | Yes |
Theorem 2.10 Let \( G \) be a group and \( a, b \in G \). 1) If \( o\left( a\right) = n \), then for \( 1 \leq k < n \), \[ o\left( {a}^{k}\right) = \frac{n}{\gcd \left( {n, k}\right) } \] In particular, \[ \langle a\rangle = \left\langle {a}^{k}\right\rangle \; \Leftrightarrow \;\gcd \left( {o\left( a\right), k}\right... | Proof. For part 1), Theorem 2.1 implies that \( {\left( {a}^{k}\right) }^{m} = 1 \) if and only if \( n \mid {km} \). But \[ n \mid {km}\; \Leftrightarrow \;\frac{n}{\gcd \left( {n, k}\right) }\left| {\frac{km}{\gcd \left( {n, k}\right) }\; \Leftrightarrow \;\frac{n}{\gcd \left( {n, k}\right) }}\right| m \] and so the ... | Yes |
Corollary 2.11 Let \( G \) be a group and let \( a \in G \) . If \( o\left( a\right) = {uv} \), where \( \left( {u, v}\right) = 1 \) , then\n\n\[ a = {a}_{1}{a}_{2} \]\n\nwhere \( {a}_{1},{a}_{2} \in \langle a\rangle \) and \( o\left( {a}_{1}\right) = u \) and \( o\left( {a}_{2}\right) = v \) . | Proof. Since \( \left( {u, v}\right) = 1 \), there exist integers \( s \) and \( t \) for which \( {su} + {tv} = 1 \) . Hence,\n\n\[ a = {a}^{{su} + {tv}} = {a}^{su}{a}^{tv} \]\n\nThen \( \left( {{su}, v}\right) = 1 \) implies that \( o\left( {a}^{su}\right) = v \) and similarly, \( o\left( {a}^{tv}\right) = u.▱ | Yes |
Let \( {S}_{n} \) be the symmetric group.\n\n1) Let \( \sigma \in {S}_{n} \) . For any \( k \) -cycle \( \left( {{a}_{1}\cdots {a}_{k}}\right) \), we have\n\n\[{\left( {a}_{1}\cdots {a}_{k}\right) }^{\sigma } = \left( {\sigma {a}_{1}\cdots \sigma {a}_{k}}\right)\]\n\nHence, if \( \tau = {c}_{1}\cdots {c}_{k} \) is a cy... | Proof. For part 1), we have\n\n\[{\left( {a}_{1}\cdots {a}_{k}\right) }^{\sigma }\left( {\sigma {a}_{i}}\right) = \left\{ \begin{array}{ll} \sigma {a}_{i + 1} & i < k \\ \sigma {a}_{1} & i = k \end{array}\right.\]\n\nAlso, if \( b \neq \sigma {a}_{i} \) for any \( i \), then \( {\sigma }^{-1}b \neq {a}_{i} \) and so\n\... | Yes |
The set \( {A}_{n} \) of all even permutations in \( {S}_{n} \) is a subgroup of \( {S}_{n} \). | To see that \( {A}_{n} \) is closed under the product, if \( \sigma \) and \( \tau \) are even, then they can each be written as a product of an even number of transpositions. Hence, \( {\sigma \tau } \) is also a product of an even number of transpositions and so is in \( {A}_{n} \). The subgroup \( {A}_{n} \) is call... | No |
Theorem 2.19 Let \( G \) be a group and let \( A, B, C \leq G \) with \( A \leq B \). 1) (Dedekind law) \[ A\left( {B \cap C}\right) = B \cap {AC} \] 2) \[ A \cap C = B \cap C\text{ and }{AC} = {BC}\; \Rightarrow \;A = B \] | Proof. We leave proof of the Dedekind law to the reader. For part 2), \[ A = A\left( {A \cap C}\right) = A\left( {B \cap C}\right) = B \cap {AC} = B \cap {BC} = B \] | No |
Theorem 2.20 Let \( X \) be a nonempty subset of a group \( G \) and let \( {X}^{-1} = \) \( \left\{ {{x}^{-1} \mid x \in X}\right\} \) . Let\n\n\[ \nW = {\left( X \cup {X}^{-1}\right) }^{ * } \n\]\n\nbe the set of all words over the alphabet \( X \cup {X}^{-1} \) .\n\n1) If we interpret juxtaposition in \( W \) as the... | Proof. It is clear that \( W \subseteq \langle X\rangle \) . It is also clear that \( W \) is closed under product. As to inverses, if \( w = {x}_{1}^{{e}_{1}}\cdots {x}_{n}^{{e}_{n}} \in W \) then \( {w}^{-1} = {x}_{n}^{-{e}_{n}}\cdots {x}_{1}^{-{e}_{1}} \in W \) . Hence, \( W \leq \langle X\rangle \) . However, since... | Yes |
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