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Corollary 4.4.18. A square complex matrix is conjugate normal if and only if it is unitarily congruent to a direct sum of nonnegative scalar multiples of real orthogonal matrices. | Proof. The previous theorem states that a conjugate-normal matrix is unitarily congruent to a direct sum of nonnegative scalar multiples of 1-by-1 and 2- by-2 real orthogonal matrices. Conversely, if \( A = {UZ}{U}^{T} \), in which \( U \) is unitary, \( Z = {\sigma }_{1}{Q}_{1} \oplus \cdots \oplus {\sigma }_{m}{Q}_{m... | Yes |
Corollary 4.4.19. Let \( A \in {M}_{n} \) be skew symmetric. Then \( r = \operatorname{rank}A \) is even, the nonzero singular values of \( A \) occur in pairs \( {\sigma }_{1} = {\sigma }_{2} = {s}_{1} \geq {\sigma }_{3} = {\sigma }_{4} = {s}_{2} \geq \cdots \geq \) \( {\sigma }_{r - 1} = {\sigma }_{r} = {s}_{r/2} \ge... | Proof. Since \( A \) is skew symmetric, it is conjugate normal and its canonical form (4.4.17) must be skew symmetric. This implies that \( \sum = 0 \) and every \( {a}_{j} = 0 \) . | Yes |
Corollary 4.4.21. Let \( V \in {M}_{n} \) be unitary. Then \( V \) is unitarily congruent to\n\n\[ \n{I}_{n - {2q}} \oplus \left\lbrack \begin{matrix} {a}_{1} & {b}_{1} \\ - {b}_{1} & {a}_{1} \end{matrix}\right\rbrack \oplus \cdots \oplus \left\lbrack \begin{matrix} {a}_{q} & {b}_{q} \\ - {b}_{q} & {a}_{q} \end{matrix}... | Proof. Since \( V \) is unitary, it is conjugate normal and its canonical form (4.4.17) must be unitary. This implies that \( \sum \) is unitary, so \( \sum = I \) . It also implies that each 2-by-2 direct summand in (4.4.17) is unitary. The blocks \( \left\lbrack \begin{matrix} {a}_{j} & {b}_{j} \\ - {b}_{j} & {a}_{j}... | Yes |
Theorem 4.4.24. Each \( A \in {M}_{n} \) is similar to a complex symmetric matrix. | Proof. Each \( A \in {M}_{n} \) is similar to a direct sum of Jordan blocks, and the preceding exercises show that each Jordan block is similar to a symmetric matrix. Thus, each \( A \in {M}_{n} \) is similar to a direct sum of symmetric matrices. | No |
Corollary 4.4.25. Let \( A \in {M}_{n} \) be given. There are symmetric matrices \( B, C \in {M}_{n} \) such that \( A = {BC} \) . Either \( B \) or \( C \) may be chosen to be nonsingular. | Proof. Use the preceding theorem to write \( A = {SE}{S}^{-1} \), in which \( E = {E}^{T} \) and \( S \) is nonsingular. Then \( A = \left( {{SE}{S}^{T}}\right) {S}^{-T}{S}^{-1} = \left( {{SE}{S}^{T}}\right) {\left( S{S}^{T}\right) }^{-1} = \left( {S{S}^{T}}\right) \left( {{S}^{-T}E{S}^{-1}}\right) \) . | Yes |
Lemma 4.4.26. Let \( X \in {M}_{n, k} \) with \( k \leq n \) . Then \( {X}^{T}X \) is nonsingular if and only if \( X = {YB} \), in which \( Y \in {M}_{n, k},{Y}^{T}Y = {I}_{k} \), and \( B \in {M}_{k} \) is nonsingular. | Proof. If \( X = {YB} \) and \( {Y}^{T}Y = {I}_{k} \), then \( {X}^{T}X = {B}^{T}{Y}^{T}{YB} = {B}^{T}B \) is nonsingular if and only if \( B \) is nonsingular. Conversely, use (4.4.4c) to factor \( {X}^{T}X = {U\sum }{U}^{T} \), in which \( U \in {M}_{k} \) is unitary and \( {U}^{ * }{X}^{T}X\bar{U} = {\left( X\bar{U}... | Yes |
Theorem 4.4.27. Let \( A \in {M}_{n} \) be symmetric. Then \( A \) is diagonalizable if and only if it is complex orthogonally diagonalizable. | Proof. If there is a complex orthogonal \( Q \) such that \( {Q}^{T}{AQ} \) is diagonal, then of course \( A \) is diagonalizable; only the converse assertion is interesting. Suppose that \( A \) is diagonalizable and let \( x, y \in {\mathbf{C}}^{n} \) be eigenvectors of \( A \) with \( {Ax} = {\lambda x} \) and \( {A... | Yes |
Theorem 4.5.5. Both \( {}^{ * } \) congruence and congruence are equivalence relations. | Proof. Reflexivity: \( A = {IA}{I}^{ * } \) . Symmetry: If \( A = {SB}{S}^{ * } \) and \( S \) is nonsingular, then \( B = {S}^{-1}A{\left( {S}^{-1}\right) }^{ * } \) . Transitivity: If \( A = {S}_{1}B{S}_{1}^{ * } \) and \( B = {S}_{2}C{S}_{2}^{ * } \), then \( A = \) \( \left( {{S}_{1}{S}_{2}}\right) C{\left( {S}_{1}... | Yes |
Theorem 4.5.8 (Sylvester). Hermitian matrices \( A, B \in {M}_{n} \) are \( {}^{ * } \) congruent if and only if they have the same inertia, that is, if and only if they have the same number of positive eigenvalues and the same number of negative eigenvalues. | Proof. Since each of \( A \) and \( B \) is \( {}^{ * } \) congruent to its inertia matrix, if they have the same inertia, they must be *congruent. The converse assertion is more interesting.\n\nSuppose that \( S \in {M}_{n} \) is nonsingular and that \( A = {SB}{S}^{ * } \) . Congruent matrices have the same rank, so ... | Yes |
Theorem 4.5.12. Let \( A, B \in {M}_{n} \) be symmetric. There is a nonsingular \( S \in {M}_{n} \) such that \( A = {SB}{S}^{T} \) if and only if \( \operatorname{rank}A = \operatorname{rank}B \) . | Proof. If \( A = {SB}{S}^{T} \) and \( S \) is nonsingular, then rank \( A = \operatorname{rank}B \) (0.4.6b). Conversely, use (4.4.4c) to write\n\n\[ A = {U}_{1}{\sum }_{1}{U}_{1}^{T} = {U}_{1}I\left( {\sum }_{1}\right) {D}_{1}^{2}{U}_{1}^{T} = \left( {{U}_{1}{D}_{1}}\right) I\left( {\sum }_{1}\right) {\left( {U}_{1}{... | Yes |
Theorem 4.5.13. Let \( A, S \in {M}_{n} \) and suppose that \( A \) is symmetric. Let \( A = {U\sum }{U}^{T} \) and \( {SA}{S}^{T} = {VM}{V}^{T} \) be factorizations (4.4.4c) of \( A \) and \( {SA}{S}^{T} \) in which \( U \) and \( V \) are unitary, \( \sum = \operatorname{diag}\left( {{\sigma }_{1},{\sigma }_{2},\ldot... | Proof. We have \( {\mu }_{k}^{2} = {\lambda }_{k}\left( {{SA}{S}^{T}\bar{S}\bar{A}{S}^{ * }}\right) = {\lambda }_{k}\left( {S\left( {A{S}^{T}\bar{S}\bar{A}}\right) {S}^{ * }}\right) = {\widehat{\theta }}_{k}{\lambda }_{k}\left( {A{S}^{T}\bar{S}\bar{A}}\right) \) in which (4.5.11) ensures that \( {\lambda }_{1}\left( {S... | Yes |
Theorem 4.5.15. Let \( A, B \in {M}_{n} \) be given.\n\n(a) Suppose that \( A \) and \( B \) are Hermitian. There is a unitary \( U \in {M}_{n} \) and real diagonal \( \Lambda, M \in {M}_{n}\left( \mathbf{R}\right) \) such that \( A = {U\Lambda }{U}^{ * } \) and \( B = {UM}{U}^{ * } \) if and only if \( {AB} \) is Herm... | Proof. (a) See (4.1.6). | No |
Theorem 4.5.17. Let \( A, B \in {M}_{n} \) be given.\n\n(a) Suppose that \( A \) and \( B \) are Hermitian and \( A \) is nonsingular. Let \( C = {A}^{-1}B \) . There is a nonsingular \( S \in {M}_{n} \) and real diagonal matrices \( \Lambda \) and \( M \) such that \( A = {S\Lambda }{S}^{ * } \) and \( B = {SM}{S}^{ *... | Proof. In each case, a computation verifies necessity of the stated conditions for simultaneous diagonalization by congruence, so we discuss only their sufficiency. The first two cases can be proved with parallel arguments, but the third case is a bit different.\n\n(a) Assume that \( A \) and \( B \) are Hermitian, \( ... | Yes |
Lemma 4.5.18. Let \( A \in {M}_{n} \) be given, and let \( A = H + {iK} \), in which \( H \) and \( K \) are Hermitian. Then \( A \) is diagonalizable by \( {}^{ * } \) congruence if and only if \( H \) and \( K \) are simultaneously diagonalizable by \( {}^{ * } \) congruence. | Proof. If there is a nonsingular \( S \in {M}_{n} \) such that \( {SH}{S}^{ * } = \Lambda \) and \( {SK}{S}^{ * } = M \) are both diagonal, then \( {SA}{S}^{ * } = {SH}{S}^{ * } + {iSK}{S}^{ * } = \Lambda + {iM} \) is diagonal. To prove the converse, it suffices to show that if \( B = \left\lbrack {b}_{jk}\right\rbrack... | Yes |
Theorem 4.5.22. Let \( A, B \in {M}_{p} \) and \( C \in {M}_{q} \) be given. Then \( A \oplus C \) and \( B \oplus C \) are *congruent if and only if \( A \) and \( B \) are *congruent. | Proof. If there is a nonsingular \( S \in {M}_{p} \) such that \( A = {SB}{S}^{ * } \), then \( S \oplus {I}_{q} \) is nonsingular and \( \left( {S \oplus {I}_{q}}\right) \left( {B \oplus C}\right) {\left( S \oplus {I}_{q}\right) }^{ * } = {SB}{S}^{ * } \oplus C = A \oplus C \) . Conversely, suppose that \( A \oplus \)... | Yes |
Proposition 4.6.6. Let \( A \in {M}_{n} \), let \( \lambda \geq 0 \) be given, let \( \sigma = \sqrt{\lambda } \geq 0 \), and suppose that there is a nonzero vector \( x \) such that \( A\bar{A}x = {\lambda x} \). There is a nonzero vector \( y \) such that \( A\bar{y} = {\sigma y}. | (a) If \( A\bar{A}x = 0 \), then \( A\bar{A} \) is singular, so \( 0 = \det A\bar{A} = {\left| \det \bar{A}\right| }^{2},\bar{A} \) is singular, there is a nonzero vector \( z \) such that \( \bar{A}z = 0 \) and \( \overline{\bar{A}}z = A\bar{z} = 0 \). (b) Suppose that \( \sigma > 0 \), so \( {\sigma x} \neq 0 \). If ... | No |
Proposition 4.6.7. Let \( A \in {M}_{n} \) be given, and let \( {x}_{1},{x}_{2},\ldots ,{x}_{k} \) be coneigenvectors of \( A \) with corresponding coneigenvalues \( {\lambda }_{1},{\lambda }_{2},\ldots ,{\lambda }_{k} \) . If \( \left| {\lambda }_{i}\right| \neq \left| {\lambda }_{j}\right| \) whenever \( 1 \leq \) \(... | Proof. Each \( {x}_{i} \) is an eigenvector of \( A\bar{A} \) with associated eigenvalue \( {\left| {\lambda }_{i}\right| }^{2} \) . Lemma 1.3.8 ensures that the vectors \( {x}_{1},\ldots ,{x}_{k} \) are linearly independent. | No |
Corollary 4.6.8. Let \( A \in {M}_{n} \) be given and suppose that \( A\bar{A} \) has \( k \) distinct nonnegative eigenvalues.\n\n(a) The matrix \( A \) has at least \( k \) linearly independent coneigenvectors.\n\n(b) If \( k = 0 \), then \( A \) has no coneigenvectors.\n\n(c) Suppose that \( k = n \) . Then \( A \) ... | Proof. Only the second assertion in (d) requires justification. Let \( \Lambda = \operatorname{diag}\left( {{\lambda }_{1},\ldots }\right. \) , \( \left. {\lambda }_{n}\right) \), let \( {\sigma }_{j}^{2} = {\lambda }_{j} \) and \( {\sigma }_{j} \geq 0 \) for each \( j = 1,\ldots, n \), let \( \sum = \operatorname{diag... | Yes |
Lemma 4.6.9. Let \( A \in {M}_{n} \) be given. Then \( A\bar{A} = I \) if and only if there is a nonsingular \( S \in {M}_{n} \) such that \( A = S{\bar{S}}^{-1} \) . | Proof. If \( A = S{\bar{S}}^{-1} \), then \( A\bar{A} = S{\bar{S}}^{-1}\bar{S}{S}^{-1} = I \) . Conversely, suppose that \( A\bar{A} = I \) , let \( {S}_{\theta } = {e}^{i\theta }A + {e}^{-{i\theta }}I,\theta \in \mathbf{R} \), and compute\n\n\[ A{\bar{S}}_{\theta } = A\left( {{e}^{-{i\theta }}\bar{A} + {e}^{i\theta }I... | Yes |
Corollary 4.6.14. Let \( A, B \in {M}_{n} \) . Then \( A \) is consimilar to \( B \) if and only if \( A\bar{A} \) is similar to \( B\bar{B} \) , \( \operatorname{rank}A = \operatorname{rank}B \), and \( \operatorname{rank}{\left( A\bar{A}\right) }^{k}A = \operatorname{rank}{\left( B\bar{B}\right) }^{k}B \) for \( k = ... | Proof. Necessity of the stated conditions is clear, so we consider only their sufficiency. The Type I and Type II blocks of the concanonical forms of \( A \) and \( B \) are determined by the Jordan canonical form of \( A\bar{A} \), since it is the same as the Jordan canonical form of \( B\bar{B} \) . The stated rank c... | Yes |
Corollary 4.6.16. Let \( A \in {M}_{n} \) be given. Then \( A\bar{A} \) is similar to the square of a real matrix. | Proof. Corollary 4.6.15 ensures that there is a nonsingular \( S \in {M}_{n} \) and a real matrix \( R \in {M}_{n}\left( \mathbf{R}\right) \) such that \( A = {SR}{\bar{S}}^{-1} \) . Then \( A\bar{A} = {SR}{\bar{S}}^{-1}\overline{{SR}{\bar{S}}^{-1}} = S{R}^{2}{S}^{-1} \) . | Yes |
Corollary 4.6.17. Let \( A \in {M}_{n} \) be given.\n\n(a) \( A = {HS} \) (as well as \( A = {SH} \) ), in which \( H \) is Hermitian, \( S \) is symmetric, and either factor may be chosen to be nonsingular.\n\n(b) \( A = {BE} \) (as well as \( A = {EB} \) ), in which \( B \) is similar to a real matrix and \( E \) is ... | Proof. (a) Use (4.6.15) to write \( A = {SH}{\bar{S}}^{-1} \), in which \( S \) is nonsingular and \( H \) is Hermitian. Then \( A = \left( {{SH}{S}^{ * }}\right) \left( {{S}^{- * }{\bar{S}}^{-1}}\right) = \left( {{SH}{S}^{ * }}\right) \left( {{\bar{S}}^{-T}{\bar{S}}^{-1}}\right) \) is a product of a Hermitian matrix a... | Yes |
If \( x \) is a complex number such that \( \left| x\right| < 1 \), we know that\n\n\[{\left( 1 - x\right) }^{-1} = 1 + x + {x}^{2} + {x}^{3} + \cdots\]\n\nThis suggests the formula\n\n\[{\left( I - A\right) }^{-1} = I + A + {A}^{2} + {A}^{3} + \cdots\]\n\nfor calculating the inverse of the square matrix \( I - A \), b... | It turns out that it is sufficient that any matrix norm of \( A \) be less than 1. | Yes |
Lemma 5.1.2. If \( \parallel \cdot \parallel \) is a vector seminorm on a real or complex vector space \( V \), then \( \left| {\parallel x\parallel - \parallel y\parallel }\right| \leq \parallel x - y\parallel \) for all \( x, y \in V. \) | Proof. Since \( y = x + \left( {y - x}\right) \), the inequality\n\n\[ \parallel y\parallel \leq \parallel x\parallel + \parallel y - x\parallel = \parallel x\parallel + \parallel x - y\parallel \]\n\nfollows from the triangle inequality (3) and the homogeneity axiom (2). It follows that\n\n\[ \parallel y\parallel - \p... | Yes |
Theorem 5.1.4 (Cauchy-Schwarz inequality). Let \( \langle \cdot , \cdot \rangle \) be an inner product on a vector space \( V \) over the field \( \mathbf{F}\left( {\mathbf{F} = \mathbf{R}\text{or}\mathbf{C}}\right) \). Then\n\n\[ \n{\left| \langle x, y\rangle \right| }^{2} \leq \langle x, x\rangle \langle y, y\rangle ... | Proof. Let \( x, y \in V \) be given. If \( x = y = 0 \), there is nothing to prove, so we may assume that \( y \neq 0 \). Let \( v = \langle y, y\rangle x - \langle x, y\rangle y \) and compute\n\n\[ \n0 \leq \langle v, v\rangle = \langle \langle y, y\rangle x - \langle x, y\rangle y,\langle y, y\rangle x - \langle x,... | Yes |
Corollary 5.1.7. If \( \langle \cdot , \cdot \rangle \) is an inner product on a real or complex vector space \( V \), then the function \( \parallel \cdot \parallel : V \rightarrow \lbrack 0,\infty ) \) defined by \( \parallel x\parallel = \langle x, x{\rangle }^{1/2} \) is a norm on \( V \) . | Exercise. Prove (5.1.7). Hint: To verify the triangle inequality, compute \( \parallel x + y{\parallel }^{2} \) = \( \langle x + y, x + y\rangle \) and use the Cauchy-Schwarz inequality. | No |
Theorem 5.1.8. Let \( \langle \cdot , \cdot \rangle \) be a semi-inner product on a vector space \( V \) over the field \( \mathbf{F}\left( {\mathbf{F} = \mathbf{R}\text{ or }\mathbf{C}}\right) \) . Then \( {\left| \langle x, y\rangle \right| }^{2} \leq \langle x, x\rangle \langle y, y\rangle \) for all \( x, y \in V \... | Proof. Let \( x, y \in V \) be given. Consider the polynomial \( p\left( t\right) = \left\langle {{tx} - {e}^{i\theta }y,{tx} - {e}^{i\theta }y}\right\rangle \) . Then \( p\left( t\right) = {t}^{2}\parallel x{\parallel }^{2} - {2t}\operatorname{Re}\left( {{e}^{-{i\theta }}\langle x, y\rangle }\right) + \parallel y{\par... | Yes |
Consider the sequence \( \\left\\{ {f}_{k}\\right\\} \) of functions in \( C\\left\\lbrack {0,1}\\right\\rbrack \) (the vector space of all real-valued or complex-valued continuous functions on \( \\left\\lbrack {0,1}\\right\\rbrack \) ) defined by\n\n\[ \n{f}_{k}\\left( x\\right) = 0,\\;0 \\leq x \\leq \\frac{1}{k} \n... | A calculation reveals that\n\n\[ \n{\\begin{Vmatrix}{f}_{k}\\end{Vmatrix}}_{1} = \\frac{1}{2}{k}^{-1/2} \\rightarrow 0\\text{ as }k \\rightarrow \\infty \n\]\n\n\[ \n{\\begin{Vmatrix}{f}_{k}\\end{Vmatrix}}_{2} = \\frac{1}{\\sqrt{3}}\\text{ for all }k = 1,2,\\ldots \n\]\n\n\[ \n{\\begin{Vmatrix}{f}_{k}\\end{Vmatrix}}_{\... | Yes |
Lemma 5.4.3. Let \( \\parallel \\cdot \\parallel \) be a norm on a vector space \( V \) over the field \( \\mathbf{F}(\\mathbf{F} = \\mathbf{R} \) or C), let \( m \\geq 1 \) be a given positive integer, let \( {x}^{\\left( 1\\right) },{x}^{\\left( 2\\right) },\\ldots ,{x}^{\\left( m\\right) } \\in V \) be given vectors... | Proof. Let \( u = {\\left\\lbrack \\begin{array}{lll} {u}_{1} & \\ldots & {u}_{m} \\end{array}\\right\\rbrack }^{T} \) and \( v = {\\left\\lbrack \\begin{array}{lll} {v}_{1} & \\ldots & {v}_{m} \\end{array}\\right\\rbrack }^{T} \) . Use (5.2.1) and the Cauchy-Schwarz inequality to calculate\n\n\\[ \\left| {g\\left( {x\... | Yes |
Theorem 5.4.4. Let \( {f}_{1} \) and \( {f}_{2} \) be real-valued functions on a finite-dimensional vector space \( V \) over the field \( \mathbf{F}\left( {\mathbf{F} = \mathbf{R}\text{or}\mathbf{C}}\right) \), let \( \mathcal{B} = \left\{ {{x}^{\left( 1\right) },\ldots ,{x}^{\left( n\right) }}\right\} \) be a basis f... | Proof. Define \( h\left( z\right) = {f}_{2}\left( {x\left( z\right) }\right) /{f}_{1}\left( {x\left( z\right) }\right) \) on the Euclidean unit sphere \( S = \left\{ {z \in {\mathbf{F}}^{n}}\right. \) : \( \left. {\parallel z{\parallel }_{2} = 1}\right\} \), a compact set in \( {\mathbf{F}}^{n} \) with respect to the E... | Yes |
Corollary 5.4.6. If \( \parallel \cdot {\parallel }_{\alpha } \) and \( \parallel \cdot {\parallel }_{\beta } \) are norms on a finite-dimensional real or complex vector space \( V \), and if \( \left\{ {x}^{\left( k\right) }\right\} \) is a given sequence of vectors in \( V \), then \( \mathop{\lim }\limits_{{k \right... | Proof. Since \( {C}_{m}{\begin{Vmatrix}{x}^{\left( k\right) } - x\end{Vmatrix}}_{\alpha } \leq {\begin{Vmatrix}{x}^{\left( k\right) } - x\end{Vmatrix}}_{\beta } \leq {C}_{M}{\begin{Vmatrix}{x}^{\left( k\right) } - x\end{Vmatrix}}_{\alpha } \) for all \( k \), it follows that \( {\begin{Vmatrix}{x}^{\left( k\right) } - ... | Yes |
Corollary 5.4.8. Let \( V = {\mathbf{F}}^{n}\left( {\mathbf{F} = \mathbf{R}\text{or}\mathbf{C}}\right) \) and let \( f\left( \cdot \right) \) be a pre-norm or norm on \( V \) . The sets \( \{ x : f\left( x\right) \leq 1\} \) and \( \{ x : f\left( x\right) = 1\} \) are compact. | Proof. It suffices to show that the respective sets are closed and bounded with respect to the Euclidean norm. Theorem 5.4.4 ensures that there is some finite \( C > 0 \) such that \( \parallel x{\parallel }_{2} \leq {Cf}\left( x\right) \) for all \( x \in V \), so both of the sets \( \{ x : f\left( x\right) \leq 1\} \... | Yes |
Theorem 5.4.10. Let \( \parallel \cdot \parallel \) be a given norm on a finite-dimensional real or complex vector space \( V \), and let \( \left\{ {x}^{\left( k\right) }\right\} \) be a given sequence of vectors in \( V \) . The sequence \( \left\{ {x}^{\left( k\right) }\right\} \) converges to a vector in \( V \) if... | Proof. By choosing a basis \( \mathcal{B} \) of \( V \) and considering the equivalent norm \( {\begin{Vmatrix}{\left\lbrack x\right\rbrack }_{\mathcal{B}}\end{Vmatrix}}_{\infty } \) , we see that there is no loss of generality if we assume that \( V = {\mathbf{R}}^{n} \) or \( {\mathbf{C}}^{n} \) for some integer \( n... | Yes |
Lemma 5.4.13. Let \( f\left( \cdot \right) \) be a pre-norm on \( V = {\mathbf{F}}^{n}\left( {\mathbf{F} = \mathbf{R}\text{or}\mathbf{C}}\right) \). Then for all \( x, y \in V \) we have\n\n\[ \left| {{y}^{ * }x}\right| \leq f\left( x\right) {f}^{D}\left( y\right) \]\n\nand\n\n\[ \left| {{y}^{ * }x}\right| \leq {f}^{D}... | Proof. If \( x \neq 0 \), then\n\n\[ \left| {{y}^{ * }\frac{x}{f\left( x\right) }}\right| \leq \mathop{\max }\limits_{{f\left( z\right) = 1}}\left| {{y}^{ * }z}\right| = {f}^{D}\left( y\right) \]\n\nand hence \( \left| {{y}^{ * }x}\right| \leq f\left( x\right) {f}^{D}\left( y\right) \). Of course, this inequality is al... | Yes |
Lemma 5.4.16. Let \( f\left( \cdot \right) \) and \( g\left( \cdot \right) \) be pre-norms on \( V = {\mathbf{F}}^{n}\left( {\mathbf{F} = \mathbf{R}\text{or}\mathbf{C}}\right) \) and let \( c > 0 \) be given. Then\n\n(a) \( {cf}\left( \cdot \right) \) is a pre-norm on \( V \), and its dual norm is \( {c}^{-1}{f}^{D}\le... | Proof. The function \( {cf}\left( \cdot \right) \) is positive, homogeneous, and continuous, so it is a pre-norm. The remaining assertions follow from (5.4.12a). | No |
Theorem 5.4.17. Let \( \parallel \cdot \parallel \) be a norm on \( V = {\mathbf{F}}^{n}\left( {\mathbf{F} = \mathbf{R}\text{or}\mathbf{C}}\right) \) and let \( c > 0 \) be given. Then \( \parallel x\parallel = c\parallel x{\parallel }^{D} \) for all \( x \in V \) if and only if \( \parallel \cdot \parallel = \sqrt{c}\... | Proof. If \( \parallel \cdot \parallel = \sqrt{c}\parallel \cdot {\parallel }_{2} \) and \( x \in V \), then (5.4.16(a)) ensures that \( \parallel \cdot {\parallel }^{D} = \frac{1}{\sqrt{c}}\parallel \cdot {\parallel }_{2}^{D} = \) \( \frac{1}{\sqrt{c}}\parallel \cdot {\parallel }_{2} = {c}^{-1}\parallel \cdot \paralle... | Yes |
Theorem 5.4.19. Let \( \parallel \cdot \parallel \) be a norm on \( V = {\mathbf{F}}^{n}\left( {\mathbf{F} = \mathbf{R}\text{or}\mathbf{C}}\right) \) . (a) If \( \parallel \cdot \parallel \) is absolute, then \[ \parallel y{\parallel }^{D} = \mathop{\max }\limits_{{x \neq 0}}\frac{{\left| y\right| }^{T}\left| x\right| ... | Proof. Suppose that \( \mathbf{F} = \mathbf{C} \) . (a) Suppose that \( \parallel \cdot \parallel \) is absolute. For a given \( y = \left\lbrack {y}_{k}\right\rbrack \in {\mathbf{C}}^{n} \), any \( x = \left\lbrack {x}_{k}\right\rbrack \in {\mathbf{C}}^{n} \), and any \( z = \left\lbrack {z}_{k}\right\rbrack \in {\mat... | Yes |
A set \( B \) in a finite-dimensional real or complex vector space \( V \) with positive dimension is the unit ball of a norm if and only if \( B \) (i) is compact,(ii) is convex, (iii) is equilibrated, and (iv) has 0 as an interior point. | The necessity of conditions (i)-(iv) has already been observed. To establish their sufficiency, consider any nonzero point \( x \in V \) . Construct a ray segment \( \{ {\alpha x} \) : \( 0 \leq \alpha \leq 1\} \) from the origin through \( x \) and define the \ | No |
Theorem 5.5.9 (Duality theorem). Let \( f \) be a pre-norm on \( V = {\mathbf{R}}^{n} \) or \( {\mathbf{C}}^{n} \), let \( {f}^{D} \) denote the dual norm of \( f \), let \( {f}^{DD} \) denote the dual norm of \( {f}^{D} \), let \( B = \{ x \in V \) : \( f\left( x\right) \leq 1\} \), and let \( {B}^{\prime \prime } = \... | Proof. (a) If \( x \in V \) is a given vector, then (5.4.13) ensures that \( \left| {{y}^{ * }x}\right| \leq f\left( x\right) {f}^{D}\left( y\right) \) for any \( y \in V \), and hence\n\n\[ \n{f}^{DD}\left( x\right) = \mathop{\max }\limits_{{{f}^{D}\left( y\right) = 1}}\left| {{y}^{ * }x}\right| \leq \mathop{\max }\li... | Yes |
Corollary 5.5.11. An absolute norm on \( {\mathbf{R}}^{n} \) or \( {\mathbf{C}}^{n} \) is monotone. | Proof. Suppose that \( \parallel \cdot \parallel \) is an absolute norm on \( {\mathbf{F}}^{n} \) . Theorem 5.4.19(b) ensures that its dual \( \parallel \cdot {\parallel }^{D} \) is absolute. The duality theorem tells us that \( \parallel \cdot \parallel \) is the dual of the absolute norm \( \parallel \cdot {\parallel... | Yes |
Theorem 5.6.2. The function \( \parallel \mid \mid \parallel \) defined in (5.6.1) has the following properties:\n\n(a) \( \parallel \mid I\parallel \mid = 1 \)\n\n(b) \( \parallel {Ay}\parallel \leq \parallel \mid A\parallel \parallel y\parallel \) for any \( A \in {M}_{n} \) and any \( y \in {\mathbf{C}}^{n} \)\n\n(c... | Proof. (a) \( \parallel \mid I\parallel \mid = \mathop{\max }\limits_{{\parallel x\parallel = 1}}\parallel {Ix}\parallel = \mathop{\max }\limits_{{\parallel x\parallel = 1}}\parallel x\parallel = 1 \) .\n\n(b) The asserted inequality is correct for \( y = 0 \), so let \( y \neq 0 \) be given and consider the unit vecto... | Yes |
The maximum column sum matrix norm \( \parallel \cdot {\parallel }_{1} \) is defined on \( {M}_{n} \) by \[ \parallel \mid A\parallel {\parallel }_{1} = \mathop{\max }\limits_{{1 \leq j \leq n}}\mathop{\sum }\limits_{{i = 1}}^{n}\left| {a}_{ij}\right| \] We claim that \( \parallel \mid \cdot {\parallel }_{1} \) is indu... | To show this, partition \( A \) according to its columns as \( A = \left\lbrack \begin{array}{lll} {a}_{1} & \ldots & {a}_{n} \end{array}\right\rbrack \). Then \( \parallel \mid A\parallel {\parallel }_{1} = \mathop{\max }\limits_{{1 \leq i \leq n}}{\begin{Vmatrix}{a}_{i}\end{Vmatrix}}_{1} \). If \( x = \left\lbrack {x... | Yes |
The maximum row sum matrix norm \( \parallel \mid \cdot {\parallel }_{\infty } \) is defined on \( {M}_{n} \) by\n\n\[ \parallel A{\parallel }_{\infty } = \mathop{\max }\limits_{{1 \leq i \leq n}}\mathop{\sum }\limits_{{j = 1}}^{n}\left| {a}_{ij}\right| \]\n\nWe claim that \( \parallel \mid \cdot {\parallel }_{\infty }... | Compute\n\n\[ \parallel {Ax}{\parallel }_{\infty } = \mathop{\max }\limits_{{1 \leq i \leq n}}\left| {\mathop{\sum }\limits_{{j = 1}}^{n}{a}_{ij}{x}_{j}}\right| \leq \mathop{\max }\limits_{{1 \leq i \leq n}}\mathop{\sum }\limits_{{j = 1}}^{n}\left| {{a}_{ij}{x}_{j}}\right| \]\n\n\[ \leq \mathop{\max }\limits_{{1 \leq i... | Yes |
The spectral norm \( \parallel \mid \cdot {\parallel }_{2} \) is defined on \( {M}_{n} \) by\n\n\[ \parallel \mid A{\parallel }_{2} = {\sigma }_{1}\left( A\right) \text{, the largest singular value of}A \] \n\nWe claim that \( \parallel \cdot {\parallel }_{2} \) is induced by the \( {l}_{2} \) -norm on \( {\mathbf{C}}^... | Let \( A = {V\sum }{W}^{ * } \) be a singular value decomposition of \( A \), in which \( V \) and \( W \) are unitary, \( \sum = \operatorname{diag}\left( {{\sigma }_{1},\ldots ,{\sigma }_{n}}\right) \), and \( {\sigma }_{1} \geq \cdots \geq {\sigma }_{n} \geq 0 \) ; see (2.6.3). Use unitary invariance and monotonicit... | Yes |
Theorem 5.6.7. Suppose that \( \parallel \cdot \parallel \mid \) is a matrix norm on \( {M}_{n} \) and \( S \in {M}_{n} \) is nonsingular. Then the function\n\n\[ \parallel \mid A\parallel {\parallel }_{S} = \begin{Vmatrix}{\mid {SA}{S}^{-1}}\end{Vmatrix}\;\text{ for all }A \in {M}_{n} \]\n\nis a matrix norm. Moreover,... | Proof. The axioms (1), (1a), (2), and (3) are verified in a straightforward manner for \( \parallel \mid \cdot {\parallel }_{S} \) . Submultiplicativity of \( \parallel \cdot {\parallel }_{S} \) follows from a calculation:\n\n\[ \parallel \mid {AB}\parallel {\parallel }_{S} = \begin{Vmatrix}{\mid {SAB}{S}^{-1}\parallel... | Yes |
Lemma 5.6.10. Let \( A \in {M}_{n} \) and \( \epsilon > 0 \) be given. There is a matrix norm \( \parallel \cdot \parallel \) such that \( \rho \left( A\right) \leq \parallel A\parallel \mid \leq \rho \left( A\right) + \epsilon \) . | Proof. Theorem 2.3.1 ensures that there is a unitary \( U \in {M}_{n} \) and an upper triangular \( \Delta \in {M}_{n} \) such that \( A = {U\Delta }{U}^{ * } \) . Set \( {D}_{t} = \operatorname{diag}\left( {t,{t}^{2},{t}^{3},\ldots ,{t}^{n}}\right) \) and compute\n\n\[ \n{D}_{t}\Delta {D}_{t}^{-1} = \left\lbrack \begi... | Yes |
Lemma 5.6.11. Let \( A \in {M}_{n} \) be given. If there is a matrix norm \( \parallel \cdot \parallel \mid \) such that \( \parallel A\parallel \) \( < 1 \), then \( \mathop{\lim }\limits_{{k \rightarrow \infty }}{A}^{k} = 0 \), that is, each entry of \( {A}^{k} \) tends to zero as \( k \rightarrow \infty \) . | Proof. If \( \parallel A\parallel \mid \mid < 1 \), then \( \begin{Vmatrix}{A}^{k}\end{Vmatrix} \mid \mid \leq \parallel A{\parallel }^{k} \rightarrow 0 \) as \( k \rightarrow \infty \) . This says that \( {A}^{k} \rightarrow 0 \) with respect to the norm \( \parallel \mid \parallel \), but since all norms on the \( {n... | Yes |
Theorem 5.6.12. Let \( A \in {M}_{n} \) . Then \( \mathop{\lim }\limits_{{k \rightarrow \infty }}{A}^{k} = 0 \) if and only if \( \rho \left( A\right) < 1 \) . | Proof. If \( {A}^{k} \rightarrow 0 \) and if \( x \neq 0 \) is a vector such that \( {Ax} = {\lambda x} \), then \( {A}^{k}x = {\lambda }^{k}x \rightarrow 0 \) only if \( \left| \lambda \right| < 1 \) . Since this inequality must hold for every eigenvalue of \( A \), we conclude that \( \rho \left( A\right) < 1 \) . Co... | Yes |
Corollary 5.6.13. Let \( A \in {M}_{n} \) and \( \epsilon > 0 \) be given. There is a constant \( C = C\left( {A,\epsilon }\right) \) such that \( \left| {\left( {A}^{k}\right) }_{ij}\right| \leq C{\left( \rho \left( A\right) + \epsilon \right) }^{k} \) for all \( k = 1,2,\ldots \) and all \( i, j = 1,\ldots, n \) . | Proof. Consider the matrix \( \widetilde{A} = {\left\lbrack \rho \left( A\right) + \epsilon \right\rbrack }^{-1}A \), whose spectral radius is strictly less than 1. We know that \( {\widetilde{A}}^{k} \rightarrow 0 \) as \( k \rightarrow \infty \) . In particular, the sequence \( \left\{ {\widetilde{A}}^{k}\right\} \) ... | Yes |
Corollary 5.6.14 (Gelfand formula). Let \( \parallel \mid \mid \parallel \) be a matrix norm on \( {M}_{n} \) and let \( A \in \) \( {M}_{n} \) . Then \( \rho \left( A\right) = \mathop{\lim }\limits_{{k \rightarrow \infty }}{\begin{Vmatrix}{A}^{k}\end{Vmatrix}}^{1/k} \) . | Proof. Since \( \rho {\left( A\right) }^{k} = \rho \left( {A}^{k}\right) \leq \begin{Vmatrix}{A}^{k}\end{Vmatrix} \), we have \( \rho \left( A\right) \leq {\begin{Vmatrix}{A}^{k}\end{Vmatrix}}^{1/k} \) for all \( k = 1,2,\ldots \) . If \( \epsilon > 0 \) is given, the matrix \( \widetilde{A} = {\left\lbrack \rho \left(... | Yes |
Corollary 5.6.16. A matrix \( A \in {M}_{n} \) is nonsingular if there is a matrix norm \( \parallel \cdot \parallel \) such that \( \parallel I - A\parallel \mid < 1 \) . If this condition is satisfied, \[ {A}^{-1} = \mathop{\sum }\limits_{{k = 0}}^{\infty }{\left( I - A\right) }^{k} \] | Proof. If \( \parallel I - A\parallel \mid < 1 \), then the series \( \mathop{\sum }\limits_{{k = 0}}^{\infty }{\left( I - A\right) }^{k} \) converges to some matrix \( C \) because the radius of convergence of the series \( \sum {z}^{k} \) is 1 . But since \[ A\mathop{\sum }\limits_{{k = 0}}^{N}{\left( I - A\right) }^... | Yes |
Corollary 5.6.17. Let \( A = \left\lbrack {a}_{ij}\right\rbrack \in {M}_{n} \) . If \( \left| {a}_{ii}\right| > \mathop{\sum }\limits_{{j \neq i}}\left| {a}_{ij}\right| \) for all \( i = 1,\ldots, n \), then \( A \) is nonsingular. | Proof. The hypothesis ensures that every main diagonal entry of \( A \) is nonzero. Set \( D = \operatorname{diag}\left( {{a}_{11},\ldots ,{a}_{nn}}\right) \) and check that \( {D}^{-1}A \) has all \( 1\mathrm{\;s} \) on the main diagonal, \( B = \left\lbrack {b}_{ij}\right\rbrack = I - {D}^{-1}A \) has all 0s on the m... | Yes |
Theorem 5.6.18. Let \( \parallel \cdot {\parallel }_{\alpha } \) and \( \parallel \cdot {\parallel }_{\beta } \) be given norms on \( {\mathbf{C}}^{n} \) . Let \( \parallel \mid \cdot {\parallel }_{\alpha } \) and \( \parallel \mid \cdot {\parallel }_{\beta } \) denote their respective induced matrix norms on \( {M}_{n... | Proof. The inequalities (5.6.10) say that \( \parallel x{\parallel }_{\alpha } \leq {R}_{\alpha \beta }\parallel x{\parallel }_{\beta } \) and \( \parallel y{\parallel }_{\beta } \leq {R}_{\beta \alpha }\parallel y{\parallel }_{\alpha } \) for all \( x, y \in {\mathbf{C}}^{n} \), with equality possible in both cases fo... | Yes |
Lemma 5.6.23. Let \( \parallel \cdot {\parallel }_{\alpha } \) and \( \parallel \cdot {\parallel }_{\beta } \) be norms on \( {\mathbf{C}}^{n} \), and let \( \parallel \cdot {\parallel }_{\alpha } \) and \( \parallel \cdot {\parallel }_{\beta } \) denote their respective induced matrix norms on \( {M}_{n} \). Then\n\n\... | Proof. Observe that\n\n\[ {R}_{\beta \alpha } = \mathop{\max }\limits_{{x \neq 0}}\frac{\parallel x{\parallel }_{\beta }}{\parallel x{\parallel }_{\alpha }} = {\left( \mathop{\min }\limits_{{x \neq 0}}\frac{\parallel x{\parallel }_{\alpha }}{\parallel x{\parallel }_{\beta }}\right) }^{-1} \geq {\left( \mathop{\max }\li... | Yes |
Corollary 5.6.25. Let \( \parallel \mid \mid {\parallel }_{\alpha } \) and \( \parallel \mid \mid {\parallel }_{\alpha } \) be induced matrix norms on \( {M}_{n} \) . Then \( \parallel \mid A\parallel {\parallel }_{\alpha } \leq \parallel \mid A\parallel {\parallel }_{\beta } \) for all \( A \in {M}_{n} \) if and only ... | Proof. If \( \parallel A{\parallel }_{\alpha } \leq \parallel A{\parallel }_{\beta } \) for all \( A \in {M}_{n} \), then (5.6.21) ensures that \( \parallel A{\parallel }_{\beta } \leq \parallel A{\parallel }_{\alpha } \) for all \( A \in {M}_{n} \) . | Yes |
Theorem 5.6.26. Let \( \parallel \cdot \parallel \) be a given matrix norm on \( {M}_{n} \), let \( \parallel \cdot {\parallel }_{\alpha } \) be a given induced matrix norm on \( {M}_{n} \), let a nonzero \( z \in {\mathbf{C}}^{n} \) be given, and define\n\n\[ \parallel x{\parallel }_{z} = \begin{Vmatrix}{x{z}^{ * }}\e... | Proof. (a) One verifies that \( \parallel \cdot {\parallel }_{z} \) satisfies the four axioms in (5.1.1); submultiplicativity of \( \parallel \cdot \parallel \) is not necessary for this purpose.\n\n(b) Use submultiplicativity of \( \parallel \cdot \parallel \mid \) to compute\n\n\[ {N}_{z}\left( A\right) = \mathop{\ma... | Yes |
Theorem 5.6.32. Let \( \parallel \cdot \parallel \) be a matrix norm on \( {M}_{n} \) . For a nonzero \( z \in {\mathbf{C}}^{n} \) let \( {N}_{z}\left( \cdot \right) \) be the induced matrix norm defined by (5.6.27) and (5.6.28). The following are equivalent:\n\n(a) \( \parallel \cdot \parallel \mid \) is an induced ma... | Proof. The implication (a) \( \Rightarrow \) (b) follows from (5.6.26(c)). The implication (b) \( \Rightarrow \) (c) is \( \left( {{5.6.26}\left( \mathrm{\;b}\right) }\right) \) . The implications \( \left( \mathrm{c}\right) \Rightarrow \left( \mathrm{d}\right) \Rightarrow \left( \mathrm{a}\right) \) are straightforwar... | No |
Theorem 5.6.33. Let \( \parallel \cdot \parallel \) be a matrix norm on \( {M}_{n} \) and let \( \parallel \cdot {\parallel }_{z} \) be the norm on \( {\mathbf{C}}^{n} \) defined by (5.6.27). The following two statements are equivalent:\n\n(a) For each pair of nonzero vectors \( y, z \in {\mathbf{C}}^{n} \) there is a ... | Proof. Assume (a). Since (b) is correct if either \( y = 0 \) or \( z = 0 \), we may assume that \( y \neq 0 \neq z \) . Then\n\n\[ \begin{Vmatrix}{\mid x{z}^{ * }}\end{Vmatrix}\begin{Vmatrix}{\mid z{y}^{ * }}\end{Vmatrix} = \parallel x{\parallel }_{z}\parallel z{\parallel }_{y} = {c}_{yz}^{-1}\parallel x{\parallel }_{... | Yes |
Theorem 5.6.34. Let \( \parallel \cdot \parallel \) be a unitarily invariant matrix norm on \( {M}_{n} \), and suppose that \( z \in {\mathbf{C}}^{n} \) is nonzero. Then\n\n(a) the vector norm \( \parallel \cdot {\parallel }_{z} \) defined by (5.6.27) is unitarily invariant\n\n(b) \( \parallel \cdot {\parallel }_{z} \)... | Proof. (a) If \( U \in {M}_{n} \) is unitary, then \( \parallel {Ux}{\parallel }_{z} = \begin{Vmatrix}{{Ux}{z}^{ * }}\end{Vmatrix} = \begin{Vmatrix}{x{z}^{ * }}\end{Vmatrix} = \parallel x{\parallel }_{z} \).\n\n(b) For each \( x \in {\mathbf{C}}^{n} \) there is a unitary \( U \in {M}_{n} \) such that \( {Ux} = \paralle... | Yes |
Theorem 5.6.35. Let \( \parallel \cdot \parallel \) be the matrix norm on \( {M}_{n} \) that is induced by a norm \( \parallel \cdot \parallel \) on \( {\mathbf{C}}^{n} \). Then\n\n(a) \( \parallel \cdot {\parallel }^{\prime } \) is induced by the norm \( \parallel \cdot {\parallel }^{D} \)\n\n(b) if \( \parallel \cdot... | Proof. (a) Use (5.6.2(d)) to compute\n\n\[ \parallel \left| A\right| {\parallel }^{\prime } = \begin{Vmatrix}\left| {A}^{ * }\right| \end{Vmatrix} = \mathop{\max }\limits_{{\parallel x\parallel = \parallel y{\parallel }^{D} = 1}}\left| {{y}^{ * }{A}^{ * }x}\right| = \mathop{\max }\limits_{{\parallel x\parallel = \paral... | Yes |
Theorem 5.6.36. Let \( \parallel \cdot \parallel \) be the matrix norm on \( {M}_{n} \) induced by a norm \( \parallel \cdot \parallel \) on \( {\mathbf{C}}^{n} \) . The following are equivalent:\n\n(a) \( \parallel \cdot \parallel \) is an absolute norm.\n\n(b) \( \parallel \cdot \parallel \) is a monotone norm.\n\n(c... | Proof. The equivalence of (a) and (b) is the assertion in (5.4.19(c)). To prove that (b) implies (c), suppose that \( \parallel \cdot \parallel \) is monotone, let \( \Lambda = \operatorname{diag}\left( {{\lambda }_{1},\ldots ,{\lambda }_{n}}\right) \), and let \( L = \) \( \max \left\{ {\left| {\lambda }_{1}\right| ,\... | Yes |
Theorem 5.6.39. Let \( \\parallel \\cdot \\parallel \) be a norm on \( {M}_{n} \). Then\n\n(a) \( \\parallel \\cdot \\parallel \) is self-adjoint if and only if \( \\parallel \\cdot {\\parallel }^{D} \) is self-adjoint\n\n(b) \( \\parallel \\cdot \\parallel \) is unitarily invariant if and only if \( \\parallel \\cdot ... | Proof. In each case, the \ | No |
Theorem 5.6.40. Let \( \parallel \cdot \parallel \mid \) be a matrix norm on \( {M}_{n} \) . Then\n\n\[ \parallel {AB}{\parallel }^{D} \leq \left\{ \begin{array}{l} \begin{Vmatrix}{A}^{ * }\end{Vmatrix}\parallel \parallel B{\parallel }^{D} \\ \parallel A{\parallel }^{D}\parallel {B}^{ * }\parallel \end{array}\right. \]... | Proof. We prove only the second upper bound. Let \( X \in {M}_{n} \) be such that \( \parallel X\parallel = 1 \) and \( \left| {\operatorname{tr}{X}^{ * }{AB}}\right| = \parallel {AB}{\parallel }^{D} \) . Use submultiplicativity of \( \parallel \cdot \parallel \) to compute\n\n\[ {\begin{Vmatrix}\left| AB\right| \end{V... | Yes |
Theorem 5.6.41. Let \( \parallel \cdot \parallel \) be a matrix norm on \( {M}_{n} \) that is induced by the norm \( \parallel \cdot \parallel \) on \( {\mathbf{C}}^{n} \). Then\n\n(a) \( \begin{Vmatrix}{A}^{ * }\end{Vmatrix} = \max \left\{ {\left| {\operatorname{tr}{B}^{ * }A}\right| : \parallel \mid B\parallel \mid =... | Proof. (a) If \( B = x{y}^{ * } \) for some nonzero \( x, y \in {\mathbf{C}}^{n} \), then (5.6.30) ensures that \( \parallel B\parallel = \) \( \begin{Vmatrix}{x{y}^{ * }}\end{Vmatrix} = \parallel x\parallel \parallel y{\parallel }^{D} \). Compute\n\n\[ \mathop{\max }\limits_{{\operatorname{rank}B = 1}}\frac{\left| \op... | Yes |
Theorem 5.6.42. Suppose that an absolute norm \( \parallel \cdot \parallel \) on \( {\mathbf{C}}^{n} \) induces the matrix norm \( \parallel \mid \cdot \parallel \mid \) on \( {M}_{n} \), and let \( \parallel \mid \cdot {\parallel }^{D} \) be its dual. Then \( \parallel \mid \cdot {\parallel }^{D} \) is a matrix norm, ... | Proof. The preceding theorem ensures that \( \parallel \cdot {\parallel }^{D} \) is a matrix norm. Write \( \Lambda = \) \( \operatorname{diag}\left( {{e}^{i{\theta }_{1}}\left| {\lambda }_{1}\right| ,\ldots ,{e}^{i{\theta }_{n}}\left| {\lambda }_{n}\right| }\right) \) and let \( U = \operatorname{diag}\left( {{e}^{i{\... | Yes |
The \( {l}_{\infty } \) norm on \( {M}_{n} \) is\n\n\[ \parallel A{\parallel }_{\infty } = \mathop{\max }\limits_{{1 \leq i, j \leq n}}\left| {a}_{ij}\right| \] | We saw in (5.6.0.3-4) that \( \parallel \cdot {\parallel }_{\infty } \) is a norm on \( {M}_{n} \) but not a matrix norm. However, \( n\parallel \cdot {\parallel }_{\infty } \) is a matrix norm. | No |
Theorem 5.7.10. If \( f \) is a pre-norm on \( {M}_{n} \), in particular, if it is a vector norm, then \( \mathop{\lim }\limits_{{k \rightarrow \infty }}{\left\lbrack f\left( {A}^{k}\right) \right\rbrack }^{1/k} = \rho \left( A\right) \) for all \( A \in {M}_{n}. \) | Proof. Let \( \parallel \cdot \parallel \mid \) be a matrix norm on \( {M}_{n} \) and consider the inequality (5.7.9), which implies that\n\n\[ \n{C}_{m}^{1/k}{\begin{Vmatrix}{A}^{k}\end{Vmatrix}}^{1/k} \leq {\left\lbrack f\left( {A}^{k}\right) \right\rbrack }^{1/k} \leq {C}_{M}^{1/k}{\begin{Vmatrix}{A}^{k}\end{Vmatrix... | Yes |
Theorem 5.7.11. Let \( G\left( \cdot \right) \) be a vector norm on \( {M}_{n} \) and let\n\n\[ c\left( G\right) = \mathop{\max }\limits_{{G\left( A\right) = 1 = G\left( B\right) }}G\left( {AB}\right) \]\n\nFor a real positive scalar \( \gamma ,{\gamma G}\left( \cdot \right) \) is a matrix norm on \( {M}_{n} \) if and ... | Proof. The value \( c\left( G\right) \) is the maximum of a positive continuous function over a compact set, so it is finite and positive. For any nonzero \( A, B \in {M}_{n} \) we have\n\n\[ c\left( G\right) \geq G\left( {\frac{A}{G\left( A\right) }\frac{B}{G\left( B\right) }}\right) = \frac{G\left( {AB}\right) }{G\le... | Yes |
Theorem 5.7.13. If \( \parallel \cdot \parallel \mid \) is a matrix norm on \( {M}_{n} \), then there is a norm on \( {\mathbf{C}}^{n} \) that is compatible with it. If \( \parallel \cdot \parallel \) is a norm on \( {\mathbf{C}}^{n} \), then there is a matrix norm on \( {M}_{n} \) that is compatible with it. | Proof. For any nonzero vector \( z \), the norm \( \parallel \cdot {\parallel }_{z} \) defined in (5.6.27) is compatible with a given matrix norm \( \parallel \cdot \parallel : \parallel {Ax}{\parallel }_{z} = \begin{Vmatrix}{{Ax}{z}^{ * }}\end{Vmatrix} \leq \parallel A\parallel \parallel \parallel x{z}^{ * }\parallel ... | Yes |
Theorem 5.7.14. Let \( G\left( \cdot \right) \) be a norm on \( {M}_{n} \) that is compatible with a norm \( \parallel \cdot \parallel \) on \( {\mathbf{C}}^{n} \) . Then\n\n\[ G\left( {A}_{1}\right) \cdots G\left( {A}_{k}\right) \geq \rho \left( {{A}_{1}\cdots {A}_{k}}\right) \text{for all}{A}_{1},\ldots ,{A}_{k} \in ... | Proof. Consider the case \( k = 2 \) and let \( x \in {\mathbf{C}}^{n} \) be a nonzero vector such that \( {A}_{1}{A}_{2}x = \) \( {\lambda x} \) with \( \left| \lambda \right| = \rho \left( {{A}_{1}{A}_{2}}\right) \) . Then\n\n\[ \rho \left( {{A}_{1}{A}_{2}}\right) \parallel x\parallel = \parallel {\lambda x}\parallel... | Yes |
Lemma 5.7.16. Let \( G\left( \cdot \right) \) be a vector norm on \( {M}_{n} \) that satisfies (5.7.15). There is a finite positive constant \( \gamma \left( G\right) \) such that\n\n\[ G\left( {A}_{1}\right) \cdots G\left( {A}_{k}\right) \geq \gamma \left( G\right) {\begin{Vmatrix}{A}_{1}\cdots {A}_{k}\end{Vmatrix}}_{... | Proof. Let \( k \) be a given positive integer, let \( {A}_{1},\ldots ,{A}_{k} \in {M}_{n} \) be given, and let \( {A}_{1}\cdots {A}_{k} = {V\sum }{W}^{ * } \) be a singular value decomposition (2.6.3). The hypothesis permits us to use (5.7.15) to compute\n\n\[ G\left( {V}^{ * }\right) G\left( {A}_{1}\right) \cdots G\l... | Yes |
Theorem 5.7.17. A vector norm \( G\left( \cdot \right) \) on \( {M}_{n} \) is compatible with some norm on \( {\mathbf{C}}^{n} \) if and only it satisfies the inequality (5.7.15). | Proof. One implication has already been proved in (5.7.14). To prove the other, we claim it is sufficient to show that there is a matrix norm \( \parallel \mid \cdot \parallel \) on \( {M}_{n} \) such that \( G\left( A\right) \geq \parallel A\parallel \mid \) for all \( A \in {M}_{n} \) . If such a matrix norm exists, ... | Yes |
Theorem 5.7.18. Every norm on \( {\mathbf{C}}^{n} \) is compatible with a vector norm on \( {M}_{n} \) that is not a matrix norm. | Proof. Let \( \parallel \cdot \parallel \) be a norm on \( {\mathbf{C}}^{n} \) and let \( P \in {M}_{n} \) be any permutation matrix with a zero main diagonal, for example, the circulant matrix (0.9.6.2). Let \( \parallel \cdot \parallel \) denote the matrix norm on \( {M}_{n} \) that is induced by \( \parallel \cdot \... | Yes |
Lemma 5.7.19. Let \( G\left( \cdot \right) \) be a spectrally dominant norm on \( {M}_{n} \), let \( A \in {M}_{n} \) be given, and let \( \lambda \) be an eigenvalue of \( A \) such that \( \left| \lambda \right| = \rho \left( A\right) \) . If \( \lambda \) is not semisimple, then \( G\left( A\right) > \rho \left( A\r... | Proof. If \( \rho \left( A\right) = 0 \), then 0 is not a semisimple eigenvalue of \( A \) if and only if \( A \neq 0 \) , in which case \( G\left( A\right) > 0 \) . We may therefore assume that \( \rho \left( A\right) \neq 0 \) . Since we may normalize any maximum-modulus eigenvalue by considering \( {e}^{i\theta }\rh... | Yes |
Theorem 5.7.20. A norm \( G\left( \cdot \right) \) on \( {M}_{n} \) is spectrally dominant if and only if for each \( A \in {M}_{n} \) there is a positive constant \( {\gamma }_{A} \) (depending only on \( A \) and \( G\left( \cdot \right) \) ) such that\n\n\[ G\left( {A}^{k}\right) \leq {\gamma }_{A}G{\left( A\right) ... | Proof. We need to show only that the stated condition is necessary. Suppose that \( A \in {M}_{n} \) and \( G\left( A\right) = 1 \geq \rho \left( A\right) \) . The preceding lemma ensures that each Jordan block of \( A \) has the form \( {J}_{m}\left( \lambda \right) \), in which \( \left| \lambda \right| \leq 1 \), an... | Yes |
Corollary 6.1.3. The eigenvalues of \( A = \left\lbrack {a}_{ij}\right\rbrack \in {M}_{n} \) are in the union of \( n \) discs | \[ \mathop{\bigcup }\limits_{{j = 1}}^{n}\left\{ {z \in \mathbf{C} : \left| {z - {a}_{jj}}\right| \leq {C}_{j}^{\prime }}\right\} = G\left( {A}^{T}\right) \] | Yes |
Corollary 6.1.5. If \( A = \left\lbrack {a}_{ij}\right\rbrack \in {M}_{n} \), then\n\n\[ \rho \left( A\right) \leq \min \left\{ {\mathop{\max }\limits_{i}\mathop{\sum }\limits_{{j = 1}}^{n}\left| {a}_{ij}\right| ,\mathop{\max }\limits_{j}\mathop{\sum }\limits_{{i = 1}}^{n}\left| {a}_{ij}\right| }\right\} \]\n\nThis res... | But it is interesting to have an essentially geometric derivation of this fact.\n\nSince \( {S}^{-1}{AS} \) has the same eigenvalues as \( A \) whenever \( S \) is nonsingular, we can apply the Geršgorin theorem to \( {S}^{-1}{AS} \) and thereby obtain additional eigenvalue inclusion sets for \( A \) . A particularly c... | No |
Let \( A = \left\lbrack {a}_{ij}\right\rbrack \in {M}_{n} \) and let \( {p}_{1},{p}_{2},\ldots ,{p}_{n} \) be positive real numbers. The eigenvalues of \( A \) are in the union of \( n \) discs \[ \mathop{\bigcup }\limits_{{i = 1}}^{n}\left\{ {z \in \mathbf{C} : \left| {z - {a}_{ii}}\right| \leq \frac{1}{{p}_{i}}\matho... | The matrix \( A = \left\lbrack \begin{array}{ll} 1 & 1 \\ 0 & 2 \end{array}\right\rbrack \) has eigenvalues 1 and 2 . A straightforward application of the Geršgorin theorem gives a rather gross estimate for the eigenvalues (Figure 6.1.7a), but the extra parameters in the preceding corollary give enough flexibility to o... | Yes |
Theorem 6.1.11. Suppose that \( A = \left\lbrack {a}_{ij}\right\rbrack \in {M}_{n} \) has nonzero diagonal entries. If \( A \) is diagonally dominant and \( \left| {a}_{ii}\right| > {R}_{i}^{\prime } \) for at least \( n - 1 \) values of \( i \in \{ 1,\ldots, n\} \), then it is nonsingular. | Proof. For some \( k \) we have \( \left| {a}_{ii}\right| > {R}_{i}^{\prime } \) for all \( i \neq k \), and \( \left| {a}_{kk}\right| \geq {R}_{k}^{\prime } \) . If \( \left| {a}_{kk}\right| > {R}_{k}^{\prime } \) , nonsingularity of \( A \) follows from (6.1.10), so we suppose that \( \left| {a}_{kk}\right| = {R}_{k}... | Yes |
Lemma 6.2.3. Let \( \lambda, x \) be an eigenpair for \( A = \left\lbrack {a}_{ij}\right\rbrack \in {M}_{n} \) and suppose that \( \lambda \) satisfies the inequalities (6.2.2a). Then\n\n(a) if \( p \in \{ 1,\ldots, n\} \) and \( \left| {x}_{p}\right| = \parallel x{\parallel }_{\infty } \), then \( \left| {\lambda - {a... | Proof. Suppose that \( \left| {x}_{p}\right| = \parallel x{\parallel }_{\infty } \) . Then (6.1.1a) ensures that\n\n\[ \left| {\lambda - {a}_{pp}}\right| \parallel x{\parallel }_{\infty } = \left| {\lambda - {a}_{pp}}\right| \left| {x}_{p}\right| = \left| {\mathop{\sum }\limits_{{j \neq p}}{a}_{pj}{x}_{j}}\right| \]\n\... | Yes |
Corollary 6.2.6. Let \( A = \left\lbrack {a}_{ij}\right\rbrack \in {M}_{n} \), and suppose that every entry of \( A \) is nonzero. If \( A \) is diagonally dominant and if there is a \( k \in \{ 1,\ldots, n\} \) such that \( \left| {a}_{kk}\right| > {R}_{k}^{\prime } \), then \( A \) is nonsingular. | Proof. Since \( A \) is diagonally dominant, \( \lambda = 0 \) satisfies the inequalities (6.2.2a). The hypothesis ensures that the \( k \) th Geršgorin circle does not pass through 0, so it follows from the preceding theorem that 0 is not an eigenvalue of \( A \) . | Yes |
Theorem 6.2.8 (Better theorem). Let \( A \in {M}_{n} \), and let \( \lambda, x = \left\lbrack {x}_{i}\right\rbrack \) be an eigenpair of A such that \( \lambda \) satisfies the inequalities (6.2.2a). If \( A \) has property \( {SC} \), then\n\n(a) every Geršgorin circle passes through \( \lambda \)\n\n(b) \( \left| {x}... | Proof. Let \( p \in \{ 1,\ldots, n\} \) be an index such that \( \left| {x}_{p}\right| = \parallel x{\parallel }_{\infty } \) . Then (6.2.3a) ensures that \( \left| {\lambda - {a}_{pp}}\right| = {R}_{p}^{\prime } \), so the \( p \) th Geršgorin circle passes through \( \lambda \) . Let \( q \in \{ 1,\ldots, n\} \) be a... | Yes |
Theorem 6.2.16. Let \( A \in {M}_{n} \), and let \( {P}_{i} \) and \( {P}_{j} \) be given nodes of \( \Gamma \left( A\right) \). The following are equivalent:\n\n(a) There is a directed path of length \( m \) in \( \Gamma \left( A\right) \) from \( {P}_{i} \) to \( {P}_{j} \).\n\n(b) The \( i, j \) entry of \( {\left| ... | Proof. We proceed by induction. For \( m = 1 \) the assertion is trivial. For \( m = 2 \) we compute\n\n\[{\left( {\left| A\right| }^{2}\right) }_{ij} = \mathop{\sum }\limits_{{k = 1}}^{n}{\left| A\right| }_{ik}{\left| A\right| }_{kj} = \mathop{\sum }\limits_{{k = 1}}^{n}\left| {a}_{ik}\right| \left| {a}_{kj}\right|\]\... | Yes |
Corollary 6.2.19. Let \( A \in {M}_{n} \) . The following are equivalent:\n\n(a) A has property \( {SC} \) .\n\n(b) \( {\left( I + \left| A\right| \right) }^{n - 1} > 0 \) .\n\n(c) \( {\left( I + M\left( A\right) \right) }^{n - 1} > 0 \) . | Proof. \( {\left( I + \left| A\right| \right) }^{n - 1} = I + \left( {n - 1}\right) \left| A\right| + \left( \begin{matrix} n - 1 \\ 2 \end{matrix}\right) {\left| A\right| }^{2} + \cdots + {\left| A\right| }^{n - 1} > 0 \) if and only if for each pair \( i, j \) of nodes with \( i \neq j \) at least one of the matrices... | Yes |
Theorem 6.3.2 (Bauer and Fike). Let \( A \in {M}_{n} \) be diagonalizable, and suppose that \( A = {S\Lambda }{S}^{-1} \), in which \( S \) is nonsingular and \( \Lambda \) is diagonal. Let \( E \in {M}_{n} \) and let \( \parallel \cdot \parallel \) be a matrix norm on \( {M}_{n} \) that is induced by an absolute norm ... | Proof. If \( \widehat{\lambda } \) is an eigenvalue of \( {S}^{-1}\left( {A + E}\right) S = \Lambda + {S}^{-1}{ES} \), then \( \widehat{\lambda }I - \Lambda - {S}^{-1}{ES} \) is singular. If \( \widehat{\lambda } \) is an eigenvalue of \( A \), the bound (6.3.3) is trivially satisfied. Suppose that \( \widehat{\lambda ... | Yes |
Theorem 6.3.5 (Hoffman and Wielandt). Let \( A, E \in {M}_{n} \), assume that \( A \) and \( A + E \) are both normal, let \( {\lambda }_{1},\ldots ,{\lambda }_{n} \) be the eigenvalues of \( A \) in some given order, and let \( {\widehat{\lambda }}_{1},\ldots ,{\widehat{\lambda }}_{n} \) be the eigenvalues of \( A + E... | Proof. Let \( \Lambda = \operatorname{diag}\left( {{\lambda }_{1},\ldots ,{\lambda }_{n}}\right) \), let \( \widehat{\Lambda } = \operatorname{diag}\left( {{\widehat{\lambda }}_{1},\ldots ,{\widehat{\lambda }}_{n}}\right) \), let \( V, W \in {M}_{n} \) be unitary matrices such that \( A = {V\Lambda }{V}^{ * } \) and \(... | Yes |
Corollary 6.3.8. Let \( A, E \in {M}_{n} \) . Assume that \( A \) is Hermitian and \( A + E \) is normal, let \( {\lambda }_{1},\ldots ,{\lambda }_{n} \) be the eigenvalues of \( A \), arranged in increasing order \( {\lambda }_{1} \leq \cdots \leq {\lambda }_{n} \), and let \( {\widehat{\lambda }}_{1},\ldots ,{\wideha... | Proof. The preceding theorem ensures that there is a permutation of the given order for the eigenvalues of \( A + E \) such that \[ \mathop{\sum }\limits_{{i = 1}}^{n}{\left| {\widehat{\lambda }}_{\sigma \left( i\right) } - {\lambda }_{i}\right| }^{2} \leq \parallel E{\parallel }_{2}^{2} \] (6.3.9) If the eigenvalues o... | Yes |
Theorem 6.3.12. Let \( A, E \in {M}_{n} \) and suppose that \( \lambda \) is a simple eigenvalue of \( A \) . Let \( x \) and \( y \) be, respectively, right and left eigenvectors of \( A \) corresponding to \( \lambda \) . Then\n\n(a) for each given \( \varepsilon > 0 \) there exists a \( \delta > 0 \) such that, for ... | Proof. Our strategy is to find a matrix that is similar to \( A + {tE} \), has \( \lambda + t{x}^{ * }{Ey}/{y}^{ * }x \) as its 1, 1 entry, and whose Geršgorin disc associated with its first row has radius at most \( \left| t\right| \varepsilon \) and is disjoint from the other \( n - 1 \) Geršgorin discs. Let \( \mu =... | Yes |
Theorem 6.3.14. Let \( A \in {M}_{n} \) be diagonalizable with \( A = {S\Lambda }{S}^{-1} \) and \( \Lambda = \) \( \operatorname{diag}\left( {{\lambda }_{1},\ldots ,{\lambda }_{n}}\right) \) . Let \( \parallel \cdot \parallel \mid \) be a matrix norm on \( {M}_{n} \) that is induced by an absolute vector norm \( \para... | Proof. If \( \widehat{\lambda } \) is an eigenvalue of \( A \), then the asserted bounds are trivially satisfied, so we suppose that it is not an eigenvalue of \( A \) . Then \( r = A\widehat{x} - \widehat{\lambda }x = S\left( {\Lambda - \widehat{\lambda }I}\right) {S}^{-1}\widehat{x} \) and \( \widehat{x} = S{\left( \... | Yes |
Theorem 6.4.7 (Brauer). Let \( A = \left\lbrack {a}_{ij}\right\rbrack \in {M}_{n} \) and assume that \( n \geq 2 \) . The eigenvalues of \( A \) are in the union of \( n\left( {n - 1}\right) /2 \) ovals of Cassini\n\n\[ \mathop{\bigcup }\limits_{{i \neq j}}\left\{ {z \in \mathbf{C} : \left| {z - {a}_{ii}}\right| \left|... | Proof. Let \( \lambda \) be an eigenvalue of \( A \), and suppose that \( {Ax} = {\lambda x} \) with \( x = \left\lbrack {x}_{i}\right\rbrack \neq 0 \) . There is an entry of \( x \) that has largest absolute value, say \( {x}_{p} \), so \( \left| {x}_{p}\right| \geq \left| {x}_{i}\right| \) for all \( i = 1,\ldots, n ... | Yes |
Lemma 6.4.17. Let \( S \) be a nonempty finite set on which a preorder \( R \) is defined. Then \( S \) contains at least one maximal element. | Proof. Arrange the elements in any order \( {s}_{1},\ldots ,{s}_{k} \) . Set \( s = {s}_{1} \) . If \( {s}_{2}{Rs} \), leave \( s \) alone; if not, then set \( s = {s}_{2} \) . If \( {s}_{3}{Rs} \), leave \( s \) alone; if not, then set \( s = {s}_{3} \) . Continue this process with \( {s}_{4},\ldots ,{s}_{k} \) . The ... | Yes |
Theorem 6.4.18 (Brualdi). Let \( A = \left\lbrack {a}_{ij}\right\rbrack \in {M}_{n} \) and suppose that \( n \geq 2 \) . If \( A \) is weakly irreducible, then every eigenvalue of \( A \) is contained in the set \[ \mathop{\bigcup }\limits_{{\gamma \in C\left( A\right) }}\left\{ {z \in \mathbf{C} : \mathop{\prod }\limi... | Proof. Weak irreducibility of \( A \) ensures that each of its deleted row sums is positive, so if \( \lambda \) is an eigenvalue of \( A \) and \( \lambda = {a}_{ii} \) for some \( i = 1,\ldots, n \), then \( \lambda \) is in the interior of the set (6.4.19). For the rest of the argument, we suppose that \( \lambda \)... | Yes |
Theorem 6.4.30 (Kolotilina). Let \( n \geq 2 \) and let \( A = \left\lbrack {a}_{ij}\right\rbrack \in {M}_{n} \) be irreducible. Every eigenvalue of \( A \) is contained in the set\n\n\[ \mathop{\bigcup }\limits_{\substack{{i \neq j} \\ {\left| {a}_{ij}\right| + \left| {a}_{ji}\right| \neq 0} }}\left\{ {z \in \mathbf{C... | (6.4.31)\n\nThe notation means that an oval corresponding to distinct rows \( i \) and \( j \) appears in the union only if either \( {a}_{ij} \) or \( {a}_{ji} \) is nonzero. | No |
Corollary 7.1.5. Let \( A \in {M}_{n} \) be positive semidefinite (respectively, positive definite). Then \( \operatorname{tr}A,\det A \), and the principal minors of \( A \) are all nonnegative (respectively, positive). Moreover, \( \operatorname{tr}A = 0 \) if and only if \( A = 0 \) . | Proof. The trace of \( A \) is the sum of its eigenvalues, which are all nonnegative (respectively, positive); if that sum is zero, then each eigenvalue is zero and hence the diagonalizable matrix \( A \) is zero; see (1.3.4). The determinant of \( A \) is the product of its eigenvalues, which are all nonnegative. Prin... | Yes |
Corollary 7.1.7. A positive semidefinite matrix is positive definite if and only if it is nonsingular. | Proof. Suppose that \( A \in {M}_{n} \) is positive semidefinite. The preceding observation ensures that the following statements are equivalent: (a) \( A \) is singular; (b) there is a nonzero vector \( x \) such that \( {Ax} = 0 \) ; (c) there is a nonzero vector \( x \) such that \( {x}^{ * }{Ax} = 0 \) ; (d) \( A \... | Yes |
Lemma 7.1.11. Suppose that \( A \in {M}_{n} \) has positive semidefinite Hermitian part \( H\left( A\right) = \frac{1}{2}\left( {A + {A}^{ * }}\right) \) . Then\n\n(a) nullspace \( A \subset \) nullspace \( H\left( A\right) \) and nullspace \( {A}^{ * } \subset \) nullspace \( H\left( A\right) \n\n(b) \( \operatorname{... | Proof. (a) Write \( A = H + {iK} \), in which \( H \) and \( K \) are Hermitian, and let \( x \in {\mathbf{C}}^{n} \) . If either \( {x}^{ * }A = 0 \) or \( {Ax} = 0 \), then \( {x}^{ * }{Ax} = 0 \), which implies that \( {x}^{ * }{Hx} = 0 \) . It follows from (7.1.6) that \( {Hx} = 0 \) .\n\n(b) Follows from (a).\n\n(... | Yes |
Corollary 7.1.13. If \( A \in {M}_{n} \) has positive definite Hermitian part, then \( A \) has the row and column inclusion properties. | Proof. If \( H\left( A\right) \) is nonsingular, then (7.1.11(b)) ensures that \( \operatorname{rank}A = n = \operatorname{rank}H\left( A\right) \) , so the assertion follows from (7.1.12). | Yes |
Corollary 7.2.2. If \( A \in {M}_{n} \) is positive semidefinite, then so is each \( {A}^{k}, k = 1,2,\ldots \) . | Proof. If the eigenvalues of \( A \) are \( {\lambda }_{1},\ldots ,{\lambda }_{n} \), then the eigenvalues of \( {A}^{k} \) are \( {\lambda }_{1}^{k},\ldots ,{\lambda }_{n}^{k} \) . The latter are nonnegative if the former are. | Yes |
Corollary 7.2.3. Suppose that \( A = \left\lbrack {a}_{ij}\right\rbrack \in {M}_{n} \) is Hermitian and strictly diagonally dominant. If \( {a}_{ii} > 0 \) for all \( i = 1,2,\ldots, n \), then \( A \) is positive definite. | Proof. This is (6.1.10(c)). The conditions imply that each Geršgorin disc for \( A \) lies in the open right half-plane. Since the eigenvalues of a Hermitian matrix are all real, the eigenvalues of \( A \) must all be positive. | Yes |
Corollary 7.2.4. Let \( A \in {M}_{n} \) be Hermitian, and let \( {p}_{A}\left( t\right) = {a}_{n}{t}^{n} + {a}_{n - 1}{t}^{n - 1} + \cdots + \) \( {a}_{n - m}{t}^{n - m} \) be its characteristic polynomial, in which \( {a}_{n} = 1,{a}_{n - m} \neq 0 \), and \( 1 \leq m \leq n \) . Then \( A \) is positive semidefinite... | Proof. The hypothesis is that the leading coefficients of \( {p}_{A}\left( t\right) \) are nonzero and alternate strictly in sign. If this condition is met, \( {p}_{A}\left( t\right) \) has no negative zeroes, so \( A \) has only nonnegative eigenvalues. Conversely, if \( A \) is positive semidefinite, denote its posit... | Yes |
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