Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Lemma 7. The map root : Unit \( {}_{n} \rightarrow {A}_{n} \) is continuous. | Proof. Let \( P \in {\mathbf{{Unit}}}_{n} \) be given. Let \( {a}_{1},\ldots ,{a}_{r} \) be the distinct roots of \( P,{m}_{j} \) their multiplicities, and \( \rho \) the minimum distance between them. We denote by \( {D}_{j} \) the open disk with center \( {a}_{j} \) and radius \( \rho /2 \), and \( {C}_{j} \) its bou... | Yes |
Theorem 5.3 Let \( {\lambda }_{0} \) be an algebraically simple eigenvalue of a matrix \( {M}_{0} \in \) \( {\mathbf{M}}_{n}\left( \mathbb{C}\right) \). Then there exists an open neighbourhood \( \mathcal{M} \) of \( {M}_{0} \) in \( {\mathbf{M}}_{n}\left( \mathbb{C}\right) \), and two analytic functions\n\n\[ \nM \map... | Proof. Let \( {X}_{0} \) be an eigenvector of \( {M}_{0} \) associated with \( {\lambda }_{0} \). We know that \( {\lambda }_{0} \) is also a simple eigenvalue of \( {M}_{0}^{T} \). Thanks to Proposition 3.15, an eigenvector \( {Y}_{0} \) of \( {M}_{0}^{T} \) (associated with \( {\lambda }_{0} \)) satisfies \( {Y}_{0}^... | Yes |
Theorem 5.4 In \( {\mathbf{M}}_{n}\left( \mathbb{C}\right) \), a matrix is normal if and only if it is unitarily diagonalizable: | Proof. A diagonal matrix is obviously normal. If \( U \) is unitary, a matrix \( M \) is normal if and only if \( {U}^{ * }{MU} \) is normal: we deduce that unitarily diagonalizable matrices are normal.\n\nWe now prove the converse. We proceed by induction on the size \( n \) of the matrix \( M \) . If \( n = 0 \), the... | Yes |
Corollary 5.2 Unitary, Hermitian, and skew-Hermitian matrices are unitarily diagonalizable. | Proof. A diagonal matrix is obviously normal. If \( U \) is unitary, a matrix \( M \) is normal if and only if \( {U}^{ * }{MU} \) is normal: we deduce that unitarily diagonalizable matrices are normal.\n\nWe now prove the converse. We proceed by induction on the size \( n \) of the matrix \( M \) . If \( n = 0 \), the... | Yes |
Theorem 5.5 Let \( M \in {\mathbf{M}}_{n}\left( \mathbb{R}\right) \) be a normal matrix. There exists an orthogonal matrix \( O \) such that \( {OM}{O}^{-1} \) is block-diagonal, the diagonal blocks being \( 1 \times 1 \) (those corresponding to the real eigenvalues of \( M \) ) or \( 2 \times 2 \), the latter being ma... | Proof. One again proceeds by induction on \( n \) . When \( n \geq 1 \), the proof is the same as in the previous section whenever \( M \) has at least one real eigenvalue.\n\nIf this is not the case, then \( n \) is even. Let us first consider the case \( n = 2 \) . Then\n\n\[ \nM = \left( \begin{array}{ll} a & b \\ c... | Yes |
Real symmetric matrices are diagonalizable over \( \mathbb{R} \), through orthogonal conjugation. In other words, given \( M \in {\operatorname{Sym}}_{n}\left( \mathbb{R}\right) \), there exists an \( O \in {\mathbf{O}}_{n}\left( \mathbb{R}\right) \) such that \( {\mathrm{{OMO}}}^{-1} \) is diagonal. | In fact, because the eigenvalues of \( M \) are real, \( {OM}{O}^{-1} \) has only \( 1 \times 1 \) blocks. We say that real symmetric matrices are orthogonally diagonalizable. | No |
Proposition 5.7 The functional calculus with holomorphic functions enjoys the following properties. Below, the domains of functions are such that the expressions make sense.\n\n- If \( f \) and \( g \) match at order \( n \) over \( \operatorname{Sp}A \), then \( f\left( A\right) = g\left( A\right) \) .\n\n- Conjugatio... | Proof. The first property follows directly from the definition, and the linearity is obvious. The conjugation formula is already true for polynomials.\n\nThe product formula is true for polynomials. Now, if \( f \) and \( g \) are interpolated by \( P \) and \( Q \), respectively, at order \( n \) at every point of \( ... | Yes |
Proposition 5.8 Let \( f \) be holomorphic over a domain \( \mathcal{U} \) containing the spectrum of \( A \in {\mathbf{M}}_{n}\left( \mathbb{C}\right) \) . Let \( \Gamma \) be a positively oriented contour around \( \mathrm{{Sp}}A \), contained in \( \mathcal{U} \) .\n\nThen we have\n\n\[ f\left( A\right) = \frac{1}{2... | Proof. Because of the conjugation property, and thanks to Proposition 3.20, it is enough to verify the formula when \( A = \lambda {I}_{n} + N \) where \( N \) is nilpotent. By translation, it suffices to treat the nilpotent case.\n\nSo we assume that \( A \) is nilpotent. Therefore \( \Gamma \) is a disjoint union of ... | Yes |
Proposition 5.11 For every \( A \in {\mathbf{M}}_{n}\left( \mathbb{C}\right) \), we have\n\n\[ w\left( A\right) \leq \parallel A{\parallel }_{2} \leq {2w}\left( A\right) \]\n | Proof. Cauchy-Schwarz gives\n\n\[ \left| {{x}^{ * }{Ax}}\right| \leq \parallel x{\parallel }_{2}\parallel {Ax}{\parallel }_{2} \leq \parallel A{\parallel }_{2}\parallel x{\parallel }_{2}^{2} \]\n\nwhich yields \( w\left( A\right) \leq \parallel A{\parallel }_{2} \) .\n\nOn the other hand, let us majorize \( \left| {{y}... | Yes |
Proposition 5.12 (Gershgorin) The spectrum of \( A \) is included in the Gershgorin domain \( \mathcal{G}\left( A\right) \), defined as the union of the Gershgorin disks\n\n\[ \n{D}_{i}\left( A\right) \mathrel{\text{:=}} D\left( {{a}_{ii};{r}_{i}}\right) ,\;{r}_{i} \mathrel{\text{:=}} \mathop{\sum }\limits_{{j \neq i}}... | Proof. For \( r \in \left\lbrack {0,1}\right\rbrack \), we define a matrix \( A\left( r\right) \) by the formula\n\n\[ \n{a}_{ij}\left( r\right) \mathrel{\text{:=}} \left\{ \begin{array}{ll} {a}_{ii}, & j = i \\ r{a}_{ij}, & j \neq i \end{array}\right. \n\]\n\nIt is clear that the Gershgorin domain \( {\mathcal{G}}_{r}... | Yes |
Theorem 5.7 There are exactly p eigenvalues of \( A \) in \( G \), counted with their multiplicities. | Proof. For \( r \in \left\lbrack {0,1}\right\rbrack \), we define a matrix \( A\left( r\right) \) by the formula\n\n\[ \n{a}_{ij}\left( r\right) \mathrel{\text{:=}} \left\{ \begin{array}{ll} {a}_{ii}, & j = i \\ r{a}_{ij}, & j \neq i \end{array}\right. \n\]\n\nIt is clear that the Gershgorin domain \( {\mathcal{G}}_{r}... | Yes |
Proposition 5.13 Let \( A \) be an irreducible matrix. If an eigenvalue of \( A \) does not belong to the interior of any Gershgorin disk, then it belongs to every circle \( S\left( {{a}_{ii};{r}_{i}}\right) \) . | Proof. Let \( \lambda \) be such an eigenvalue and \( x \) an associated eigenvector. By assumption, one has \( \left| {\lambda - {a}_{ii}}\right| \geq \mathop{\sum }\limits_{{j \neq i}}\left| {a}_{ij}\right| \) for every \( i \) . Let \( I \) be the set of indices for which \( \left| {x}_{i}\right| = \) \( \parallel x... | Yes |
Corollary 5.4 Let \( A \) be a square matrix. If \( A \) is strictly diagonally dominant, or if \( A \) is irreducible and strongly diagonally dominant, then \( A \) is invertible. | In fact, either zero does not belong to the Gershgorin domain, or it is not interior to the disks. In the latter case, \( A \) is assumed to be irreducible, and there exists a disk \( {D}_{j} \) that does not contain zero. | No |
Proposition 6.1 Let \( H \in {\mathbf{{HPD}}}_{n} \) and \( K \in {\mathbf{H}}_{n} \) be given. Then the product \( {HK} \) (or KH as well) is diagonalizable with real eigenvalues. The number of positive (respectively, negative) eigenvalues of \( {HK} \) equals that for \( K \) . | Proof. Recall that \( H \) is unitary diagonalizable with real eigenvalues: there is a \( U \in \) \( {\mathbf{U}}_{n} \) such that \( H = {U}^{ * }\operatorname{diag}\left( {{\mu }_{1},\ldots ,{\mu }_{n}}\right) U \) . Because \( H \) is positive-definite, we have \( {\mu }_{j} > 0 \) . Setting \( h = {U}^{ * }\operat... | Yes |
Theorem 6.1 Given \( H \in {\mathbf{{HPD}}}_{n} \), there exists one and only one \( h \in {\mathbf{{HPD}}}_{n} \) such that \( {h}^{2} = H \) . The matrix \( h \) is denoted \( \sqrt{H} \), and called the square root of \( H \) . | Proof. Such a square root was constructed in the proof of Proposition 6.1. There remains to prove uniqueness. So let \( h,{h}^{\prime } \in {HP}{D}_{n} \) be such that \( {h}^{\prime 2} = {h}^{2} \) . Set \( U \mathrel{\text{:=}} \) \( {h}^{\prime }{h}^{-1} \) . We have \( {U}^{ * }U = {h}^{-1}{h}^{\prime 2}{h}^{-1} = ... | Yes |
Theorem 6.2 Let \( M \) be an \( n \times n \) Hermitian matrix and \( {\lambda }_{1},\ldots ,{\lambda }_{n} \) its eigenvalues arranged in increasing order, counted with multiplicity. Then\n\n\[{\lambda }_{k} = \mathop{\min }\limits_{{\dim F = {kx}}}\mathop{\max }\limits_{{x \in F\smallsetminus \{ 0\} }}\frac{{x}^{ * ... | Proof. There remains to prove (6.4). There are two possible strategies. Either we start from nothing and argue again about the intersection of subspaces whose dimensions sum up to \( n + 1 \), or we remark that \( {\lambda }_{k}\left( M\right) = - {\lambda }_{n - k + 1}\left( {-M}\right) \), because \( M \) and \( - M ... | Yes |
Theorem 6.3 Let \( A \) and \( B \) be \( n \times n \) Hermitian matrices. Let \( 1 \leq i, j, k \leq n \) be indices.\n\n- If \( i + j = k + 1 \), we have\n\n\[ \n{\lambda }_{k}\left( {A + B}\right) \geq {\lambda }_{i}\left( A\right) + {\lambda }_{j}\left( B\right) \n\]\n\n- If \( i + j = k + n \), we have\n\n\[ \n{\... | Proof. Once again, any inequality can be deduced from the other ones by means of \( \left( {A, B}\right) \leftrightarrow \left( {-A, - B}\right) \) . Thus it is sufficient to treat the case where \( i + j = k + n \) .\n\nFrom (6.4), we know that there exists an \( \left( {n - k + 1}\right) \) -dimensional subspace \( H... | Yes |
Proposition 6.2 The eigenvalues \( {\lambda }_{k} \) over \( {\mathbf{H}}_{n} \) are Lipschitz functions, with Lipschitz ratio equal to one:\n\n\[ \left| {{\lambda }_{k}\left( M\right) - {\lambda }_{k}\left( N\right) }\right| \leq \left\lbrack {M - N}\right\rbrack \] | Proof. Taking \( i = k \) and either \( j = 1 \) or \( j = n \), then \( A = N \) and \( B = M - N \), we obtain\n\n\[ {\lambda }_{1}\left( {M - N}\right) \leq {\lambda }_{k}\left( M\right) - {\lambda }_{k}\left( N\right) \leq {\lambda }_{n}\left( {M - N}\right) \] | Yes |
Proposition 6.3 The square root is Hölderian with exponent \( \frac{1}{2} \) over \( {\mathbf{{HPD}}}_{n} \) :\n\n\[ \parallel \sqrt{H} - \sqrt{K}{\parallel }_{2} \leq \sqrt{\parallel H - K{\parallel }_{2}},\;\forall H, K \in {\mathbf{{HPD}}}_{n}. \] | Proof. Let \( A, B \in {\mathbf{{HPD}}}_{n} \) be given. Let us develop\n\n\[ {B}^{2} - {A}^{2} = {\left( B - A\right) }^{2} + \left( {B - A}\right) A + A\left( {B - A}\right) . \]\n\n(6.8)\n\nUp to exchanging the roles of \( A \) and \( B \), we may assume that \( {\lambda }_{n}\left( {B - A}\right) \geq {\lambda }_{n... | Yes |
Theorem 6.4 The square root is operator monotone over the positive-semidefinite Hermitian matrices: if \( {0}_{n} \leq H \leq K \), then \( \sqrt{H} \leq \sqrt{K} \) . | Proof. Let \( B \) and \( A \) be Hermitian positive-semidefinite. If \( {B}^{2} \leq {A}^{2} \), then (6.8) yields\n\n\[ \n{\left( B - A\right) }^{2} + \left( {B - A}\right) A + A\left( {B - A}\right) \leq {0}_{n}.\n\]\n\nLet \( {\lambda }_{1} \) be the smallest eigenvalue of \( A - B \), and \( x \) an associate eige... | Yes |
Theorem 6.5 Let \( H \in {\mathbf{H}}_{n - 1}, x \in {\mathbb{C}}^{n - 1} \), and \( a \in \mathbb{R} \) be given. Let \( {\lambda }_{1} \leq \cdots \leq {\lambda }_{n - 1} \) be the eigenvalues of \( H \) and \( {\mu }_{1} \leq \cdots \leq {\mu }_{n} \) those of the Hermitian matrix \[ {H}^{\prime } = \left( \begin{ar... | Proof. The inequality \( {\mu }_{j} \leq {\lambda }_{j} \) follows from (6.3), because the infimum concerns the same quantity \[ \mathop{\max }\limits_{{x \in F, x \neq 0}}\frac{{x}^{ * }{H}^{\prime }x}{\parallel x{\parallel }_{2}^{2}} \] but is taken over a smaller set in the case of \( {\lambda }_{j} \) : that of sub... | Yes |
Theorem 6.6 Let \( {\lambda }_{1} \leq \cdots \leq {\lambda }_{n - 1} \) and \( {\mu }_{1} \leq \cdots \leq {\mu }_{n} \) be real numbers satisfying \( {\mu }_{1} \leq {\lambda }_{1} \leq \cdots \leq {\mu }_{j} \leq {\lambda }_{j} \leq {\mu }_{j + 1} \leq \cdots \) . Then there exist a vector \( x \in {\mathbb{R}}^{n} ... | Proof. Let us compute the characteristic polynomial of \( H \) from Schur’s complement formula \( {}^{1} \) (see Proposition 3.9):\n\n\[ \n{p}_{n}\left( X\right) = \left( {X - a - {x}^{T}{\left( X{I}_{n - 1} - \Lambda \right) }^{-1}x}\right) \det \left( {X{I}_{n - 1} - \Lambda }\right)\n\]\n\n\[ \n= \left( {X - a - \ma... | No |
Corollary 6.1 Let \( H \in {\mathbf{H}}_{n - 1}\left( \mathbb{R}\right) \) be given, with eigenvalues \( {\lambda }_{1} \leq \cdots \leq {\lambda }_{n - 1} \) . Let \( {\mu }_{1},\ldots ,{\mu }_{n} \) be real numbers satisfying \( {\mu }_{1} \leq {\lambda }_{1} \leq \cdots \leq {\mu }_{j} \leq {\lambda }_{j} \leq {\mu ... | The proof consists in diagonalizing \( H \) through a unitary matrix \( U \in {\mathbf{U}}_{n - 1} \), then applying Theorem 6.6, and conjugating the resulting matrix by \( \operatorname{diag}\left( {{U}^{ * },1}\right) \) . | No |
Proposition 6.4 Let \( x, y \in {\mathbb{R}}^{n} \) be given. Then \( x \prec y \) if and only if for every real number \( t \) , | Proof. We may assume that \( x \) and \( y \) are nondecreasing. If the inequality (6.9) holds, we write it first for \( t \) outside the interval \( I \) containing the \( {x}_{j}\mathrm{\;s} \) and the \( {y}_{j}\mathrm{\;s} \) . This gives \( {s}_{n}\left( x\right) = {s}_{n}\left( y\right) \) . Then we write it for ... | Yes |
Theorem 6.7 (Schur) Let \( H \) be an Hermitian matrix with diagonal a and spectrum \( \lambda \) . Then \( a \succ \lambda \) . | Proof. Let \( n \) be the size of \( H \) . We argue by induction on \( n \) . We may assume that \( {a}_{n} \) is the largest component of \( a \) . Because \( {s}_{n}\left( \lambda \right) = \operatorname{Tr}A \), one has \( {s}_{n}\left( \lambda \right) = {s}_{n}\left( a\right) \) . In particular, the theorem holds ... | Yes |
Theorem 6.8 Let \( a \) and \( \lambda \) be two sequences of \( n \) real numbers such that \( a \succ \lambda \) . Then there exists a real symmetric matrix of size \( n \times n \) whose diagonal is a and spectrum is \( \lambda \) . | Proof. We proceed by induction on \( n \) . The statement is trivial if \( n = 1 \) . If \( n \geq 2 \), we use the following lemma, which is proved afterwards.\n\nLemma 9. Let \( n \geq 2 \) and \( \alpha ,\beta \) be two nondecreasing sequences of \( n \) real numbers, satisfying \( \alpha \prec \beta \) . Then there... | Yes |
Lemma 9. Let \( n \geq 2 \) and \( \alpha ,\beta \) be two nondecreasing sequences of \( n \) real numbers, satisfying \( \alpha \prec \beta \) . Then there exists a sequence \( \gamma \) of \( n - 1 \) real numbers such that\n\n\[{\alpha }_{1} \leq {\gamma }_{1} \leq {\alpha }_{2} \leq \cdots \leq {\gamma }_{n - 1} \l... | We now prove Lemma 9. Let \( \Delta \) be the set of sequences \( \delta \) of \( n - 1 \) real numbers satisfying\n\n\[{\alpha }_{1} \leq {\delta }_{1} \leq {\alpha }_{2} \leq \cdots \leq {\delta }_{n - 1} \leq {\alpha }_{n}\]\n\n(6.10)\n\ntogether with\n\n\[\mathop{\sum }\limits_{{j = 1}}^{k}{\delta }_{j} \leq \matho... | Yes |
Proposition 6.5 Let \( H \in {\mathbf{H}}_{n} \) be a positive-semidefinite Hermitian matrix. Then\n\n\[ \det H \leq \mathop{\prod }\limits_{{j = 1}}^{n}{h}_{jj} \]\n\nIf \( H \in {\mathbf{{HPD}}}_{n} \), the equality holds only if \( H \) is diagonal. | Proof. If \( \det H = 0 \), there is nothing to prove, because the \( {h}_{jj} \) are nonnegative (these are numbers \( {\left( {\mathbf{e}}^{j}\right) }^{ * }H{\mathbf{e}}^{j} \) ). Otherwise, \( H \) is positive-definite and one has \( {h}_{jj} > 0 \) . We restrict our attention to the case with a constant diagonal b... | Yes |
Theorem 6.10 The map\n\n\\[ \nH \mapsto {\\left( \\det H\\right) }^{1/n}\n\\]\n\nis concave over the cone of positive semidefinite \\( n \\times n \\) Hermitian matrices. | Proof. Let \\( H, K \\in {\\mathbf{{HPD}}}_{n} \\) be given. From Proposition 6.1, the eigenvalues \\( {\\mu }_{1},\\ldots ,{\\mu }_{n} \\) of \\( {HK} \\) are real and positive, even though \\( {HK} \\) is not Hermitian. We thus have\n\n\\[ \n{\\left( \\det H\\right) }^{1/n}{\\left( \\det K\\right) }^{1/n} = {\\left( ... | Yes |
Proposition 7.2 For conjugate exponents \( p,{p}^{\prime } \), one has\n\n\[ \parallel x{\parallel }_{p} = \mathop{\sup }\limits_{{y \neq 0}}\frac{\Re \langle x, y\rangle }{\parallel y{\parallel }_{{p}^{\prime }}} = \mathop{\sup }\limits_{{y \neq 0}}\frac{\left| \langle x, y\rangle \right| }{\parallel y{\parallel }_{{p... | Proof. The inequality \( \geq \) is a consequence of Hölder’s. The reverse inequality is obtained by taking \( {y}_{j} = {\bar{x}}_{j}{\left| {x}_{j}\right| }^{p - 2} \) if \( p < \infty \) . If \( p = \infty \), choose \( {y}_{j} = {\bar{x}}_{j} \) for an index \( j \) such that \( \left| {x}_{j}\right| = \parallel x{... | Yes |
Proposition 7.3 All norms on \( E = {K}^{n} \) are equivalent. For example,\n\n\[ \parallel x{\parallel }_{\infty } \leq \parallel x{\parallel }_{p} \leq {n}^{1/p}\parallel x{\parallel }_{\infty } \] | Proof. It is sufficient to show that every norm is equivalent to \( \parallel \cdot {\parallel }_{1} \) .\n\nLet \( N \) be a norm on \( E \) . If \( x \in E \), the triangle inequality gives\n\n\[ N\left( x\right) \leq \mathop{\sum }\limits_{i}\left| {x}_{i}\right| N\left( {\mathbf{e}}^{i}\right) \]\n\nwhere \( \left(... | Yes |
Proposition 7.4 The bi-dual (dual of the dual norm) of a norm is this norm itself: | Proof. From (7.3), one has \( {\left( \parallel \cdot {\parallel }^{\prime }\right) }^{\prime } \leq \parallel \cdot \parallel \) . The converse is a consequence of the Hahn-Banach theorem: the unit ball \( B \) of \( \parallel \cdot \parallel \) is convex and compact. If \( x \) is a point of its boundary (i.e., \( \p... | Yes |
Proposition 7.5 For an induced norm, the condition \( \parallel B\parallel < 1 \) implies that \( {I}_{n} - B \) is invertible, with the inverse given by the sum of the series\n\n\[ \mathop{\sum }\limits_{{k = 0}}^{\infty }{B}^{k} \] | Proof. The series \( \mathop{\sum }\limits_{k}{B}^{k} \) is normally convergent, because \( \mathop{\sum }\limits_{k}\begin{Vmatrix}{B}^{k}\end{Vmatrix} \leq \mathop{\sum }\limits_{k}\parallel B{\parallel }^{k} \), where the latter series converges because \( \parallel B\parallel < 1 \) . Because \( {\mathbf{M}}_{n}\le... | Yes |
Proposition 7.6 For every induced norm, one has\n\n\[ \rho \left( A\right) \leq \parallel A\parallel \] | Proof. The case \( K = \mathbb{C} \) is easy, because there exists an eigenvector \( X \in E \) associated with an eigenvalue of modulus \( \rho \left( A\right) \) :\n\n\[ \rho \left( A\right) \parallel X\parallel = \parallel {\lambda X}\parallel = \parallel {AX}\parallel \leq \parallel A\parallel \parallel X\parallel ... | Yes |
Proposition 7.7 Let \( \parallel \cdot \parallel \) be a norm on \( {K}^{n} \) and \( P \in {\mathbf{{GL}}}_{n}\left( K\right) \) . Hence, \( N\left( x\right) \mathrel{\text{:=}} \parallel {Px}\parallel \) defines a norm on \( {K}^{n} \) . Denoting still by \( \parallel \cdot \parallel \) and \( N \) the induced norms ... | Proof. Using the change of dummy variable \( y = {Px} \), we have\n\n\[ N\left( A\right) = \mathop{\sup }\limits_{{x \neq 0}}\frac{\parallel {PAx}\parallel }{\parallel {Px}\parallel } = \mathop{\sup }\limits_{{y \neq 0}}\frac{\begin{Vmatrix}PA{P}^{-1}y\end{Vmatrix}}{\parallel y\parallel } = \begin{Vmatrix}{{PA}{P}^{-1}... | Yes |
Theorem 7.1 For every \( B \in {\mathbf{M}}_{n}\left( \mathbb{C}\right) \) and all \( \varepsilon > 0 \), there exists a norm on \( {\mathbb{C}}^{n} \) such that for the induced norm,\n\n\[ \parallel B\parallel \leq \rho \left( B\right) + \varepsilon \]\n\nIn other words, \( \rho \left( B\right) \) is the infimum of \(... | Proof. From Theorem 3.5 there exists \( P \in {\mathbf{{GL}}}_{n}\left( \mathbb{C}\right) \) such that \( T \mathrel{\text{:=}} {PB}{P}^{-1} \) is upper-triangular. From Proposition 7.7, one has\n\n\[ \inf \parallel B\parallel = \inf \begin{Vmatrix}{{PB}{P}^{-1}}\end{Vmatrix} = \inf \parallel T\parallel \]\n\nwhere the... | Yes |
Proposition 7.8 If \( A \in {\mathbf{M}}_{n}\left( k\right) \) (with \( k = \mathbb{R} \) or \( \mathbb{C} \) ), then\n\n\[ \rho \left( A\right) = \mathop{\lim }\limits_{{m \rightarrow \infty }}{\begin{Vmatrix}{A}^{m}\end{Vmatrix}}^{1/m} \]\n\nfor every matrix norm. | Proof. Let \( \parallel \cdot \parallel \) be a matrix norm over \( {\mathbf{M}}_{n}\left( k\right) \) . From Proposition 7.6 and the fact that\n\n\[ \operatorname{Sp}\left( {A}^{m}\right) = \left\{ {{\lambda }^{m} \mid \lambda \in \operatorname{Sp}A}\right\} \]\n\nwe have\n\n\[ \rho \left( A\right) = \rho {\left( {A}^... | Yes |
Theorem 7.3 (Riesz-Thorin) Let \( \Omega \) be an open set in \( {\mathbb{R}}^{D} \) and \( \omega \) an open set in \( {\mathbb{R}}^{d} \) . Let \( {p}_{0},{p}_{1},{q}_{0},{q}_{1} \) be four numbers in \( \left\lbrack {1, + \infty }\right\rbrack \) . Let \( \theta \in \left\lbrack {0,1}\right\rbrack \) and \( p, q \) ... | Proof. (Due to Riesz) Let us fix \( x \) and \( y \) in \( {K}^{n} \) . We have to bound \[ \left| {\langle y,{Ax}\rangle }\right| = \left| {\mathop{\sum }\limits_{{j, k}}{a}_{jk}{x}_{j}{\bar{y}}_{k}}\right| . \] Let \( B \) be the strip in the complex plane defined by \( \Re z \in \left\lbrack {0,1}\right\rbrack \) . ... | Yes |
Proposition 8.1 A matrix is nonnegative if and only if \( x \geq 0 \) implies \( {Ax} \geq 0 \) . It is positive if and only if \( x \geq 0 \) and \( x \neq 0 \) imply \( {Ax} > 0 \) . | Proof. Let us assume that \( {Ax} \geq 0 \) (respectively, \( > 0 \) ) for every \( x \geq 0 \) (respectively, \( \geq 0 \) and \( \neq 0 \) ). Then the \( i \) th column \( {A}^{\left( i\right) } \) is nonnegative (respectively, positive), since it is the image of the \( i \) th vector of the canonical basis. Hence \(... | Yes |
Proposition 8.2 If \( A \in {\mathbf{M}}_{n}\left( \mathbb{R}\right) \) is nonnegative and irreducible, then \( {\left( I + A\right) }^{n - 1} > 0 \) . | Proof. Let \( x \neq 0 \) be nonnegative, and define \( {x}^{m} = {\left( I + A\right) }^{m}x \), which is nonnegative too. Let us denote by \( {P}_{m} \) the set of indices of the nonzero components of \( {x}^{m} \) : \( {P}_{0} \) is nonempty. Because \( {x}_{i}^{m + 1} \geq {x}_{i}^{m} \), one has \( {P}_{m} \subset... | Yes |
Theorem 8.1 Let \( A \in {\mathbf{M}}_{n}\left( \mathbb{R}\right) \) be a nonnegative matrix. Then \( \rho \left( A\right) \) is an eigenvalue of \( A \) associated with a nonnegative eigenvector. | Proof. Let \( \lambda \) be an eigenvalue of maximal modulus and \( v \) an eigenvector, normalized by \( \parallel v{\parallel }_{1} = 1 \) . Then\n\n\[ \rho \left( A\right) \left| v\right| = \left| {\lambda v}\right| = \left| {Av}\right| \leq A\left| v\right| \]\n\nLet us denote by \( C \) the subset of \( {\mathbb{R... | Yes |
Theorem 8.2 Let \( A \in {\mathbf{M}}_{n}\left( \mathbb{R}\right) \) be a nonnegative irreducible matrix. Then \( \rho \left( A\right) \) is a simple eigenvalue of \( A \), associated with a positive eigenvector: Moreover, \( \rho \left( A\right) > 0 \) . | Proof. For \( r \geq 0 \), we denote by \( {C}_{r} \) the set of vectors of \( {\mathbb{R}}^{n} \) defined by the conditions\n\n\[ x \geq 0,\;\parallel x{\parallel }_{1} = 1,\;{Ax} \geq {rx}. \]\n\nEach \( {C}_{r} \) is a convex compact set. We saw in the previous section that if \( \lambda \) is an eigenvalue associat... | Yes |
Lemma 10. Let \( r \geq 0 \) and \( x \geq 0 \) such that \( {Ax} \geq {rx} \) and \( {Ax} \neq {rx} \) . Then there exists \( {r}^{\prime } > r \) such that \( {C}_{{r}^{\prime }} \) is nonempty. | Proof. Set \( y \mathrel{\text{:=}} {\left( {I}_{n} + A\right) }^{n - 1}x \) . Because \( A \) is irreducible and \( x \geq 0 \) is nonzero, one has \( y > 0 \) . Likewise, \( {Ay} - {ry} = {\left( {I}_{n} + A\right) }^{n - 1}\left( {{Ax} - {rx}}\right) > 0 \) . Let us define \( {r}^{\prime } \mathrel{\text{:=}} \) \( ... | Yes |
Lemma 11. The nonnegative eigenvectors of \( A \) are positive. The corresponding eigenvalue is positive too. | Proof. Given such a vector \( x \) with \( {Ax} = {\lambda x} \), we observe that \( \lambda \in {\mathbb{R}}^{ + } \) . Then\n\n\[ x = \frac{1}{{\left( 1 + \lambda \right) }^{n - 1}}{\left( {I}_{n} + A\right) }^{n - 1}x \]\n\nand the right-hand side is strictly positive, from Proposition 8.2.\n\nInasmuch as \( A \) is... | Yes |
Lemma 12. Let \( M, B \in {\mathbf{M}}_{n}\left( \mathbb{C}\right) \) be matrices, with \( M \) irreducible and \( \left| B\right| \leq M \) . Then \( \rho \left( B\right) \leq \rho \left( M\right) \) . | In order to establish the inequality, we proceed as above. If \( \lambda \) is an eigenvalue of \( B \), of modulus \( \rho \left( B\right) \), and if \( x \) is a normalized eigenvector, then \( \rho \left( B\right) \left| x\right| \leq \left| B\right| \cdot \left| x\right| \leq \) \( M\left| x\right| \), so that \( {... | Yes |
Theorem 8.3 Under the assumptions of Theorem 8.2, let \( p \) be the cardinality of the set \( {\operatorname{Sp}}_{\max }\left( A\right) \) of eigenvalues of \( A \) of maximal modulus \( \rho \left( A\right) \) . Then we have \( {\operatorname{Sp}}_{\max }\left( A\right) = \rho \left( A\right) {\mathcal{U}}_{p} \), w... | Proof. Let us denote by \( X \) the unique nonnegative eigenvector of \( A \) normalized by \( \parallel X{\parallel }_{1} = 1 \) . If \( Y \) is a unitary eigenvector, associated with an eigenvalue \( \mu \) of maximal modulus \( \rho \left( A\right) \), the inequality \( \rho \left( A\right) \left| Y\right| = \left| ... | Yes |
Theorem 8.4 (Birkhoff) A matrix \( M \in {\mathbf{M}}_{n}\left( \mathbb{R}\right) \) is bistochastic if and only if it is a center of mass (i.e., a barycenter with nonnegative weights) of permutation matrices. | Proof. Let \( M \in {\mathbf{{DS}}}_{n} \) be given. If \( M \) is not a permutation matrix, there exists an entry \( {m}_{{i}_{1}{j}_{1}} \in \left( {0,1}\right) \) . Inasmuch as \( M \) is stochastic, there also exists \( {j}_{2} \neq {j}_{1} \) such that \( {m}_{{i}_{1}{j}_{2}} \in \left( {0,1}\right) \) . Because \... | Yes |
Corollary 8.1 Let \( \parallel \cdot \parallel \) be a norm on \( {\mathbb{R}}^{n} \), invariant under permutation of the coordinates. Then \( \parallel M\parallel = 1 \) for every bistochastic matrix (where as usual we have denoted \( \parallel \cdot \parallel \) the induced norm on \( {\mathbf{M}}_{n}\left( \mathbb{R... | Proof. To begin with, \( \parallel P\parallel = 1 \) for every permutation matrix, by assumption. Because the induced norm is convex (true for every norm), one deduces from Birkhoff's theorem that \( \parallel M\parallel \leq 1 \) for every bistochastic matrix. Furthermore, \( M\mathbf{e} = \mathbf{e} \) implies \( \pa... | Yes |
Theorem 8.5 A matrix \( A \) is bistochastic if and only if \( {Ax} \succ x \) for every \( x \in {\mathbb{R}}^{n} \) . | Proof. If \( A \) is bistochastic, then \( \parallel {Ax}{\parallel }_{1} \leq \parallel A{\parallel }_{1}\parallel x{\parallel }_{1} = \parallel x{\parallel }_{1} \), because \( {A}^{T} \) is stochastic. Because \( A \) is stochastic, \( A\mathbf{e} = \mathbf{e} \). Applying the inequality to \( x - t\mathbf{e} \), on... | Yes |
Theorem 8.6 Let \( x, y \in {\mathbb{R}}^{n} \) . Then \( x \prec y \) if and only if there exists a bistochastic matrix \( A \) such that \( y = {Ax} \) . | Proof. From the previous theorem, it is enough to show that if \( x \prec y \), there exists \( A \), a bistochastic matrix, such that \( y = {Ax} \) . To do so, one applies Theorem 6.8: there exists an Hermitian matrix \( H \) whose diagonal and spectrum are \( y \) and \( x \), respectively. Let us diagonalize \( H \... | Yes |
Proposition 9.1 In a principal ideal domain, every pair of elements has a greatest common divisor. The gcd satisfies the Bézout identity: for every \( a, b \in A \), there exist \( u, v \in A \) such that\n\n\[ \gcd \left( {a, b}\right) = {ua} + {vb}. \]\n\nSuch \( u \) and \( v \) are coprime. | Proof. Let \( A \) be a principal ideal domain. If \( a, b \in A \), the ideal \( \mathcal{I} = : \left( {a, b}\right) \) is principal: \( \mathcal{I} = \left( d\right) \). Because \( a, b \in \mathcal{I}, d \) divides \( a \) and \( b \). Furthermore, \( d = {ua} + {vb} \) because \( d \in \mathcal{I} \). If \( c \) d... | Yes |
Proposition 9.2 The principal ideal domains are Noetherian. | Proof. Let \( A \) be a principal ideal domain and let \( {\left( {I}_{j}\right) }_{j \geq 0} \) be a nondecreasing sequence of ideals in \( A \) . Let \( \mathcal{I} \) be their union, which happens to be an ideal. Let \( a \) be a generator: \( \mathcal{I} = \left( a\right) \) . Then \( a \) belongs to one of the ide... | Yes |
Proposition 9.3 Euclidean domains are principal ideal domains. | Proof. Let \( \mathcal{I} \) be an ideal of a Euclidean domain \( A \) . If \( \mathcal{I} = \left( 0\right) \), there is nothing to show. Otherwise, let us select an element \( a \) for which \( N\left( a\right) \) is minimal in \( \mathcal{I} \smallsetminus \{ 0\} \) . If \( b \in \mathcal{I} \), the remainder \( r \... | Yes |
Theorem 9.1 A square invertible matrix of size \( n \) with entries in a Euclidean domain \( A \) is a product of elementary matrices with entries in \( A \) . | Proof. We prove the theorem for \( n = 2 \) . The general case is deduced from that particular one and from the proof of Theorem 9.2 below, which uses multiplications by block-diagonal matrices with \( 1 \times 1 \) and \( 2 \times 2 \) diagonal blocks.\n\nLet\n\n\[ M = \left( \begin{matrix} a & {a}_{1} \\ c & d \end{m... | Yes |
Lemma 13. There exists a map \( T : {\mathbf{M}}_{n \times m}\left( A\right) \rightarrow {\mathbf{M}}_{n \times m}\left( A\right) \) with the following properties.\n\n- \( {N}^{\prime } \mathrel{\text{:=}} T\left( N\right) \) is equivalent to \( N \) .\n\n- \( {n}_{11}^{\prime } \mid {n}_{11} \),\n\n- This divisibility... | Proof. (of the lemma).\n\nWe intentionally write this proof in the algorithmic style. Given the matrix \( N \in \) \( {\mathbf{M}}_{n \times m}\left( A\right) \), we distinguish four cases.\n\n1. IF \( {n}_{11} \) does not divide some \( {n}_{1j} \), THEN take the smallest such index \( j | No |
Theorem 9.5 Two matrices in \( {\mathbf{M}}_{n}\left( k\right) \) are similar if and only if they have the same list of invariant polynomials (counted with their multiplicities). | Proof. We prove Theorem 9.6. The condition is clearly sufficient.\n\nConversely, if \( X{A}_{0} + {B}_{0} \) and \( X{A}_{1} + {B}_{1} \) are equivalent, there exist matrices \( P, Q \in \) \( {\mathbf{{GL}}}_{n}\left( A\right) \), such that \( P\left( {X{A}_{0} + {B}_{0}}\right) = \left( {X{A}_{1} + {B}_{1}}\right) Q ... | Yes |
Theorem 9.6 Let \( {A}_{0},{A}_{1},{B}_{0},{B}_{1} \) be matrices in \( {\mathbf{M}}_{n}\left( k\right) \), with \( {A}_{0},{A}_{1} \in {\mathbf{{GL}}}_{n}\left( k\right) \). Then the matrices \( X{A}_{0} + {B}_{0} \) and \( X{A}_{1} + {B}_{1} \) are equivalent (in \( {\mathbf{M}}_{n}\left( A\right) \)) if and only if ... | Proof. We prove Theorem 9.6. The condition is clearly sufficient.\n\nConversely, if \( X{A}_{0} + {B}_{0} \) and \( X{A}_{1} + {B}_{1} \) are equivalent, there exist matrices \( P, Q \in \) \( {\mathbf{{GL}}}_{n}\left( A\right) \), such that \( P\left( {X{A}_{0} + {B}_{0}}\right) = \left( {X{A}_{1} + {B}_{1}}\right) Q ... | Yes |
Theorem 9.7 If \( B \in {\mathbf{M}}_{n}\left( k\right) \), then \( B \) and \( {B}^{T} \) are similar. | Indeed, \( X{I}_{n} - B \) and \( X{I}_{n} - {B}^{T} \) are transposes of each other, and hence have the same list of minors, and the same invariant factors. | Yes |
Theorem 9.9 Let \( k \) be a field, \( M \in {\mathbf{M}}_{n}\left( k\right) \) a square matrix, and \( {p}_{1},\ldots ,{p}_{n} \) its similarity invariants. Then \( {p}_{n} \) is the minimal polynomial of \( M \) . In particular, the minimal polynomial does not depend on the field under consideration, as long as it co... | Proof. We use the first canonical form \( {M}^{\prime } \) of \( M \) . Because \( {M}^{\prime } \) and \( M \) are similar, they have the same minimal polynomial. One thus can assume that \( M \) is in the canonical form \( M = \operatorname{diag}\left( {{M}_{1},\ldots ,{M}_{n}}\right) \), where \( {M}_{j} \) is the c... | Yes |
Theorem 10.1 For every \( M \in {\mathbf{{GL}}}_{n}\left( \mathbb{C}\right) \), there exists a unique pair\n\n\[ \n\left( {H, Q}\right) \in {\mathbf{{HPD}}}_{n} \times {\mathbf{U}}_{n}\n\]\n\nsuch that \( M = {HQ} \) . If \( M \in {\mathbf{{GL}}}_{n}\left( \mathbb{R}\right) \), then \( \left( {H, Q}\right) \in {\mathbf... | Proof. Existence. Because \( M{M}^{ * } \in {\mathbf{{HPD}}}_{n} \), we can set \( H \mathrel{\text{:=}} \sqrt{M{M}^{ * }} \) (the square root was defined in Section 6.1). Then \( Q \mathrel{\text{:=}} {H}^{-1}M \) satisfies \( {Q}^{ * }Q = {M}^{ * }{H}^{-2}M = \) \( {M}^{ * }{\left( M{M}^{ * }\right) }^{-1}M = {I}_{n}... | Yes |
Proposition 10.1 Let \( A, B \in {\mathbf{M}}_{n}\left( \mathbb{C}\right) \) be commuting matrices; that is, \( {AB} = {BA} \) . Then \( \exp \left( {A + B}\right) = \left( {\exp A}\right) \left( {\exp B}\right) . | Proof. The proof is exactly the same as for the exponential of complex numbers. We observe that because the series defining the exponential of a matrix is normally convergent, we may compute the product \( \left( {\exp A}\right) \left( {\exp B}\right) \) by multiplying term by term the series\n\n\[ \left( {\exp A}\righ... | Yes |
For every \( A \in {\mathbf{M}}_{n}\left( \mathbb{C}\right) \), exp \( A \) is invertible, and its inverse is \( \exp \left( {-A}\right) \) . | Given two conjugate matrices \( B = {P}^{-1}{AP} \), we have \( {B}^{k} = {P}^{-1}{A}^{k}P \) for each integer \( k \) and thus\n\n\[ \exp \left( {{P}^{-1}{AP}}\right) = {P}^{-1}\left( {\exp A}\right) P. \]\n\n(10.1) | No |
Proposition 10.2 For every \( A \in {\mathbf{M}}_{n}\left( \mathbb{C}\right) \) , \[ \det \exp A = \exp \operatorname{Tr}A. \] | Proof. We begin with a reduction of \( A \) of the form \( A = {P}^{-1}{TP} \), where \( T \) is upper-triangular. Because \( {T}^{k} \) is still triangular, with diagonal entries equal to \( {t}_{jj}^{k} \) , \( \exp T \) is triangular too, with diagonal entries equal to \( \exp {t}_{jj} \) . Hence \[ \det \exp T = \m... | Yes |
Proposition 10.3 The map \( \exp : {\mathbf{H}}_{n} \rightarrow {\mathbf{{HPD}}}_{n} \) is a homeomorphism. | Proof. Injectivity: Let \( A, B \in {\mathbf{H}}_{n} \) with \( \exp A = \exp B = : H \) . Then\n\n\[ \exp \frac{1}{2}A = \sqrt{H} = \exp \frac{1}{2}B \]\n\nBy induction, we have\n\n\[ \exp {2}^{-m}A = \exp {2}^{-m}B,\;m \in \mathbb{Z}. \]\n\nSubtracting \( {I}_{n} \), multiplying by \( {2}^{m} \), and passing to the l... | Yes |
Proposition 10.4 Let \( G \) be a subgroup of \( {\mathbf{{GL}}}_{n}\left( \mathbb{C}\right) \). We assume that \( G \) is stable under the map \( M \mapsto {M}^{ * } \) and that for every \( M \in G \cap {\mathbf{{HPD}}}_{n} \), the square root \( \sqrt{M} \) is an element of \( G \). Then \( G \) is stable under pola... | Proof. Let \( M \in G \) be given and let \( {HQ} \) be its polar decomposition. Because \( M{M}^{ * } \in \) \( G \cdot G = G \), we have \( {H}^{2} \in G \), hence \( H \in G \) by assumption. Finally, we have \( Q = \) \( {H}^{-1}M \in {G}^{-1} \cdot G = G \). An application of Theorem 10.1 finishes the proof. | Yes |
Proposition 10.5 Let \( J \) be a complex \( n \times n \) matrix satisfying \( {J}^{2} = \pm {I}_{n} \) . The subgroup \( G \) of \( {\mathbf{M}}_{n}\left( \mathbb{C}\right) \) defined by the equation \( {M}^{ * }{JM} = J \) is invariant under polar decomposition. If \( M \in G \), then \( \left| {\det M}\right| = 1 \... | Proof. Let \( M \in G \) . Then \( \det J = \det {M}^{ * }\det M\det J \) ; that is, \( {\left| \det M\right| }^{2} = 1 \) . In particular \( M \) is nonsingular. Then we have\n\n\[ \n{M}^{- * }J{M}^{-1} = {M}^{- * }\left( {{M}^{ * }{JM}}\right) {M}^{-1} = J, \n\] \n\nthus \( {M}^{-1} \in G \) . If \( M, N \in G \), we... | Yes |
Theorem 10.2 Under the hypotheses of Proposition 10.5, the group \( G \) is homeomorphic to \( \left( {G \cap {\mathbf{U}}_{n}}\right) \times {\mathbb{R}}^{d} \), for a suitable integer \( d \) . | Proof. (of Theorem 10.2.)\n\nAccording to Proposition 10.4, the proof amounts to showing that the factor \( G \cap \) \( {\mathbf{{HPD}}}_{n} \) is homeomorphic to some \( {\mathbb{R}}^{d} \) . To do this, we define\n\n\[ \mathcal{G} \mathrel{\text{:=}} \left\{ {N \in {\mathbf{M}}_{n}\left( k\right) \mid \exp {tN} \in ... | Yes |
Lemma 14. The set \( \mathcal{G} \) defined above satisfies\n\n\[ \mathcal{G} = \left\{ {N \in {\mathbf{M}}_{n}\left( k\right) \mid {N}^{ * }J + {JN} = {0}_{n}}\right\} . \] | Proof. If \( {N}^{ * }J + {JN} = {0}_{n} \), let us set \( M\left( t\right) = \exp {tN} \) . Then \( M\left( 0\right) = {I}_{n} \) and, thanks to (10.3)\n\n\[ \frac{d}{dt}M{\left( t\right) }^{ * }{JM}\left( t\right) = {M}^{ * }\left( t\right) \left( {{N}^{ * }J + {JN}}\right) M\left( t\right) = {0}_{n}, \]\n\nso that \... | Yes |
Lemma 15. The map \( \exp : \mathcal{G} \cap {\mathbf{H}}_{n} \rightarrow G \cap {\mathbf{{HPD}}}_{n} \) is a homeomorphism. | Proof. We must show that \( \exp : \mathcal{G} \cap {\mathbf{H}}_{n} \rightarrow G \cap {\mathbf{{HPD}}}_{n} \) is onto. Let \( M \in G \cap {\mathbf{{HPD}}}_{n} \) and let \( N \) be the Hermitian matrix such that \( \exp N = M \) . Let \( p \in \mathbb{R}\left\lbrack X\right\rbrack \) be a polynomial with real entrie... | No |
Proposition 10.6 The unitary group \( \mathbf{U}\left( {p, q}\right) \) is homeomorphic to \( {\mathbf{U}}_{p} \times {\mathbf{U}}_{q} \times {\mathbb{R}}^{2pq} \) . In particular, \( \mathbf{U}\left( {p, q}\right) \) is connected. | There remains to show connectedness. It is a straightforward consequence of the following lemma.\n\nLemma 16. The unitary group \( {\mathbf{U}}_{n} \) is connected.\n\nInasmuch as \( {\mathbf{{GL}}}_{n}\left( \mathbb{C}\right) \) is homeomorphic to \( {\mathbf{U}}_{n} \times {\mathbf{{HPD}}}_{n} \) (via polar decomposi... | No |
Lemma 17. The linear group \( {\mathbf{{GL}}}_{n}\left( \mathbb{C}\right) \) is connected. | Proof. Let \( M \in {\mathbf{{GL}}}_{n}\left( \mathbb{C}\right) \) be given. Define \( A \mathrel{\text{:=}} \mathbb{C} \smallsetminus \left\{ {{\left( 1 - \lambda \right) }^{-1} \mid \lambda \in \operatorname{Sp}\left( M\right) }\right\} \), which is arcwise connected because its complement is finite. The set \( A \) ... | Yes |
Lemma 18. Given \( M \in {\mathbf{O}}_{n} \), there exists \( Q \in {\mathbf{O}}_{n} \) such that the matrix \( {Q}^{-1}{MQ} \) has the form\n\n\[ \left( \begin{matrix} \left( \cdot \right) & 0 & \cdots & 0 \\ 0 & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & 0 \\ 0 & \cdots & 0 & \left( \cdot \right) \end{mat... | We now prove Lemma 18. As an orthogonal matrix, \( M \) is normal. From Theorem 5.5, it decomposes into a matrix of the form (10.8), the \( 1 \times 1 \) diagonal blocks being the real eigenvalues. These eigenvalues are \( \pm 1 \), inasmuch as \( M \) is orthogonal. The diagonal blocks \( 2 \times 2 \) are direct simi... | Yes |
Proposition 10.8 \( \left( {p, q \geq 1}\right) \) The connected components of \( G = \mathbf{O}\left( {p, q}\right) \) are the sets \( {G}_{\alpha } \mathrel{\text{:=}} {\sigma }^{-1}\left( \alpha \right) \), defined by \( {\alpha }_{1}\det A > 0 \) and \( {\alpha }_{2}\det D > 0 \), when a matrix \( M \) is written b... | 1. \( {G}_{\alpha }^{-1} = {G}_{\alpha } \). 2. \( {G}_{\alpha } \cdot {G}_{{\alpha }^{\prime }} = {G}_{\alpha {\alpha }^{\prime }} \). | No |
Proposition 10.9 The symplectic group \( {\mathbf{{Sp}}}_{n} \) is homeomorphic to \( {\mathbf{U}}_{n} \times {\mathbb{R}}^{n\left( {n + 1}\right) } \) . | Corollary 10.2 In particular, every symplectic matrix has determinant +1 . Indeed, Proposition 10.9 implies that \( {\mathbf{{Sp}}}_{n} \) is connected. Because the determinant is continuous, with values in \( \{ - 1,1\} \), it is constant, equal to +1 . | No |
Corollary 10.2 In particular, every symplectic matrix has determinant +1 . | Indeed, Proposition 10.9 implies that \( {\mathbf{{Sp}}}_{n} \) is connected. Because the determinant is continuous, with values in \( \{ - 1,1\} \), it is constant, equal to +1 . | Yes |
Theorem 11.1 The matrix \( M \in {\mathbf{{GL}}}_{n}\left( k\right) \) admits an LU factorization if and only if its leading principal minors are nonzero. When this condition is fulfilled, the LU factorization is unique. | Proof. Let us begin with uniqueness: if \( {LU} = {L}^{\prime }{U}^{\prime } \), then \( {\left( {L}^{\prime }\right) }^{-1}L = {U}^{\prime }{U}^{-1} \), which reads \( {L}^{\prime \prime } = {U}^{\prime \prime } \), where \( {L}^{\prime \prime } \) and \( {U}^{\prime \prime } \) are triangular of opposite types, the d... | Yes |
Corollary 11.1 Let \( M \in {\mathbf{{GL}}}_{n}\left( k\right) \), with \( n = {2m} \), read blockwise\n\n\[ M = \left( \begin{array}{ll} A & B \\ C & D \end{array}\right) ,\;A, B, C, D \in {\mathbf{{GL}}}_{m}\left( k\right) .\n\]\n\nThen\n\n\[ {M}^{-1} = \left( \begin{array}{ll} {\left( A - B{D}^{-1}C\right) }^{-1} & ... | Proof. We can verify the formula by multiplying by \( M \) . The only point to show is that the inverses are meaningful: \( A - B{D}^{-1}C,\ldots \) are invertible. Because of the symmetry of the formulæ, it is enough to check it for a single term, namely \( D - \) \( C{A}^{-1}B \) . Schur’s complement formula gives \(... | Yes |
Lemma 19. If \( {P}_{n} \leq {c}_{\alpha }{n}^{\alpha } \) (with \( 2 \leq \alpha \leq 3 \) ), then \( {j}_{\ell } \leq {C}_{\alpha }{\pi }_{\ell } \), where \( {C}_{\alpha } = 1 + \) \( 3{c}_{\alpha }/\left( {{2}^{\alpha - 1} - 1}\right) \) . | Proof. It is enough to sum (11.3) from \( k = 1 \) to \( l \) and use the inequality \( 1 + q + \cdots + \) \( {q}^{l - 1} \leq {q}^{\ell }/\left( {q - 1}\right) \) for \( q > 1 \) . | No |
Proposition 11.2 If the complexity \( {P}_{n} \) of the product in \( {\mathbf{M}}_{n}\left( \mathbb{C}\right) \) is bounded by \( {c}_{\alpha }{n}^{\alpha } \) , then the complexity \( {J}_{n} \) of inversion in \( {\mathbf{{GL}}}_{n}\left( \mathbb{C}\right) \) is bounded by \( {d}_{\alpha }{n}^{\alpha } \), where\n\n... | Proof. There remains to prove the second part. We notice that if \( A, B \in {\mathbf{M}}_{n}\left( \mathbb{C}\right) \) are given, then the matrix\n\n\[ \nM = \left( \begin{matrix} {I}_{n} & - A & {0}_{n} \\ {0}_{n} & {I}_{n} & - B \\ {0}_{n} & {0}_{n} & {I}_{n} \end{matrix}\right) \in {\mathbf{M}}_{3n}\left( \mathbb{... | Yes |
Proposition 11.3 The complexity of the multiplication of \( n \times n \) matrices is \( O\left( {n}^{\alpha }\right) \), with \( \alpha = \log 7/\log 2 = {2.807}\ldots \) More precisely,\n\n\[ \n{P}_{n} \leq \frac{147}{2}{n}^{\log 7/\log 2}.\n\] | Here is Strassen’s formula [37]. Let \( M, N \in {\mathbf{M}}_{2}\left( A\right) \), with\n\n\[ \nM = \left( \begin{array}{ll} a & b \\ c & d \end{array}\right) ,\;N = \left( \begin{array}{ll} x & y \\ z & t \end{array}\right) .\n\]\n\nOne first forms the expressions \( {x}_{1} = \left( {a + d}\right) \left( {x + t}\ri... | Yes |
Theorem 11.2 Let \( M \in {\mathbf{{SPD}}}_{n} \) . Then there exists a unique lower-triangular matrix \( L \in {\mathbf{M}}_{n}\left( \mathbb{R}\right) \), with strictly positive diagonal entries, satisfying \( M = L{L}^{T} \) . | Proof. Uniqueness. If \( {L}_{1} \) and \( {L}_{2} \) have the properties stated above, then \( {I}_{n} = L{L}^{T} \) , for \( L = {L}_{2}^{-1}{L}_{1} \), which still has the same form. In other words, \( L = {L}^{-T} \), where both sides are triangular matrices, but of opposite types (lower and upper). This equality s... | Yes |
Proposition 11.4 Let \( M \in {\mathbf{{GL}}}_{n}\left( \mathbb{C}\right) \) be given. Then there exist a unitary matrix \( Q \) and an upper-triangular matrix \( R \) ; the diagonal entries of the latter real positive, such that \( M = {QR} \) . This factorization is unique. | Proof. Uniqueness. If \( \left( {{Q}_{1},{R}_{1}}\right) \) and \( \left( {{Q}_{2},{R}_{2}}\right) \) give two factorizations, then \( Q = R \) with \( Q \mathrel{\text{:=}} {Q}_{2}^{-1}{Q}_{1} \) and \( R \mathrel{\text{:=}} {R}_{2}{R}_{1}^{-1} \) . Because \( Q \) is unitary, that is, \( {Q}^{ * } = {Q}^{-1} \), this... | Yes |
Theorem 11.5 Let \( A \in {M}_{n \times m}\left( \mathbb{C}\right) \) be given. There exists a unique matrix \( {A}^{ \dagger } \in \) \( {M}_{m \times n}\left( \mathbb{C}\right) \), called the Moore-Penrose generalized inverse, satisfying the following four properties.\n\n1. \( A{A}^{ \dagger }A = A \) .\n\n2. \( {A}^... | Proof. We first remark that if \( X \) satisfies these four properties, and if \( U \in {\mathbf{U}}_{n} \) , \( V \in {\mathbf{U}}_{m} \), then \( {V}^{ * }X{U}^{ * } \) is a generalized inverse of \( {UAV} \) . Therefore, existence and uniqueness need to be proved for only a single representative \( D \) of the equiv... | Yes |
Proposition 11.5 The following equalities hold for the generalized inverse:\n\n\[ \n{\left( \lambda A\right) }^{ \dagger } = \frac{1}{\lambda }{A}^{ \dagger }\;\left( {\lambda \neq 0}\right) ,\;{\left( {A}^{ \dagger }\right) }^{ \dagger } = A,\;{\left( {A}^{ \dagger }\right) }^{ * } = {\left( {A}^{ * }\right) }^{ \dagg... | Because \( {\left( A{A}^{ \dagger }\right) }^{2} = A{A}^{ \dagger } \), the matrix \( A{A}^{ \dagger } \) is a projector, which can therefore be described in terms of its range and kernel. Because \( A{A}^{ \dagger } \) is Hermitian, these subspaces are orthogonal to each other. Obviously, \( R\left( {A{A}^{ \dagger }}... | Yes |
Proposition 11.6 The system (11.6) is solvable if and only if \( b = M{M}^{ \dagger }b \) . When it is solvable, its general solution is \( x = {M}^{ \dagger }b + \left( {{I}_{m} - {M}^{ \dagger }M}\right) z \), where \( z \) ranges \( {\mathbb{C}}^{m} \) . Finally, the special solution \( {x}_{0} \mathrel{\text{:=}} {... | There remains to prove that \( {x}_{0} \) has the smallest norm among the solutions. That comes from the Pythagorean theorem and from the fact that \( R\left( {M}^{ \dagger }\right) = R\left( {{M}^{ \dagger }M}\right) = \) \( {\left( \ker M\right) }^{ \bot } \) . | No |
Proposition 12.1 An iterative method is convergent if and only if \( \rho \left( {{M}^{-1}N}\right) < 1 \) . | Proof. If the method is convergent, then for \( b = 0 \) ,\n\n\[ \mathop{\lim }\limits_{{m \rightarrow + \infty }}{\left( {M}^{-1}N\right) }^{m}{x}^{0} = 0 \]\n\nfor every \( {x}^{0} \in {K}^{n} \) . In other words,\n\n\[ \mathop{\lim }\limits_{{m \rightarrow + \infty }}{\left( {M}^{-1}N\right) }^{m} = 0 \]\n\nFrom Pro... | Yes |
Proposition 12.2 We have \( \rho \left( {\mathcal{L}}_{\omega }\right) \geq \left| {\omega - 1}\right| \) . In particular, if the relaxation method converges for a matrix \( A \in {\mathbf{M}}_{n}\left( \mathbb{C}\right) \) and a parameter \( \omega \in \mathbb{C} \), then\n\n\[ \left| {\omega - 1}\right| < 1\text{.} \... | Proof. If the method is convergent, we have \( \rho \left( {\mathcal{L}}_{\omega }\right) < 1 \) . However,\n\n\[ \det {\mathcal{L}}_{\omega } = \frac{\det \left( {\left( {1 - \omega }\right) D + {\omega F}}\right) }{\det \left( {D - {\omega E}}\right) } = \frac{\det \left( {\left( {1 - \omega }\right) D}\right) }{\det... | Yes |
Proposition 12.3 Under Hypothesis (1) or (2), the Jacobi method converges, as well as the relaxation method, for every \( \omega \in (0,1\rbrack \) . | Proof. Jacobi method: The matrix \( J = {D}^{-1}\left( {E + F}\right) \) is clearly irreducible when \( A \) is so. Furthermore, \[ \mathop{\sum }\limits_{{j = 1}}^{n}\left| {J}_{ij}\right| \leq 1,\;i = 1,\ldots, n \] in which all inequalities are strict if (1) holds, and at least one inequality is strict under the hyp... | Yes |
Lemma 20. If \( A \) and \( {M}^{ * } + N \) are Hermitian positive-definite (in a decomposition \( A = M - N) \), then \( \rho \left( {{M}^{-1}N}\right) < 1 \) . | Proof. Let us remark first that \( {M}^{ * } + N = {M}^{ * } + M - A \) is necessarily Hermitian when \( A \) is so.\n\nIt is therefore enough to show that \( {\begin{Vmatrix}{M}^{-1}Nx\end{Vmatrix}}_{A} < \parallel x{\parallel }_{A} \) for every nonzero \( x \in {\mathbb{C}}^{n} \) , where \( \parallel \cdot {\paralle... | Yes |
Theorem 12.1 If \( A \) is Hermitian positive-definite, then the relaxation method converges if and only if \( \left| {\omega - 1}\right| < 1 \) . | Proof. We have seen in Proposition 12.2 that the convergence implies \( \left| {\omega - 1}\right| < 1 \) . Let us see the converse. We have \( {E}^{ * } = F \) and \( {D}^{ * } = D \) . Thus\n\n\[ \n{M}^{ * } + N = \left( {\frac{1}{\omega } + \frac{1}{\bar{\omega }} - 1}\right) D = \frac{1 - {\left| \omega - 1\right| ... | Yes |
Lemma 21. Let \( \mu \) be a nonzero complex number and \( C \) a tridiagonal matrix, of diagonal \( {C}_{0} \), of upper-triangular part \( {C}_{ + } \) and lower-triangular part \( {C}_{ - } \) . Then\n\n\[ \det C = \det \left( {{C}_{0} + \frac{1}{\mu }{C}_{ - } + \mu {C}_{ + }}\right) . \] | Proof. It is enough to observe that the matrix \( C \) is conjugate to\n\n\[ {C}_{0} + \frac{1}{\mu }{C}_{ - } + \mu {C}_{ + } \]\n\nby the nonsingular matrix\n\n\[ {Q}_{\mu } = \left( \begin{matrix} \mu & & & & \\ & {\mu }^{2} & & 0 & \\ & & \ddots & & \\ & 0 & & \ddots & \\ & & & & {\mu }^{n} \end{matrix}\right) . \] | Yes |
Proposition 12.4 If \( A \) is tridiagonal and \( D \) invertible, then:\n\n1. \( {P}_{G}\left( {X}^{2}\right) = {X}^{n}{P}_{J}\left( X\right) \), where \( {P}_{G} \) is the characteristic polynomial of the Gauss-Seidel matrix \( G \) ,\n\n2. \( \rho \left( G\right) = \rho {\left( J\right) }^{2} \) ,\n\n3. The Gauss-Se... | Proof. Point 1 comes from Lemma 22. The spectrum of \( G \) is thus formed of \( \lambda = 0 \) (which is of multiplicity \( \left\lbrack {\left( {n + 1}\right) /2}\right\rbrack \) at least) and of squares of the eigenvalues of \( J \) , which proves Point 2. Point 3 follows immediately. Finally, if \( \mu \in \mathrm{... | Yes |
Theorem 12.2 (See Figure 12.1) Suppose that \( A \) is tridiagonal, \( D \) is invertible, and that the eigenvalues of \( J \) are real and belong to \( \left( {-1,1}\right) \) . Assume also that \( \omega \in \mathbb{R} \) .\n\nThen the relaxation method converges if and only if \( \omega \in \left( {0,2}\right) \) . ... | \n\nFig. 12.1 \( \rho \left( {\mathcal{L}}_{\omega }\right) \) in the tridiagonal case. | Yes |
Lemma 23. Let us denote by \( {\lambda }_{n} \geq \cdots \geq {\lambda }_{1}\left( { > 0}\right) \) the eigenvalues of \( A \) . If \( k \leq \ell \), then\n\n\[ E\left( {{x}_{k} - \bar{x}}\right) \leq E\left( {e}_{0}\right) \cdot \mathop{\min }\limits_{{\deg Q \leq k - 1}}\mathop{\max }\limits_{j}{\left| 1 + {\lambda ... | Proof. Let us compute\n\n\[ E\left( {{x}_{k} - \bar{x}}\right) = \min \left\{ {E\left( {x - \bar{x}}\right) \mid x \in {x}_{0} + {\mathcal{H}}_{k}}\right\} \]\n\n\[ = \min \left\{ {E\left( {{e}_{0} + y}\right) \mid y \in {\mathcal{H}}_{k}}\right\} \]\n\n\[ = \min \left\{ {E\left( {\left( {{I}_{n} + {AQ}\left( A\right) ... | Yes |
For every matrix \( M \in {\mathbf{M}}_{n}\left( \mathbb{C}\right) \) there exists a unitary transformation \( U \) such that \( {U}^{-1}{MU} \) is a Hessenberg matrix. If \( M \in {\mathbf{M}}_{n}\left( \mathbb{R}\right) \), one may take \( U \in {\mathbf{O}}_{n} \) . | Proof. Let \( X \in {\mathbb{C}}^{m} \) be a unit vector: \( {X}^{ * }X = 1 \) . The matrix of the unitary (orthogonal) symmetry with respect to the hyperplane \( {X}^{ \bot } \) is \( S = {I}_{m} - {2X}{X}^{ * } \) . In fact, \( {SX} = X - \) \( {2X} = - X \), and \( Y \in {X}^{ \bot } \) (i.e., \( {X}^{ * }Y = 0 \) )... | Yes |
Lemma 24. Let \( A \in {\mathbf{{GL}}}_{n}\left( K\right) \) be given, with \( K = \mathbb{R} \) or \( \mathbb{C} \) . Let \( {A}_{k} = {Q}_{k}{R}_{k} \) be the sequence of matrices given by the QR algorithm. Let us define \( {P}_{k} = {Q}_{0}\cdots {Q}_{k - 1} \) and \( {U}_{k} = {R}_{k - 1}\cdots {R}_{0} \) . Then \(... | Proof. From (13.2), we have \( {A}_{k} = {P}_{k}^{-1}A{P}_{k} \) ; that is, \( {P}_{k}{A}_{k} = A{P}_{k} \) . Then\n\n\[ {P}_{k + 1}{U}_{k + 1} = {P}_{k}{Q}_{k}{R}_{k}{U}_{k} = {P}_{k}{A}_{k}{U}_{k} = A{P}_{k}{U}_{k}. \]\n\nBy induction, \( {P}_{k}{U}_{k} = {A}^{k} \) . However, \( {P}_{k} \in {\mathbf{U}}_{n} \) and \... | Yes |
Theorem 13.2 Let \( A \in {\mathbf{{GL}}}_{n}\left( \mathbb{C}\right) \) be given. Assume that the moduli of the eigenvalues of \( A \) are distinct:\n\n\[ \n\left| {\lambda }_{1}\right| > \left| {\lambda }_{2}\right| > \cdots > \left| {\lambda }_{n}\right| \;\left( { > 0}\right) .\n\]\n\nIn particular, the eigenvalues... | Proof. Let \( Y = {LU} \) be the factorization of \( Y \) . We also make use of the \( {QR} \) factorization of \( {Y}^{-1} : {Y}^{-1} = {QR} \) . Because \( {A}^{k} = {Y}^{-1}{D}^{k}Y \), we have \( {P}_{k}{U}_{k} = {Y}^{-1}{D}^{k}Y = \) \( {QR}{D}^{k}{LU} \) .\n\nThe matrix \( {D}^{k}L{D}^{-k} \) is lower-triangular ... | Yes |
Theorem 13.3 Let \( A \in {\mathbf{{GL}}}_{n}\left( \mathbb{C}\right) \) be an irreducible Hessenberg matrix whose eigenvalues are of distinct moduli:\n\n\[ \left| {\lambda }_{1}\right| > \cdots > \left| {\lambda }_{n}\right| \;\left( { > 0}\right) \]\n\nThen the QR method converges; that is, the lower-triangular part ... | Proof. In the light of Theorem 13.2, it is enough to show that the matrix \( Y \) in the previous proof admits an \( {LU} \) factorization. We have \( {YA} = \operatorname{diag}\left( {{\lambda }_{1},\ldots ,{\lambda }_{n}}\right) Y \) . The rows of \( Y \) are thus the left eigenvectors: \( {\ell }_{j}A = {\lambda }_{... | Yes |
Corollary 13.1 If \( A \in {\mathbf{{HPD}}}_{n} \) and if \( {A}_{0} \) is a Hessenberg matrix, unitarily similar to \( A \) (e.g., a matrix obtained by Householder’s method), then the sequence \( {A}_{k} \) defined by the QR method converges to a diagonal matrix whose diagonal entries are the eigenvalues of \( A \) . | Indeed, the lower-triangular part converges, hence the whole matrix, because it is Hermitian. | No |
Lemma 25. We have\n\n\[ \n{\begin{Vmatrix}{E}_{k + 1}\end{Vmatrix}}^{2} = {\begin{Vmatrix}{E}_{k}\end{Vmatrix}}^{2} - 2{\left( {a}_{pq}^{\left( k\right) }\right) }^{2}. \n\] | Proof. It suffices to redo the calculations of Section 13.4.1, noting that\n\n\[ \n{k}_{ip}^{2} + {k}_{iq}^{2} = {h}_{ip}^{2} + {h}_{iq}^{2} \n\]\n\nwhenever \( i \neq p, q \), whereas \( {k}_{pq}^{2} = 0 \) . | Yes |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.