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Theorem 6.\n\n\\[ \n\\dim {R}_{\\mathrm{T}} = \\dim {R}_{{\\mathrm{T}}^{\\prime }} \n\\]\n\n(13) | Proof. We apply Theorem 4 of Chapter 2 to \\( U \\) and its subspace \\( {R}_{\\mathrm{T}} \\) :\n\n\\[ \n\\dim {R}_{\\mathrm{T}}^{ \\bot } + \\dim {R}_{\\mathrm{T}} = \\dim U.\n\\]\n\nNext we use Theorem 2 of this chapter applied to \\( {T}^{\\prime } : {U}^{\\prime } \\rightarrow {X}^{\\prime } \\) :\n\n\\[ \n\\dim {... | Yes |
Theorem 7. (a) Every similarity transformation is an automorphism of \( L\left( {X, X}\right) \) , maps sums into sums, products into products, scalar multiples into scalar multiples:\n\n\[{\left( k\mathrm{M}\right) }_{\mathrm{S}} = k{\mathrm{M}}_{\mathrm{S}}\]\n\n(15)\n\n\[{\left( \mathrm{M} + \mathrm{K}\right) }_{\ma... | Proof. (15) and \( {\left( {15}\right) }^{\prime } \) are obvious; to verify \( {\left( {15}\right) }^{\prime \prime } \) we use the definition (14):\n\n\[{\mathrm{M}}_{\mathrm{S}}{\mathrm{K}}_{\mathrm{S}} = {\mathrm{{SMS}}}^{-1}{\mathrm{{SKS}}}^{-1} = {\mathrm{{SMKS}}}^{-1} = {\left( \mathrm{{MK}}\right) }_{\mathrm{S}... | Yes |
Theorem 8. Similarity is an equivalence relation; that is, it is:\n\n(i) Reflexive. M is similar to itself.\n\n(ii) Symmetric. If \( M \) is similar to \( K \), then \( K \) is similar to \( M \) .\n\n(iii) Transitive. If \( M \) is similar to \( K \), and \( K \) is similar to \( L \), then \( M \) is similar to L. | Proof. (i) is true because we can in the definition (14) choose \( \mathrm{S} = \mathrm{I} \) .\n\n(ii) \( \mathrm{M} \) similar to \( \mathrm{K} \) means that\n\n\[ \mathrm{K} = {\mathrm{{SMS}}}^{-1} \]\n\n\( {\left( {14}\right) }^{\prime } \)\n\nMultiply both sides by \( \mathrm{S} \) on the right and \( {\mathrm{S}}... | Yes |
Theorem 1. Every linear map \( \mathrm{T}x = u \) from \( {\mathbb{R}}^{n} \) to \( {\mathbb{R}}^{m} \) can be written in form (1). | Proof. The vector \( x \) can be expressed as a linear combination of the unit vectors \( {e}_{1},\ldots ,{e}_{n} \) where \( {e}_{j} \) has \( j \) th component 1, all others 0 :\n\n\[ x = \sum {x}_{j}{e}_{j} \]\n\n(2)\n\nSince \( T \) is linear\n\n\[ u = \mathrm{T}x = \sum {x}_{j}\mathrm{T}{e}_{j} \]\n\n(3)\n\nDenote... | Yes |
Example 6. \( \left( \begin{array}{ll} 1 & 2 \end{array}\right) \left( \begin{array}{ll} 3 & 4 \\ 5 & 6 \end{array}\right) \left( \begin{array}{l} 1 \\ 2 \end{array}\right) = \left( \begin{array}{ll} 1 & 2 \end{array}\right) \left( \begin{array}{l} {11} \\ {17} \end{array}\right) = \left( {45}\right) \) ; | \( \left( \begin{array}{ll} 1 & 2 \end{array}\right) \left( \begin{array}{ll} 3 & 4 \\ 5 & 6 \end{array}\right) \left( \begin{array}{l} 1 \\ 2 \end{array}\right) = \left( \begin{array}{ll} {13} & {16} \end{array}\right) \left( \begin{array}{l} 1 \\ 2 \end{array}\right) = \left( {45}\right) ; \) | Yes |
Theorem 2. The range of T consists of all linear combinations of the columns of the matrix \( T \) . | The dimension of this space is called in old-fashioned texts the column rank of T. The row rank is defined similarly; \( {\left( {13}\right) }^{\prime \prime } \) shows that the row rank of \( T \) is the dimension of the range of \( {\mathrm{T}}^{T} \) . Since according to Theorem 6 of Chapter 3.\n\n\[ \dim {R}_{\math... | No |
The permutation \( p = \frac{12345}{24513} \) is the product of three transpositions \( {t}_{1} = \frac{12345}{12543},\;{t}_{2} = \frac{12345}{21345},\;{t}_{3} = \frac{12345}{42315}. \) | \[ p = {t}_{3} \circ {t}_{2} \circ {t}_{1} \] | Yes |
Corollary 3. An \( n \times n \) matrix \( \mathrm{A} \) is invertible iff det \( \mathrm{A} \neq 0 \) . | Proof. Suppose \( \mathrm{A} \) is not invertible; then its range is a proper subspace of \( {\mathbb{R}}^{n} \) . The range of \( A \) consists of all linear combinations of the columns of \( A \) ; therefore the columns are linearly dependent. According to property (v), this implies that \( \det A = 0 \) .\n\nSuppose... | Yes |
Lemma 4. Let \( \mathrm{A} \) be an \( n \times n \) matrix whose first column is \( {e}_{1} \) :\n\n\[ A = \left( \begin{matrix} 1 & \times & \times & \times \\ 0 & & & \\ \vdots & & {A}_{11} & \\ 0 & & & \end{matrix}\right) ; \]\n\nhere \( {\mathrm{A}}_{11} \) denotes the \( \left( {n - 1}\right) \times \left( {n - 1... | Proof. As first step we show that\n\n\[ \det A = \det \left( \begin{matrix} 1 & 0\ldots 0 \\ 0 & {A}_{11} \\ 0 & \end{matrix}\right) . \]\n\nFor it follows from Properties (i) and (ii) that if we alter a matrix by adding a multiple of one of its columns to another, the altered matrix has the same determinant as the ori... | Yes |
Theorem 6. Let \( \mathrm{A} \) be any \( n \times n \) matrix and \( j \) any index between 1 and \( n \) . Then\n\n\[ \det \mathrm{A} = \mathop{\sum }\limits_{i}{\left( -1\right) }^{i + j}{a}_{ij}\det {\mathrm{A}}_{ij} \] | Proof. To simplify notation, we take \( j = 1 \) . We write \( {a}_{1} \) as a linear combination of standard unit vectors:\n\n\[ {a}_{1} = {a}_{11}{e}_{1} + \cdots + {a}_{n1}{e}_{n} \]\n\nUsing multilinearity, we get\n\n\[ \det A = D\left( {{a}_{1},\ldots ,{a}_{n}}\right) = D\left( {{a}_{11}{e}_{1} + \cdots + {a}_{n1}... | Yes |
Theorem 7. The inverse matrix \( {A}^{-1} \) of an invertible matrix \( A \) has the form\n\n\[ \n{\left( {\mathrm{A}}^{-1}\right) }_{ki} = {\left( -1\right) }^{i + k}\frac{\det {\mathrm{A}}_{ik}}{\det \mathrm{A}}. \n\] | Proof. Since \( \mathrm{A} \) is invertible, \( \det \mathrm{A} \neq 0.{\mathrm{A}}^{-1} \) acts on the vector \( u \) ; see formula (1) of Chapter 4,\n\n\[ \n{\left( {\mathrm{A}}^{-1}u\right) }_{k} = \mathop{\sum }\limits_{i}{\left( {\mathrm{A}}^{-1}\right) }_{ki}{u}_{i} \n\]\n\nUsing (30) in (31) and comparing it to ... | Yes |
Theorem 9. Similar matrices have the same determinat and the same trace. | Proof. Using Theorem 2, we get from (37)\n\n\[ \det A = \left( {\det S}\right) \left( {\det B}\right) \left( {\det {S}^{-1}}\right) = \left( {\det B}\right) \left( {\det S}\right) \det \left( {S}^{-1}\right) \]\n\n\[ = \det B\det \left( {S{S}^{-1}}\right) = \left( {\det B}\right) \left( {\det I}\right) = \det B. \]\n\n... | Yes |
We shall construct a formula for \( {f}_{n} \) that displays its rate of growth. We start by rewriting the recurrence relation (10) in matrix-vector form: | \[ \left( \begin{matrix} {f}_{n} \\ {f}_{n + 1} \end{matrix}\right) = {\mathrm{A}}^{n}\left( \begin{array}{l} {f}_{0} \\ {f}_{1} \end{array}\right) ,\;\mathrm{A} = \left( \begin{array}{ll} 0 & 1 \\ 1 & 1 \end{array}\right) ,\] \[ \det \left( {\mathrm{A} - a\mathrm{I}}\right) = \det \left( \begin{matrix} - a & 1 \\ 1 & ... | Yes |
Theorem 1. Eigenvectors of a matrix A corresponding to distinct eigenvalues are linearly independent. | Proof. Suppose \( {a}_{i} \neq {a}_{k} \) for \( i \neq k \) and\n\n\[ \mathrm{A}{h}_{i} = {a}_{i}{h}_{i},\;{h}_{i} \neq 0. \]\n\n(12)\n\nSuppose now that there were a nontrivial linear relation among the \( {h}_{i} \) . There may be several; since all \( {h}_{i} \neq 0 \), all involve at least two eigenvectors. Among ... | Yes |
Theorem 2. If the characteristic polynomial of the \( n \times n \) matrix A has \( n \) distinct roots, then A has \( n \) linearly independent eigenvectors. | In this case the \( n \) eigenvectors form a basis; therefore every vector \( h \) in \( {\mathbb{C}}^{n} \) can be expressed as a linear combination of the eigenvectors:\n\n\[ h = \mathop{\sum }\limits_{1}^{n}{b}_{j}{h}_{j} \]\n\n(14) | No |
Theorem 3. Denote by \( {a}_{1},\ldots ,{a}_{n} \) the eigenvalues of \( \mathrm{A} \), with the same multiplicity they have as roots of the characteristic equation of \( A \) . Then\n\n\[ \n\sum {a}_{i} = \operatorname{tr}\mathrm{A},\;\prod {a}_{i} = \det \mathrm{A}.\n\]\n\n(15) | Proof. We claim that the characteristic polynomial of \( \mathrm{A} \) has the form\n\n\[ \n{p}_{\mathrm{A}}\left( s\right) = {s}^{n} - \left( {\operatorname{tr}\mathrm{A}}\right) {s}^{n - 1} + \cdots + {\left( -1\right) }^{n}\det \mathrm{A}.\n\]\n\n\( {\left( {15}\right) }^{\prime } \)\n\nAccording to elementary algeb... | Yes |
Theorem 4. (a) Let \( q \) be any polynomial, \( \mathrm{A} \) a square matrix, \( a \) an eigenvalue of A. Then \( q\left( a\right) \) is an eigenvalue of \( q\left( \mathrm{\;A}\right) \) . | Proof. Part (a) is merely a verbalization of relation (17), which shows also that A and \( q\left( \mathrm{\;A}\right) \) have \( h \) as common eigenvector. | No |
Theorem 5 (Cayley-Hamilton). Every matrix A satisfies its own characteristic equation:\n\n\[ \n{p}_{\mathrm{A}}\left( \mathrm{A}\right) = 0.\n\] | Proof. If A has distinct eigenvalues, then according to Theorem 2 it has \( n \) linearly independent eigenvectors \( {h}_{j}, j = 1,\ldots, n \) . Using (4) we apply \( {p}_{\mathrm{A}}\left( \mathrm{A}\right) \) :\n\n\[ \n{p}_{\mathrm{A}}\left( \mathrm{A}\right) h = \sum {p}_{\mathrm{A}}\left( {a}_{j}\right) {b}_{j}{... | No |
Lemma 6. Let \( \\mathrm{P} \) and \( \\mathrm{Q} \) be two polynomials with matrix coefficients\n\n\[ \n\\mathrm{P}\\left( s\\right) = \\sum {\\mathrm{P}}_{j}{s}^{j},\\;\\mathrm{Q}\\left( s\\right) = \\sum {\\mathrm{Q}}_{k}{s}^{k}.\n\]\n\nThe product \( \\mathrm{{PQ}} = \\mathrm{R} \) is then\n\n\[ \n\\mathrm{R}\\left... | The proof is self-evident. | No |
Example 3. \( \mathrm{A} = \mathrm{I} \) | \[ {p}_{\mathrm{A}}\left( s\right) = \det \left( {s\mathrm{I} - \mathrm{I}}\right) = {\left( s - 1\right) }^{n}; \] 1 is an \( n \) -fold zero. In this case every nonzero vector \( h \) is an eigenvector of \( \mathrm{A} \). | Yes |
Example 4. \( \mathrm{A} = \left( \begin{matrix} 3 & 2 \\ - 2 & - 1 \end{matrix}\right) \) , \( \operatorname{tr}\mathrm{A} = 2,\det \mathrm{A} = 1 \) ; therefore by Theorem 3, | \[ {p}_{\mathrm{A}}\left( s\right) = {s}^{2} - {2s} + 1 \] whose roots are one, with multiplicity two. The equation \[ \mathrm{A}h = \left( \begin{matrix} 3{h}_{1} + 2{h}_{2} \\ - 2{h}_{1} - {h}_{2} \end{matrix}\right) = \left( \begin{array}{l} {h}_{1} \\ {h}_{2} \end{array}\right) \] has as solution all vectors \( h \... | Yes |
Theorem 7 (Spectral Theorem). Let \( \mathrm{A} \) be an \( n \times n \) matrix with complex entries. Every vector in \( {\mathbb{C}}^{n} \) can be written as a sum of eigenvectors of \( A \), genuine or generalized. | For the proof, we need the following results of algebra.\n\nLemma 8. Let \( p \) an | No |
Lemma 8. Let \( p \) and \( q \) be a pair of polynomials with complex coefficients and assume that \( p \) and \( q \) have no common zero. Then there are two other polynomials \( a \) and \( b \) such that\n\n\[ {ap} + {bq} \equiv 1\text{.} \] | Proof. Denote by \( \mathcal{I} \) all polynomials of the form \( {ap} + {bq} \) . Among them there is one, nonzero, of lowest degree; call it \( d \) . We claim that \( d \) divides both \( p \) and \( q \) ; for suppose not; then the division algorithm yields a remainder \( r \), say\n\n\[ r = p - {md}. \]\n\nSince \... | Yes |
Theorem 17. Let \( X \) be a finite-dimensional linear space over \( \mathbb{C}, A \) a linear mapping of \( X \) into \( X \) . Denote by \( {X}^{\prime } \) the dual of \( X,{\mathrm{\;A}}^{\prime } : {X}^{\prime } \rightarrow {X}^{\prime } \) the transpose of \( \mathrm{A} \) . Let \( a \) and \( b \) denote two dis... | Proof. The transpose of \( \mathrm{A} \) is defined in equation (9) of Chapter 3 by requiring that for every \( x \) in \( X \) and every \( l \) in \( {X}^{\prime } \)\n\n\[ \left( {{\mathrm{A}}^{\prime }l, x}\right) = \left( {l,\mathrm{A}x}\right) . \]\n\nIf in particular we take \( x \) to be an eigenvector of \( \m... | Yes |
Theorem 1 (Schwarz Inequality). For all \( x, y \) ,\n\n\[ \left| \left( {x, y}\right) \right| \leq \parallel x\parallel \parallel y\parallel \] | Proof. Consider the function \( q\left( t\right) \) of the real variable \( t \) defined by\n\n\[ q\left( t\right) = \parallel x + {ty}{\parallel }^{2}. \]\n\nUsing the definition (8) and properties (i) and (ii) we can write\n\n\[ q\left( t\right) = \parallel x{\parallel }^{2} + {2t}\left( {x, y}\right) + {t}^{2}\paral... | Yes |
Theorem 3 (Triangle Inequality). For all \( x, y \)\n\n\[ \parallel x + y\parallel \leq \parallel x\parallel + \parallel y\parallel \text{.} \] | Proof. Using the algebraic properties of scalar product, we derive, analogously to (4), the identity\n\n\[ \parallel x + y{\parallel }^{2} = \parallel x{\parallel }^{2} + 2\left( {x, y}\right) + \parallel y{\parallel }^{2} \]\n\nand estimate the middle term by the Schwarz inequality. | Yes |
Theorem 4 (Gram-Schmidt). Given an arbitrary basis \( {y}^{\left( 1\right) },\ldots ,{y}^{\left( n\right) } \) in a finite-dimensional linear space equipped with a Euclidean structure, there is a related basis \( {x}^{\left( 1\right) },\ldots ,{x}^{\left( n\right) } \) with the following properties:\n\n(i) \( {x}^{\lef... | Proof. We proceed recursively; suppose \( {x}^{\left( 1\right) },\ldots ,{x}^{\left( k - 1\right) } \) have already been constructed. We set\n\n\[ \n{x}^{\left( k\right) } = c\left( {{y}^{\left( k\right) } - \mathop{\sum }\limits_{1}^{{k - 1}}{c}_{j}{x}^{\left( j\right) }}\right) \n\]\n\nSince \( {x}^{\left( 1\right) }... | Yes |
Theorem 5. Every linear function \( l\left( x\right) \) on a finite-dimensional linear space \( X \) with Euclidean structure can be written in the form\n\n\[ l\left( x\right) = \left( {x, y}\right) \]\n\n\( y \) some element of \( X \) . | Proof. Introduce an orthonormal basis \( {x}^{\left( 1\right) },\ldots ,{x}^{\left( n\right) } \) in \( X \) ; denote the value of \( l \) on \( {x}^{\left( k\right) } \) by\n\n\[ l\left( {x}^{\left( k\right) }\right) = {b}_{k} \]\n\nSet\n\n\[ y = \sum {b}_{k}{x}^{\left( k\right) } \]\n\nIt follows from orthonormality ... | Yes |
Theorem 6. For any subspace \( Y \) of \( X \) ,\n\n\[ X = Y \oplus {Y}^{ \bot } \] | Proof. We show first that a decomposition of form \( {\left( {20}\right) }^{\prime } \) is unique. Suppose we could write\n\n\[ x = z + {z}^{ \bot },\;z\text{ in }Y,{z}^{ \bot }\text{ in }{Y}^{ \bot }.\]\n\nComparing this with \( {\left( {20}\right) }^{\prime } \) gives\n\n\[ y - z = {z}^{ \bot } - {y}^{ \bot }\]\n\nIt... | Yes |
Theorem 7. (i) The mapping \( {P}_{Y} \) is linear. | Proof. Let \( w \) be any vector in \( X \), unrelated to \( x \), and let its decomposition \( {\left( {20}\right) }^{\prime } \) be\n\n\[ w = z + {z}^{ \bot },\;z\text{ in }Y,{z}^{ \bot }\text{ in }{Y}^{ \bot }.\]\n\nAdding this to \( {\left( {20}\right) }^{\prime } \) gives\n\n\[ x + w = \left( {y + z}\right) + \lef... | Yes |
Theorem 8. Let \( Y \) be a linear subspace of the Euclidean space \( X, x \) some vector in \( X \) . Then among all elements \( z \) of \( Y \), the one closest in Euclidean distance to \( x \) is \( {\mathrm{P}}_{Y}x \) . | Proof. Using the decomposition \( {\left( {20}\right) }^{\prime } \) of \( x \) we have\n\n\[ x - z = y - z + {y}^{ \bot },\;y = {\mathbf{P}}_{y}x. \]\n\nSince \( y \) and \( z \) both belong to \( Y \), so does \( y - z \) .\n\nTherefore by the Pythagorean theorem \( {\left( {13}\right) }^{\prime } \) ,\n\n\[ \paralle... | Yes |
Theorem 9. (i) If \( \\mathrm{A} \) and \( \\mathrm{B} \) are linear mappings of \( X \) into \( U \), then\n\n\[ \n{\\left( \\mathrm{A} + \\mathrm{B}\\right) }^{ * } = {\\mathrm{A}}^{ * } + {\\mathrm{B}}^{ * }.\n\] | Proof. (i) is an immediate consequence of \( {\\left( {23}\\right) }^{\\prime } \); | No |
Theorem 13. Let \( \mathrm{A} \) be a linear mapping from the Euclidean space \( X \) into the Euclidean space \( U \), where \( \parallel \mathrm{A}\parallel \) is its norm.\n\n(i)\n\n\[ \parallel \mathrm{A}z\parallel \leq \parallel \mathrm{A}\parallel \parallel z\parallel \;\text{ for all }z\text{ in }X. \]\n\n(33)\n... | Proof. (i) follows for unit vectors \( z \) from the definition (32) of \( \parallel \mathrm{A}\parallel \) . For any \( z \neq 0 \), write \( z = {kx}, x \) a unit vector; since \( \parallel \mathrm{A}{kx}\parallel = \parallel k\mathrm{A}x\parallel = \left| k\right| \parallel \mathrm{A}x\parallel \) and \( \parallel {... | Yes |
Theorem 14. For A as in Theorem 13, we have the following:\n\n(i) \( \parallel k\mathrm{\;A}\parallel = \left| k\right| \parallel \mathrm{A}\parallel \; \) for any scalar \( k \) . | Proof. (i) follows from the observation that \( \parallel k\mathrm{\;A}x\parallel = \left| k\right| \parallel \mathrm{A}x\parallel \) . | Yes |
Every Cauchy sequence in a finite-dimensional Euclidean space converges to a limit. | Proof. (i) Let \( x \) and \( y \) be two vectors in \( X,{a}_{j} \) and \( {b}_{j} \) their \( j \) th component; then\n\n\[ \left| {{a}_{j} - {b}_{j}}\right| \leq \parallel x - y\parallel \]\n\nDenote by \( {a}_{k, j} \) the \( j \) th component of \( {x}_{k} \) . Since \( \left\{ {x}_{k}\right\} \) is a Cauchy seque... | Yes |
Theorem 17. Let \( X \) be a linear space with a Euclidean structure, and suppose that it is locally compact-that is, that every bounded sequence \( \left\{ {x}_{k}\right\} \) of vectors in \( X \) has a convergent subsequence. Then \( X \) is finite dimensional. | Proof. We shall show that if \( X \) is not finite dimensional, then it is not locally compact. Not being finite dimensional means that given any linearly independent set of vectors \( {y}_{1},\ldots ,{y}_{k} \), there is a vector \( {y}_{k + 1} \) that is not a linear combination of them. In this way we obtain an infi... | Yes |
Theorem 1. (a) Given a real quadratic form (6) it is possible to change variables as in (10) so that in terms of the new variables, \( z, q \) is diagonal, that is, of the form\n\n\[ q\left( {{\mathrm{\;L}}^{-1}z}\right) = \mathop{\sum }\limits_{1}^{n}{d}_{i}{z}_{i}^{2} \] | Proof. Part (a) is entirely elementary and constructive. Suppose that one of the diagonal elements of \( q \) is nonzero, say \( {h}_{11} \neq 0 \) . We then group together all terms containing \( {y}_{1} \) :\n\n\[ q\left( y\right) = {h}_{11}{y}_{1}^{2} + \mathop{\sum }\limits_{2}^{n}{h}_{1j}{y}_{1}{y}_{j} + \mathop{\... | Yes |
Theorem 5. Let \( X \) be a complex Euclidean space, \( \mathrm{H} : X \rightarrow X \) a self-adjoint linear map. Then there is a resolution of the identity, in the sense of (29), (31), and (32) that gives a spectral resolution (30) of \( \mathrm{H} \) . | The restated form of the spectral theorem is very useful for defining functions of self-adjoint operators. We remark that its greatest importance is as the model for the infinite-dimensional version.\n\nSquaring relation (30) and using properties (31) of the \( {\mathrm{P}}_{j} \) we get\n\n\[ \n{\mathrm{H}}^{2} = \sum... | Yes |
Theorem 6. Suppose \( \mathrm{H} \) and \( \mathrm{K} \) are a pair of self-adjoint matrices that commute:\n\n\[ \n{\mathrm{H}}^{ * } = \mathrm{H},\;{\mathrm{K}}^{ * } = \mathrm{K},\;\mathrm{{HK}} = \mathrm{{KH}}.\n\]\n\nThen they have a common spectral resolution, that is, there exist orthogonal projections satisfying... | Proof. Denote by \( N \) one of the eigenspaces of \( \mathrm{H} \) ; then for every \( x \) in \( N \)\n\n\[ \n\mathrm{H}x = {ax}\n\]\n\nApplying \( \mathrm{K} \), we get\n\n\[ \n\mathrm{{KH}}x = a\mathrm{\;K}x.\n\]\n\nSince \( \mathrm{H} \) and \( \mathrm{K} \) commute, we can rewrite this as\n\n\[ \n\mathrm{{HK}}x =... | Yes |
Theorem 8. A normal map \( \mathrm{N} \) has an orthonormal basis consisting of eigenvectors. | Proof. If \( \mathrm{N} \) and \( {\mathrm{N}}^{ * } \) commute, so do\n\n\[ \mathrm{H} = \frac{\mathrm{N} + {\mathrm{N}}^{ * }}{2}\;\text{ and }\;\mathrm{A} = \frac{\mathrm{N} - {\mathrm{N}}^{ * }}{2}. \]\n\n(35)\n\nClearly, \( \mathrm{H} \) is adjoint and \( \mathrm{A} \) is anti-self-adjoint. According to Theorem 6 ... | Yes |
Theorem 9. Let \( U \) be a unitary map of a complex Euclidean space into itself, that is, an isometric linear map.\n\n(a) There is an orthonormal basis consisting of genuine eigenvectors of \( U \) .\n\n(b) The eigenvalues of \( U \) are complex numbers of absolute value \( = 1 \) . | Proof. According to equation (42) of Chapter 7, an isometric map U satisfies \( {\mathrm{U}}^{ * }\mathrm{U} = 1 \) . This relation says that \( {\mathrm{U}}^{ * } \) is a left inverse for \( \mathrm{U} \) . We have shown in Chapter 3 (see Corollary B of Theorem 1 there) that a mapping that has a left inverse is invert... | Yes |
Theorem 10. Let \( \mathrm{H} \) be a real symmetric linear map of a real Euclidean space \( X \) of finite dimension. Denote the eigenvalues of \( \mathrm{H} \), arranged in increasing order, by \( {a}_{1},\ldots ,{a}_{n} \) . Then\n\n\[ \n{a}_{j} = \mathop{\min }\limits_{{\dim S = j}}\mathop{\max }\limits_{{x\operato... | Proof. We shall show that for any linear subspace \( S \) of \( X \) of \( \dim S = j \) .\n\n\[ \n\mathop{\max }\limits_{{x\text{ in }S}}\frac{\left( x,\mathrm{H}x\right) }{\left( x, x\right) } \geq {a}_{j}\n\]\n\n(41)\n\nTo prove this it suffices to display a single vector \( x \neq 0 \) in \( S \) for which\n\n\[ \n... | Yes |
Theorem 12. Suppose \( \mathrm{N} \) is a normal mapping of a Euclidean space \( X \) into itself. Then\n\n\[ \parallel \mathrm{N}\parallel = \max \left| {n}_{j}\right| ,\]\n\nwhere the \( {n}_{j} \) are the eigenvalues of \( \mathrm{N} \) . | EXERCISE 10. Prove Theorem 12. (Hint: Use Theorem 8.) | No |
Theorem 13. Let \( A \) be a linear mapping of a finite-dimensional Euclidean space \( X \) into another finite-dimensional Euclidean space \( U \) . The norm \( \parallel \mathrm{A}\parallel \) of \( \mathrm{A} \) equals the square root of the largest eigenvalue of A*A. | Proof. \( \parallel \mathrm{A}x{\parallel }^{2} = \left( {\mathrm{A}x,\mathrm{A}x}\right) = \left( {x,{\mathrm{\;A}}^{ * }\mathrm{\;A}x}\right) \) . According to the Schwarz inequality, the right-hand side is \( \leq \parallel x\parallel \begin{Vmatrix}{{\mathrm{A}}^{ * }\mathrm{A}x}\end{Vmatrix} \) . It follows that f... | Yes |
Theorem 2. Let \( \mathrm{A}\left( t\right) \) be a matrix-valued function, differentiable and invertible. Then \( {\mathrm{A}}^{-1}\left( t\right) \) also is differentiable, and\n\n\[ \frac{d}{dt}{\mathrm{\;A}}^{-1} = - {\mathrm{A}}^{-1}\left( {\frac{d}{dt}\mathrm{\;A}}\right) {\mathrm{A}}^{-1}. \] | Proof. The following identity is easily verified:\n\n\[ {\mathrm{A}}^{-1}\left( {t + h}\right) - {\mathrm{A}}^{-1}\left( t\right) = {\mathrm{A}}^{-1}\left( {t + h}\right) \left\lbrack {\mathrm{A}\left( t\right) - \mathrm{A}\left( {t + h}\right) }\right\rbrack {\mathrm{A}}^{-1}\left( t\right) . \]\n\nDividing both sides... | No |
If for a particular value of \( t \) the matrices \( \mathrm{A}\left( t\right) \) and \( \dot{\mathrm{A}}\left( t\right) \) commute, then the chain rule in form (4) holds as \( t \) : | Proof. Suppose A and A commute; then (5) can be rewritten as\n\n\[ \frac{d}{dt}{\mathrm{\;A}}^{k} = k{\mathrm{\;A}}^{k - 1}\dot{\mathrm{\;A}} \]\n\nThis is formula (6) for \( p\left( s\right) = {s}^{k} \) ; since all polynomials are linear combinations of powers, using the linearity of differentiation we deduce (6) for... | No |
Theorem 4. Let \( Y\\left( t\\right) \) be a differentiable square matrix-valued function. Then for those values of \( t \) for which \( \\mathrm{Y}\\left( t\\right) \) is invertible, | \[ \frac{d}{dt}\\log \\det \\mathrm{Y} = \\operatorname{tr}\\left( {{\\mathrm{Y}}^{-1}\\frac{d}{dt}\\mathrm{Y}}\\right) \] | No |
Theorem 5. (a) If \( A \) and \( B \) are commuting square matrices,\n\n\[ \n{e}^{\mathrm{A} + \mathrm{B}} = {e}^{\mathrm{A}}{e}^{\mathrm{B}} \n\] | Proof. Part (a) follows from the definition \( {\left( {11}\right) }^{\prime } \) of \( {e}^{\mathrm{A} + \mathrm{B}} \), after \( {\left( \mathrm{A} + \mathrm{B}\right) }^{k} \) is expressed as \( \sum \left( \begin{array}{l} k \\ j \end{array}\right) {\mathrm{A}}^{j}{\mathrm{B}}^{k - j} \), valid for commuting variab... | No |
Theorem 6. The eigenvalues depend continuously on the matrix in the following sense: If \( \left\{ {A}_{m}\right\} \) is a convergent sequence of square matrices, in the sense that all entries of \( {\mathrm{A}}_{m} \) converge to the corresponding entry of \( \mathrm{A} \), then the set of eigenvalues of \( {\mathrm{A... | Proof. The eigenvalues of \( {\mathrm{A}}_{m} \) are the roots of the characteristic polynomial \( {p}_{m}\left( s\right) = \det \left( {s\mathbf{I} - {\mathbf{A}}_{m}}\right) \) . Since \( {\mathbf{A}}_{m} \) tends to \( \mathbf{A} \), all entries of \( {\mathbf{A}}_{m} \) tend to the corresponding entries of \( \math... | Yes |
Theorem 7. Let \( \\mathrm{A}\\left( t\\right) \) be a differentiable square matrix-valued function of the real variable \( t \) . Suppose that \( \\mathrm{A}\\left( 0\\right) \) has an eigenvalue \( {a}_{0} \) of multiplicity one, in the sense that \( {a}_{0} \) is a simple root of the characteristic polynomial of \( ... | Proof. The characteristic polynomial of \( \\mathrm{A}\\left( t\\right) \) is\n\n\[ \n\\det \\left( {s\\mathrm{I} - \\mathrm{A}\\left( t\\right) }\\right) = p\\left( {s, t}\\right) \n\]\n\na polynomial of degree \( n \) in \( s \) whose coefficients are differentiable functions of \( t \) . The assumption that \( {a}_{... | Yes |
Theorem 8. Let \( \mathrm{A}\left( t\right) \) be a differentiable matrix-valued function of \( t, a\left( t\right) \) an eigenvalue of \( \mathrm{A}\left( t\right) \) of multiplicity one. Then we can choose an eigenvector \( h\left( t\right) \) of \( \mathrm{A}\left( t\right) \) pertaining to the eigenvalue \( a\left(... | Proof. We need the following lemma. | No |
Lemma 9. Let \( \mathrm{A} \) be an \( n \times n \) matrix, \( p \) its characteristic polynomial, \( a \) some simple root of \( p \) . Then at least one of the \( \left( {n - 1}\right) \times \left( {n - 1}\right) \) principal minors of \( \mathrm{A} - a\mathrm{I} \) has nonzero determinant, where the \( i \) th pri... | Proof. We may, at the cost of subtracting al from A, take the eigenvalue to be zero. The condition that 0 is a simple root of \( p\left( s\right) \) means that \( p\left( 0\right) = 0 \) ; \( \left( {{dp}/{ds}}\right) \left( 0\right) \neq 0 \) . To compute the derivative of \( p \) we denote by \( {c}_{1},\ldots ,{c}_{... | Yes |
Theorem 12. Let \( \mathrm{A}\left( t\right) \) be a differentiable function of the real variable \( t \) whose values are self-adjoint matrices. Suppose that at \( t = 0,\mathrm{\;A}\left( 0\right) \) has an eigenvalue \( {a}_{0} \) of multiplicity \( k > 1 \) . Denote by \( N \) the eigenspace of \( \mathrm{A}\left( ... | Proof. For \( t \) small enough the characteristic polynomial of \( \mathrm{A}\left( t\right) \) differs little from that of \( \mathrm{A}\left( 0\right) \) . By hypothesis, the latter has a \( k \) -fold root at \( {a}_{0} \) ; it follows that the former have exactly \( k \) roots that approach \( {a}_{0} \) as \( t \... | Yes |
Lemma 13. As \( t \rightarrow 0 \), the distance of each of the normalized eigenvectors \( {h}_{j}\left( t\right) \) from the eigenspace \( N \) tends to zero. | Proof. Using the orthogonal projection \( \mathrm{P} \) onto \( N \), we can reformulate the conclusion as follows:\n\n\[ \mathop{\lim }\limits_{{t \rightarrow 0}}\begin{Vmatrix}{\left( {\mathrm{I} - \mathrm{P}}\right) {h}_{j}\left( t\right) }\end{Vmatrix} = 0,\;j = 1,\ldots, k. \]\n\n(31)\n\nTo show this, we use the f... | Yes |
Theorem 2. Let \( M \) and \( N \) denote positive mappings that satisfy\n\n\[ \mathrm{O} < \mathrm{M} < \mathrm{N}\text{.}\]\n\nThen\n\n\[ {\mathrm{M}}^{-1} > {\mathrm{N}}^{-1} \] | First Proof. We start with the case when \( \mathrm{N} = \mathrm{I} \) . By definition, \( \mathrm{M} < \mathrm{I} \) means that \( \mathrm{I} - \mathrm{M} \) is positive. According to part (iv) of Theorem 1, that means that the eigenvalues of \( \mathrm{I} - \mathrm{M} \) are positive, that is, that the eigenvalues of... | Yes |
Lemma 3. Let \( \mathrm{A}\left( t\right) \) be a differentiable function of the real variable whose values are self-adjoint mappings; the derivative \( \left( {d/{dt}}\right) \mathrm{A} \) is then also self-adjoint. Suppose that \( \left( {d/{dt}}\right) \mathrm{A} \) is positive; then \( \mathrm{A}\left( t\right) \) ... | Proof. Let \( x \) be any nonzero vector, independent of \( t \) . Then by the assumption that the derivative of \( A \) is positive, we obtain\n\n\[ \frac{d}{dt}\left( {x,\mathrm{\;A}x}\right) = \left( {x,\frac{d}{dt}\mathrm{\;A}x}\right) > 0. \]\n\nSo by ordinary calculus, \( \left( {x,\mathbf{A}\left( t\right) x}\ri... | Yes |
Theorem 4. Let \( A \) and \( B \) denote two self-adjoint maps with the following properties:\n\n(i) \( \mathrm{A} \) is positive.\n\n(ii) The symmetrized product \( \mathrm{S} = \mathrm{{AB}} + \mathrm{{BA}} \) is positive.\n\nThen \( \mathrm{B} \) is positive. | Proof. Define \( \mathrm{B}\left( t\right) \) as \( \mathrm{B}\left( t\right) = \mathrm{B} + t\mathrm{\;A} \) . We claim that for \( t \geq 0 \) the symmetrized product of \( \mathrm{A} \) and \( \mathrm{B}\left( t\right) \) is positive. For\n\n\[ \mathrm{S}\left( t\right) = \mathrm{A}\mathrm{B}\left( t\right) + \mathr... | Yes |
Theorem 5. Let \( M \) and \( N \) denote positive mappings that satisfy\n\n\[ \mathrm{O} < \mathrm{M} < \mathrm{N} \]\n\nthen\n\n\[ \sqrt{\mathbf{M}} < \sqrt{\mathbf{N}} \]\n\nwhere \( \sqrt{} \) denotes the positive square root. | Proof. Define the function \( \mathrm{A}\left( t\right) \) as in (12):\n\n\[ \mathrm{A}\left( t\right) = \mathrm{M} + t\left( {\mathrm{\;N} - \mathrm{M}}\right) \]\n\nWe have shown that \( \mathrm{A}\left( t\right) \) is positive when \( 0 \leq t \leq 1 \) ; so we can define\n\n\[ R\left( t\right) = \sqrt{\mathrm{A}\le... | Yes |
Theorem 6. (i) Every Gram matrix is nonnegative.\n\n(ii) The Gram matrix of a set of linearly independent vectors is positive.\n\n(iii) Every positive matrix can be represented as a Gram matrix. | Proof. The quadratic form associated with a Gram matrix can be expressed as follows:\n\n\[ \left( {x,\mathrm{\;G}x}\right) = \mathop{\sum }\limits_{{i, j}}{x}_{i}{\bar{\mathrm{G}}}_{ij}{\bar{x}}_{j} = \sum \left( {{f}_{i},{f}_{j}}\right) {x}_{i}{\bar{x}}_{j} \]\n\n\[ = \left( {\mathop{\sum }\limits_{i}{x}_{i}{f}_{i},\s... | Yes |
Theorem 8. The determinant of every positive matrix is positive. | Proof. According to Theorem 3 of Chapter 6, the determinant of a matrix is the product of its eigenvalues. According to Theorem 1 of this chapter, the eigenvalues of a positive matrix are positive. Then so is their product. | Yes |
Theorem 10. The determinant of a positive matrix \( \mathrm{H} \) does not exceed the product of its diagonal elements:\n\n\[ \det \mathrm{H} \leq \prod {h}_{ii} \] | Proof. Since \( \mathrm{H} \) is positive, so are its diagonal entries. Define \( {d}_{i} = 1/\sqrt{{h}_{ii}} \), and denote by \( \mathrm{D} \) the diagonal matrix with diagonal entries \( {d}_{i} \) . Define the matrix \( \mathrm{B} \) by\n\n\[ \mathrm{B} = \mathrm{{DHD}}\text{.} \]\n\nClearly, B is symmetric and pos... | Yes |
Theorem 11. Let \( \\mathrm{T} \) be any \( n \\times n \) matrix whose columns are \( {c}_{1},{c}_{2},\\ldots ,{c}_{n} \) . Then the determinant of \( T \) is in absolute value not greater than the product of the length of its columns:\n\n\[ \n\\left| {\\det \\mathrm{T}}\\right| \\leq \\prod \\begin{Vmatrix}{c}_{j}\\e... | Proof. Define \( \\mathrm{H} = {\\mathrm{T}}^{ * }\\mathrm{\\;T} \) ; its diagonal elements are\n\n\[ \n{h}_{ii} = \\mathop{\\sum }\\limits_{j}{t}_{ij}^{ * }{t}_{ji} = \\mathop{\\sum }\\limits_{j}{\\bar{t}}_{ji}{t}_{ji} = \\mathop{\\sum }\\limits_{j}{\\left| {t}_{ji}\\right| }^{2} = {\\begin{Vmatrix}{c}_{i}\\end{Vmatri... | Yes |
Theorem 12. Let \( \mathrm{H} \) be an \( n \times n \) real, symmetric, positive matrix. Then\n\n\[ \n\frac{{\pi }^{n/2}}{\sqrt{\det \mathrm{H}}} = {\int }_{{\mathrm{R}}^{n}}{e}^{-\left( {x,\mathrm{\;H}x}\right) }{dx}.\n\]\n\n(31) | Proof. It follows from inequality \( {\left( 5\right) }^{\prime } \) that the integral (31) converges. To evaluate it, we appeal to the spectral theorem for self-adjoint mappings, see Theorem \( {4}^{\prime } \) of Chapter 8, and introduce new coordinates\n\n\[ \nx = \mathrm{M}y\n\]\n\n(32)\n\nM an orthogonal matrix so... | Yes |
Lemma 13. Let \( A \) be a self-adjoint map of a Euclidean space \( U \) into itself. We denote by \( {p}_{ + }\left( \mathrm{A}\right) \) the number of positive eigenvalues of \( \mathrm{A} \), and denote by \( {p}_{ - }\left( \mathrm{A}\right) \) the number of its negative eigenvalues.\n\n\( {p}_{ + }\left( \mathrm{A... | Proof. This follows from the minmax characterization of the eigenvalues of A; see Theorem 10, as well as Lemma 2 of Chapter 8. | No |
Theorem 14. Let \( U \) and \( A \) be as in Lemma 13, and let \( V \) be a subspace of \( U \) whose dimension is one less than the dimension of \( U \) :\n\n\[ \n\dim \mathrm{V} = \dim \mathrm{U} - 1.\n\]\n\nDenote by P orthogonal projection onto V. Then PAP is a self-adjoint map of U into \( \mathrm{U} \) that maps ... | Proof. Let T denote a subspace of V of dimension \( {p}_{ + }\left( \mathrm{B}\right) \) on which \( \mathrm{B} \) is positive:\n\n\[ \n\left( {\mathrm{B}v, v}\right) > 0,\;v\text{ in }\mathrm{T},\;v \neq 0.\n\]\n\nBy definition of \( B \), we can write this as\n\n\[ \n0 < \left( {\mathrm{{PAP}}v, v}\right) = \left( {\... | Yes |
Theorem 15. Let U, V, A, and B be as in Theorem 14. Denote the eigenvalues of \( \mathrm{A} \) as \( {a}_{1},\ldots ,{a}_{n} \), and denote those of \( \mathrm{B} \) as \( {b}_{1},\ldots ,{b}_{n - 1} \) . The eigenvalues of \( \mathrm{B} \) separate the eigenvalues of A : \n\n\[ \n{a}_{1} \leq {b}_{1} \leq {a}_{2} \leq... | Proof. Apply Theorem 14 to \( \mathrm{A} - c \) and \( \mathrm{B} - c \) . We conclude that the number of \( {b}_{i} \) less than \( c \) is not greater than the number of \( {a}_{i} \) less than \( c \), and at most one less. We claim that \( {a}_{i} \leq {b}_{ij} \) if not, we could choose \( {b}_{i} < c < {a}_{i} \)... | Yes |
Theorem 16. Let \( M \) and \( N \) denote self-adjoint \( k \times k \) matrices satisfying\n\n\[ \mathrm{M} < \mathrm{N}\text{.}\]\n\nDenote the eigenvalues of \( \mathrm{M} \), arranged in increasing order, by \( {m}_{1} \leq \cdots \leq {m}_{k} \), and those of \( \mathrm{N} \) by \( {n}_{1} \leq \cdots \leq {n}_{k... | First Proof. We appeal to the minmax principle, Theorem 10 in Chapter 8, formula (40), according to which\n\n\[ {m}_{j} = \mathop{\min }\limits_{{\dim S = j}}\mathop{\max }\limits_{{x\text{ in }S}}\frac{\left( x,\mathrm{M}x\right) }{\left( x, x\right) }\]\n\n\[ {n}_{j} = \mathop{\min }\limits_{{\dim S = j}}\mathop{\max... | Yes |
Theorem 17. Let \( \mathrm{M} \) and \( \mathrm{N} \) be self-adjoint \( \mathrm{k} \times \mathrm{k} \) matrices \( {m}_{j} \) and \( {n}_{j} \) their eigenvalues arrayed in increasing order. Then\n\n\[ \left| {{n}_{j} - {m}_{j}}\right| \leq \parallel \mathrm{M} - \mathrm{N}\parallel . \] | Proof. Denote \( \parallel \mathrm{M} - \mathrm{N}\parallel \) by \( d \) . It is easy to see that\n\n\[ \mathrm{N} - d\mathrm{I} \leq \mathrm{M} \leq \mathrm{N} + d\mathrm{I} \]\n\nInequality (44) follows from \( {\left( {44}\right) }^{\prime } \) and (41). | No |
Theorem 19. Denote by \( {e}_{\min }\left( \mathrm{H}\right) \) the smallest eigenvalue of a self-adjoint mapping \( \mathrm{H} \) in a Euclidean space. We claim that \( {e}_{\min } \) is a concave function of \( \mathrm{H} \), that is, that for \( 0 \leq t \leq 1 \) ,\n\n\[ \n{e}_{\min }\left( {t\mathrm{\;L} + \left( ... | Proof. We have shown in Chapter 8, equation (37), that the smallest eigenvalue of a mapping can be characterized as a minimum: \n\n\[ \n{e}_{\min }\left( \mathrm{H}\right) = \mathop{\min }\limits_{{\parallel x\parallel = 1}}\left( {x,\mathrm{H}x}\right) \n\] \n\n(53) \n\nLet \( y \) be a unit vector where \( \left( {x,... | Yes |
Theorem 20. Suppose the self-adjoint part \( Z \) is positive:\n\n\[ \n\mathrm{Z} + {\mathrm{Z}}^{ * } > 0\text{.}\n\]\n\nThen the eigenvalues of \( Z \) have positive real part. | Proof. Using the conjugate symmetry of scalar product in a complex Euclidean space, and the definition of adjoint, we have the following identity for any vector \( h \) :\n\n\[ \n2\;\mathrm{{Re}}\left( {\mathrm{Z}h, h}\right) = \left( {\mathrm{Z}h, h}\right) + \overline{\left( \mathrm{Z}h, h\right) } = \left( {\mathrm{... | Yes |
Theorem 21. Let \( a \) be a complex number with \( \operatorname{Re}a > 0 \) . Let \( \mathrm{Z} \) be a mapping whose self-adjoint part \( Z + {Z}^{ * } \) is positive. Then \[ \mathrm{W} = \left( {\mathrm{I} - a\mathrm{Z}}\right) {\left( \mathrm{I} + \bar{a}\mathrm{Z}\right) }^{-1} \] is a mapping of norm less than ... | Proof. According to Theorem 20 the eigenvalues \( z \) of \( Z \) have positive real part. It follows that the eigenvalues of \( \mathrm{I} + \bar{a}\mathrm{Z},\mathrm{I} + a\mathrm{Z} \) are \( \neq 0 \) ; therefore \( \mathrm{I} + \bar{a}\mathrm{Z} \) is invertible. For any vector \( x \), denote \( {\left( \mathrm{I... | Yes |
Theorem 22. Let \( A \) be a linear mapping of a complex Euclidean space into itself. Then A can be factored as\n\n\[ A = {RU}, \]\n\nwhere \( R \) is a nonnegative self-adjoint mapping, and \( U \) is unitary. When \( A \) is invertible, \( \mathrm{R} \) is positive. | Proof. Take first the case that \( \mathrm{A} \) is invertible; then so is \( {\mathrm{A}}^{ * } \) . For any \( x \neq 0 \) ,\n\n\[ \left( {{\mathrm{{AA}}}^{ * }x, x}\right) = \left( {{\mathrm{A}}^{ * }x,{\mathrm{A}}^{ * }x}\right) = {\begin{Vmatrix}{\mathrm{A}}_{x}^{ * }\end{Vmatrix}}^{2} > 0. \]\n\nThis proves that ... | Yes |
Theorem 1 (Euler). An isometry \( M \) of three-dimensional real Euclidean space with determinant plus 1 that is nontrivial, that is not equal to I, is a rotation; it has a uniquely defined axis of rotation and angle of rotation \( \theta \) . | Proof. Points \( f \) on the axis of rotation remain fixed, so they satisfy\n\n\[ \mathrm{M}f = f \]\n\n(2)\n\n---\n\nLinear Algebra and Its Applications. Second Edition, by Peter D. Lax\n\nCopyright (C) 2007 John Wiley & Sons, Inc.\n\n---\n\nthat is, they are eigenvectors of \( M \) with eigenvalue 1 . We claim that a... | Yes |
Theorem 2. Solutions of the differential equation (32)' are uniquely determined by their initial data \( x\left( 0\right) \) and \( \dot{x}\left( 0\right) \) . That is, two solutions that have the same initial data are equal for all time. | Proof. Since equation \( {\left( {32}\right) }^{\prime } \) is linear, the difference of two solutions is again a solution. Therefore it is sufficient to prove that if a solution \( x \) has zero initial data, then \( x\left( t\right) \) is zero for all \( t \) . To see this, we observe that if \( x\left( 0\right) = 0,... | Yes |
Theorem 3. Every solution of the differential equation \( {\left( {32}\right) }^{\prime } \) is of form (36). | Proof. Solutions of the form \( {\left( {36}\right) }^{\prime } \) form a \( {2n} \) -dimensional space. We have shown that the set of all solutions is a linear space of dimension \( \leq {2n} \) . It follows that all solutions are of form \( {\left( {36}\right) }^{\prime } \) . | No |
Consider two differential equations of form (32)':\n\n\[ \mathrm{M}\ddot{x} + \mathrm{K}x = 0,\;\mathrm{\;N}\ddot{y} + \mathrm{L}y = 0, \]\n\n\( \mathrm{M},\mathrm{K},\mathrm{N},\mathrm{L} \) positive, real \( n \times n \) matrices. Suppose that\n\n\[ \mathrm{M} \geq \mathrm{N}\text{ and }\mathrm{K} \leq \mathrm{L}\te... | Proof. We introduce an intermediate differential equation\n\n\[ \mathrm{M}\ddot{z} + \mathrm{L}z = 0, \]\n\nDenote its natural frequencies by \( {f}_{j}/{2\pi } \) . In analogy with equation (35), the \( {f}_{j} \) satisfy\n\n\[ {f}^{2}\mathbf{M}h = \mathbf{L}h \]\n\nwhere \( h \) is an eigenvector. In analogy with equ... | Yes |
Theorem 5. Suppose that the numbers \( {k}_{i} \) and \( {k}_{ij}, i \neq j \) are positive. Then the symmetric matrix \( \mathrm{K} \) ,\n\n\[ \n{\mathrm{K}}_{ij} = - {k}_{ij},\;i \neq j;\;{\mathrm{\;K}}_{ii} = {k}_{i} + \mathop{\sum }\limits_{{i \neq j}}{k}_{ij} \n\]\n\n(40)\n\nis positive. | Proof. It suffices to show that every eigenvalue \( a \) of \( \mathrm{K} \) is positive:\n\n\[ \n\mathbf{K}u = {au}.\n\]\n\n(41)\n\nNormalize the eigenvector \( u \) of \( \mathrm{K} \) so that the largest component, say \( {u}_{i} \), equals 1, and all others are \( \leq 1 \) . The \( i \) th component of (41) is\n\n... | Yes |
Theorem 2. (a) The gauge function \( p \) of an open convex set \( K \) that contains the origin is well-defined for every \( x \) . | Proof. Call the set of \( r > 0 \) for which \( x/r \) is in \( K \) admissible for \( x \) . To prove (a) we have to show that for any \( x \) the set of admissible \( r \) is nonempty. This follows from the assumption that 0 is an interior point of \( K \) . | Yes |
Theorem 3. Let \( K \) be an open convex set, and let \( y \) be a point not in \( K \) . Then there is an open half-space containing \( K \) but not \( y \) . | Proof. An open half-space is by definition a set of points satisfying inequality \( l\left( x\right) < c \) ; see (3). So we have to construct a linear function \( l \) and a number \( c \) such that\n\n\[ l\left( x\right) < c\;\text{ for all }x\text{ in }K \]\n\n(11)\n\n\[ l\left( y\right) = c \]\n\n(12)\n\nWe assume ... | Yes |
Theorem 4 (Hahn-Banach). Let \( p \) be a real-valued positive homogeneous subadditive function defined on a linear space \( X \) over \( \mathbb{R} \). Let \( U \) be a subspace of \( X \) on which a linear function is defined, satisfying (13):\n\n\[ l\left( u\right) \leq p\left( u\right) \;\text{ for all }u\text{ in ... | Proof. Proof is by induction; we show that \( l \) can be extended to a subspace \( V \) spanned by \( U \) and any vector \( z \) not in \( U \). That is, \( V \) consists of all vectors of form\n\n\[ v = u + {tz},\;u\text{in}U, t\text{any real number}. \]\n\nSince \( l \) is linear\n\n\[ l\left( v\right) = l\left( u\... | Yes |
Theorem 5. Let \( K \) and \( H \) be open convex sets that are disjoint. Then there is a hyperplane that separates them. That is, there is a linear function \( l \) and a constant \( d \) such that\n\n\[ l\left( x\right) < d\text{ on }K,\;l\left( y\right) > d\text{ on }H. \] | Proof. Define the difference \( K - H \) to consist of all differences \( x - y, x \) in \( K, y \) in \( H \) . It is easy to verify that this is an open, convex set. Since \( K \) and \( H \) are disjoint, \( K - H \) does not contain the origin. Then by Theorem 3, with \( y = 0 \), and therefore \( c = 0 \), there i... | Yes |
Theorem 6. Let \( K \) be an open convex set, \( {q}_{K} \) its support function. Then \( x \) belongs to \( K \) iff\n\n\[ l\left( x\right) < {q}_{K}\left( l\right) \]\n\n(17)\n\nfor all \( l \) in \( {X}^{\prime } \) . | Proof. It follows from definition (16) that for every \( x \) in \( {Kl}\left( x\right) \leq {q}_{K}\left( l\right) \) for every \( l \) ; therefore the strict inequality (17) holds for all interior points \( x \) in \( K \) . To see the converse, suppose that \( y \) is not in \( K \) . Then by Theorem 3 there is an \... | Yes |
Theorem 7. Let \( K \) be a closed, convex set, and \( y \) a point not in \( K \) . Then there is a closed half-space that contains \( K \) but not \( y \) . | Sketch of Proof. Suppose \( K \) contains the origin. If \( K \) has no interior points, it lies in a lower-dimensional subspace. If it has an interior point, we choose it to be the origin. Then the gauge function \( {p}_{K} \) of \( K \) can be defined as before. If \( x \) belongs to \( K \) , we may choose in the de... | No |
Theorem 9. The closed convex hull of any set \( S \) is the set of points \( x \) satisfying \( l\left( x\right) \leq {q}_{S}\left( l\right) \) for all \( l \) in \( {X}^{\prime }. | Let \( {x}_{1},\ldots ,{x}_{m} \) denote \( m \) points in \( X \), and \( {p}_{1},\ldots ,{p}_{m} \) denote \( m \) nonnegative numbers whose sum is 1 .\n\n\[ {p}_{j} \geq 0,\;\mathop{\sum }\limits_{1}^{m}{p}_{j} = 1 \]\n\n(20)\n\nThen\n\n\[ x = \sum {p}_{j}{x}_{j} \]\n\n\( {\left( {20}\right) }^{\prime } \)\n\nis cal... | No |
Theorem 12 (Helly). Let \( X \) be a linear space of dimension \( n \) over the reals. Let \( \left\{ {{K}_{1},\ldots ,{K}_{N}}\right\} \) be a collection of \( N \) convex sets in \( X \) . Suppose that every subcollection of \( n + 1 \) sets \( K \) has a nonempty intersection. Then all \( K \) in the whole collectio... | Proof (Radon). We argue by induction on \( N \), the number of sets, starting with the trivial situation \( N = n + 1 \) . Suppose that \( N > n + 1 \) and that the assertion is true for \( N - 1 \) sets. It follows that if we omit any one of the sets, say \( {K}_{i} \), the rest have a point \( {x}_{i} \) in common:\n... | Yes |
A vector \( y \) can be written as a linear combination of given vectors \( {y}_{j} \) with nonnegative coefficients as in (5) iff every \( \xi \) that satisfies\n\n\[ \xi {y}_{j} \geq 0,\;j = 1,\ldots, m \]\n\nalso satisfies\n\n\[ {\xi y} \geq 0\text{.} \] | The necessity of condition \( {\left( 6\right) }^{\prime } \) is evident upon multiplying (5) on the left by \( \xi \) . To show the sufficiency we consider the set \( K \) of all points \( y \) of form (5). Clearly, this is a convex set; we claim it is closed. To see this we first note that any vector \( y \) which ma... | Yes |
Theorem 2. Given an \( n \times m \) matrix \( \mathrm{Y} \) and a column vector \( y \) with \( n \) components, the inequality\n\n\[ y \geq \mathrm{Y}p,\;p \geq 0 \]\n\n(13)\n\ncan be satisfied iff every \( \xi \) that satisfies\n\n\[ \xi \mathrm{Y} \geq 0,\;\xi \geq 0 \]\n\n(14)\n\nalso satisfies\n\n\[ {\xi y} \geq ... | Proof. To prove necessity, multiply (13) by \( \xi \) on the left and use (14) to deduce (15). Conversely by definition of \( \geq 0 \) for vectors,(13) means that there is a column vector \( z \) with \( n \) components such that\n\n\[ y = \mathrm{Y}p + z,\;z \geq 0,\;p \geq 0. \]\n\n\( {\left( {13}\right) }^{\prime }... | Yes |
Theorem 3 (Duality Theorem). Let \( \mathrm{Y} \) be a given \( n \times m \) matrix, \( y \) a given column vector with \( n \) components, and \( \gamma \) a given row vector with \( m \) components.\n\nWe define two quantities, \( S \) and \( s \), as follows:\n\nDefinition\n\n\[ S = \mathop{\sup }\limits_{p}{\gamma... | Proof. Let \( p \) and \( \xi \) be admissible vectors. Multiply (17) by \( \xi \) on the left,(19) by \( p \) on the right. Using Exercise 2 we conclude that\n\n\[ {\xi y} \geq \xi \mathrm{Y}p \geq {\gamma p}. \]\n\nThis shows that any \( {\gamma p} \) is bounded from above by every \( {\xi y} \) ; therefore\n\n\[ s \... | No |
Theorem 2. In a finite-dimensional linear space, all norms are equivalent; that is, any two satisfy (14) with some \( c \), depending on the pair of norms. | Proof. Any finite-dimensional linear space \( X \) over \( \mathbb{R} \) is isomorphic to \( {\mathbb{R}}^{n} \) , \( n = \dim X \) ; so we may take \( X \) to be \( {\mathbb{R}}^{n} \) . In Chapter 7 we introduced the Euclidean norm:\n\n\[ \parallel x\parallel = {\left( \mathop{\sum }\limits_{1}^{n}{a}_{j}^{2}\right) ... | Yes |
Theorem 3. (i) In a finite-dimensional normed linear space \( X \), every Cauchy sequence converges to a limit. | Proof. (i) Introduce a Euclidean structure in \( X \) . According to Theorem 2, the Euclidean norm and the norm in \( X \) are equivalent. Therefore a Cauchy sequence in the norm of \( X \) is also a Cauchy sequence in the Euclidean norm. According to Theorem 16 in Chapter 7, a Cauchy sequence in a finite-dimensional E... | Yes |
Theorem 4. Let \( X \) be a normed linear space that is locally compact-that is, in which every bounded sequence has a convergent subsequence. Then \( X \) is finite-dimensional. | Proof. We need the following result. | No |
Lemma 5. Let \( Y \) be a finite-dimensional subspace of a normed linear space \( X \) . Let \( x \) be a vector in \( X \) that does not belong to \( Y \) . Then\n\n\[ d = \mathop{\inf }\limits_{{y\text{ in }Y}}\left| {x - y}\right| \]\n\nis positive. | Proof. Suppose not; then there would be a sequence of vectors \( \left\{ {y}_{n}\right\} \) in \( Y \) such that\n\n\[ \lim \left| {x - {y}_{n}}\right| = 0 \]\n\nIn words, \( {y}_{n} \) tends to \( x \) . It follows that \( \left\{ {y}_{n}\right\} \) is a Cauchy sequence; according to part (i) of Theorem \( 3,{y}_{n} \... | Yes |
Theorem 7. \( X \) is a finite-dimensional normed linear space.\n\n(i) Given a linear function \( l \) defined on \( X \), there is an \( x \) in \( X, x \neq 0 \), for which equality holds in (24).\n\n(ii) Given a vector \( x \) in \( X \), there is a linear function \( l \) defined on \( X, l \neq 0 \), for which equ... | Proof. (i) We shall show that the supremum definition (23) of \( {\left| l\right| }^{\prime } \) is a maximum. We note that the ratio \( \left| {l\left( x\right) }\right| /\left| x\right| \) doesn’t change if we replace \( x \) by any multiple of \( x \) . Therefore it suffices to take the supremum (23) over the unit s... | Yes |
Theorem 8. The norm induced in \( {X}^{\prime \prime } \) by the induced norm in \( {X}^{\prime } \) is the same as the original norm in \( X \) . | Proof. The norm of a linear function of on \( {X}^{\prime } \) is, according to formula (23), \[ {\left| f\right| }^{\prime \prime } = \mathop{\sup }\limits_{{l \neq 0}}\frac{\left| f\left( l\right) \right| }{{\left| l\right| }^{\prime }} \] (27) The linear functions \( f \) on \( {X}^{\prime } \) are of the form (26);... | Yes |
Theorem 9. Let \( Z \) be a subspace of \( X, y \) any vector in \( X \) . The distance \( d\left( {y, Z}\right) \) of \( y \) to \( Z \) is defined to be\n\n\[ d\left( {y, z}\right) = \mathop{\inf }\limits_{{z\text{ in }Z}}\left| {y - z}\right| \]\n\nThen\n\n\[ d\left( {y, Z}\right) = \max l\left( y\right) \]\n\nover ... | Proof. By definition of distance, for any \( \epsilon > 0 \) there is a \( {z}_{0} \) in \( Z \) such that\n\n\[ \left| {y - {z}_{0}}\right| < d\left( {y, Z}\right) + \epsilon \]\n\nFor any \( l \) satisfying (32) we get, using (33) that\n\n\[ l\left( y\right) = l\left( y\right) - l\left( {z}_{0}\right) = l\left( {y - ... | Yes |
Theorem 10. Definition (38) is a norm, that is, has all three properties (1). | Proof. Every member \( x \) of a given congruence class \{\} can be described as \( x = {x}_{0} - z,{x}_{0} \) some vector in \( \{ \}, z \) any vector in \( Z \) . We claim that property (i), positivity, holds: for \( \{ \} \neq 0 \) ,\n\n\[ \left| {\{ \} }\right| > 0\text{.} \]\n\n\( {\left( {38}\right) }^{\prime } \... | Yes |
Theorem 11. \( X \) is a linear space over \( \mathbb{C} \), and \( p \) is a real-valued function defined on \( X \) with the following properties:\n\n(i) \( p \) is absolute homogeneous; that is, it satisfies\n\n\[ p\left( {ax}\right) = \left| a\right| p\left( x\right) \]\n\nfor all complex numbers \( a \) and all \(... | Proof. The complex linear space \( X \) can also be regarded as a linear space over \( \mathbb{R} \) . Any linear function on complex \( X \) can be split into its real and imaginary part:\n\n\[ l\left( u\right) = {l}_{1}\left( u\right) + i{l}_{2}\left( u\right) \]\n\nwhere \( {l}_{1} \) and \( {l}_{2} \) are real-valu... | Yes |
Theorem 12. This identity characterizes Euclidean space. That is, if in a real normed linear space \( X \)\n\n\[{\left| u + v\right| }^{2} + {\left| u - v\right| }^{2} = 2{\left| u\right| }^{2} + 2{\left| v\right| }^{2}\]\n\nfor all pairs of vectors \( u, v \), then the norm \( \left| \right| \) is Euclidean. | Proof. We define a scalar product in \( X \) as follows:\n\n\[4\left( {x, y}\right) = {\left| x + y\right| }^{2} - {\left| x - y\right| }^{2}.\](45)\n\nThe following properties of a scalar product follow immediately from definition (45):\n\n\[\left( {x, x}\right) = {\left| x\right| }^{2}\](46)\n\nSymmetry:\n\n\[\left( ... | Yes |
For any linear map \( T : X \rightarrow Y \), there is a constant \( c \) such that for all \( x \) in \( X \) , \[ \left| {\mathrm{T}x}\right| \leq c\left| x\right| \] | Proof. Express \( x \) with respect to a basis \( \left\{ {x}_{j}\right\} \) : \[ x = \sum {a}_{j}{x}_{j} \] then \[ \mathrm{T}x = \sum {a}_{j}\mathrm{T}{x}_{j}. \] By properties of the norm in \( Y \) , \[ \left| {\mathrm{T}x}\right| \leq \sum \left| {a}_{j}\right| \left| {\mathrm{T}{x}_{j}}\right| . \] From this we d... | Yes |
Theorem 2. |T| as defined in (4) and \( {\left( 4\right) }^{\prime } \) is a norm in the linear space of all linear mappings of \( X \) into \( Y \) . | Proof. Suppose \( \mathrm{T} \) is nonzero; that means that for some vector \( {x}_{0} \neq 0,\mathrm{\;T}{x}_{0} \neq 0 \) . Then by (4),\n\n\[ \left| \mathrm{T}\right| \geq \frac{\left| \mathrm{T}{x}_{0}\right| }{\left| {x}_{0}\right| } \]\n\nsince the norms in \( X \) and \( Y \) are positive, the positivity of \( \... | Yes |
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