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For example, if we reconsider the previous example, \( {A}^{\top }{u}^{\top } = 0 \) becomes\n\n\[ \left( \begin{matrix} 1 & 1 & 1 & 1 \\ 1 & 1 & - 1 & - 1 \end{matrix}\right) \left( \begin{array}{l} {u}_{1} \\ {u}_{2} \\ {u}_{3} \\ {u}_{4} \end{array}\right) = \left( \begin{array}{l} 0 \\ 0 \end{array}\right) \] | Since the rref of \( {A}^{\top } \) is\n\n\[ \left( \begin{array}{llll} 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \end{array}\right) \]\nthe above system is equivalent to\n\n\[ \left( \begin{array}{llll} 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \end{array}\right) \left( \begin{array}{l} {u}_{1} \\ {u}_{2} \\ {u}_{3} \\ {u}_{4} \end{array}\... | Yes |
Let us now consider the problem of finding a basis of the hyperplane \( H \) in \( {\mathbb{R}}^{n} \) defined by the equation\n\n\[ \n{c}_{1}{x}_{1} + \cdots + {c}_{n}{x}_{n} = 0.\n\]\n\nMore precisely, if \( {u}^{ * }\left( {{x}_{1},\ldots ,{x}_{n}}\right) \) is the linear form in \( {\left( {\mathbb{R}}^{n}\right) }... | Since \( {u}^{ * } \) is not the linear form which is identically zero, there is a smallest positive index \( j \leq n \) such that \( {c}_{j} \neq 0 \), so our linear form is really \( {u}^{ * }\left( {{x}_{1},\ldots ,{x}_{n}}\right) = {c}_{j}{x}_{j} + \cdots + {c}_{n}{x}_{n} \) . We claim that the following \( n - 1 ... | Yes |
Conversely, given a hyperplane \( H \) in \( {\mathbb{R}}^{n} \) given as the span of \( n - 1 \) linearly vectors \( \left( {{u}_{1},\ldots ,{u}_{n - 1}}\right) \), it is possible using determinants to find a linear form \( \left( {{\lambda }_{1},\ldots ,{\lambda }_{n}}\right) \) that vanishes on \( H \). | In the case \( n = 3 \), we are looking for a row vector \( \left( {{\lambda }_{1},{\lambda }_{2},{\lambda }_{3}}\right) \) such that if\n\n\[ u = \left( \begin{array}{l} {u}_{1} \\ {u}_{2} \\ {u}_{3} \end{array}\right) \;\text{ and }\;v = \left( \begin{array}{l} {v}_{1} \\ {v}_{2} \\ {v}_{3} \end{array}\right) \]\n\na... | No |
Let \( E = {\mathrm{M}}_{n}\left( \mathbb{R}\right) \), and consider the equations asserting that the sum of the entries in every row of a matrix \( A \in {\mathrm{M}}_{n}\left( \mathbb{R}\right) \) is equal to the same number. We have \( n - 1 \) equations \[ \mathop{\sum }\limits_{{j = 1}}^{n}\left( {{a}_{ij} - {a}_{... | It is not so obvious to find a basis for this space. | No |
Proposition 11.5. Let \( E \) be a vector space. The following properties hold:\n\n(a) The linear map \( {\operatorname{eval}}_{E} : E \rightarrow {E}^{* * } \) defined such that\n\n\[ \n{\operatorname{eval}}_{E}\left( v\right) = {\operatorname{eval}}_{v}\;\text{ for all }v \in E, \]\n\nthat is, \( {\operatorname{eval}... | Proof. (a) Let \( {\left( {u}_{i}\right) }_{i \in I} \) be a basis of \( E \), and let \( v = \mathop{\sum }\limits_{{i \in I}}{v}_{i}{u}_{i} \) . If \( {\operatorname{eval}}_{E}\left( v\right) = 0 \), then in particular \( {\operatorname{eval}}_{E}\left( v\right) \left( {u}_{i}^{ * }\right) = 0 \) for all \( {u}_{i}^{... | Yes |
Proposition 11.8. If \( f : E \rightarrow F \) is any linear map, then the following properties hold:\n\n(1) If \( f \) is injective, then \( {f}^{\top } \) is surjective.\n\n(2) If \( f \) is surjective, then \( {f}^{\top } \) is injective. | Proof. If \( f : E \rightarrow F \) is injective, then it has a retraction \( r : F \rightarrow E \) such that \( r \circ f = {\operatorname{id}}_{E} \) , and if \( f : E \rightarrow F \) is surjective, then it has a section \( s : F \rightarrow E \) such that \( f \circ s = {\operatorname{id}}_{F} \) . Now if \( f : E... | Yes |
Proposition 11.9. For any subspace \( U \) of a vector space \( E \), if \( p : E \rightarrow E/U \) is the canonical surjection onto \( E/U \), then \( {p}^{\top } \) is injective and\n\n\[ \operatorname{Im}\left( {p}^{\top }\right) = {U}^{0} = {\left( \operatorname{Ker}\left( p\right) \right) }^{0}. \]\n\nTherefore, ... | Proof. Since \( p \) is surjective, by Proposition 11.8, the map \( {p}^{\top } \) is injective. Obviously, \( U = \) \( \operatorname{Ker}\left( p\right) \) . Observe that \( \operatorname{Im}\left( {p}^{\top }\right) \) consists of all linear forms \( \psi \in {E}^{ * } \) such that \( \psi = \varphi \circ p \) for s... | Yes |
Given a linear map \( f : E \rightarrow F \), for any subspace \( V \) of \( E \), we have\n\n\[ f{\left( V\right) }^{0} = {\left( {f}^{\top }\right) }^{-1}\left( {V}^{0}\right) = \left\{ {{w}^{ * } \in {F}^{ * } \mid {f}^{\top }\left( {w}^{ * }\right) \in {V}^{0}}\right\} . \] | Proof. We have\n\n\[ \left\langle {{w}^{ * }, f\left( v\right) }\right\rangle = \left\langle {{f}^{\top }\left( {w}^{ * }\right), v}\right\rangle \]\n\nfor all \( v \in E \) and all \( {w}^{ * } \in {F}^{ * } \), and thus, we have \( \left\langle {{w}^{ * }, f\left( v\right) }\right\rangle = 0 \) for every \( v \in V \... | Yes |
Given a linear map \( f : E \rightarrow F \), the following properties hold.\n\n(a) The dual \( {\left( \operatorname{Im}f\right) }^{ * } \) of \( \operatorname{Im}f \) is isomorphic to \( \operatorname{Im}{f}^{\top } = {f}^{\top }\left( {F}^{ * }\right) \) ; that is,\n\n\[{\left( \operatorname{Im}f\right) }^{ * } \con... | Proof. (a) Consider the linear maps\n\n\[E\overset{p}{ \rightarrow }\operatorname{Im}f\overset{j}{ \rightarrow }F\]\n\nwhere \( E\overset{p}{ \rightarrow }\operatorname{Im}f \) is the surjective map induced by \( E\overset{f}{ \rightarrow }F \), and \( \operatorname{Im}f\overset{j}{ \rightarrow }F \) is the injective i... | Yes |
Proposition 11.13. If \( f : E \rightarrow F \) is any linear map, then the following identities hold:\n\n\[ \operatorname{Im}{f}^{\top } = {\left( \operatorname{Ker}\left( f\right) \right) }^{0} \]\n\n\[ \operatorname{Ker}\left( {f}^{\top }\right) = {\left( \operatorname{Im}f\right) }^{0} \]\n\n\[ \operatorname{Im}f =... | Proof. The equation \( \operatorname{Ker}\left( {f}^{\top }\right) = {\left( \operatorname{Im}f\right) }^{0} \) has already been proven in Proposition 11.11.\n\nBy the duality theorem \( {\left( \operatorname{Ker}\left( f\right) \right) }^{00} = \operatorname{Ker}\left( f\right) \), so from \( \operatorname{Im}{f}^{\to... | Yes |
Proposition 11.14. Let \( E \) and \( F \) be two vector spaces, and let \( \left( {{u}_{1},\ldots ,{u}_{n}}\right) \) be a basis for \( E \) and \( \left( {{v}_{1},\ldots ,{v}_{m}}\right) \) be a basis for \( F \) . Given any linear map \( f : E \rightarrow F \), if \( M\left( f\right) \) is the \( m \times n \) - mat... | Proof. Recall that the entry \( {a}_{ij} \) in row \( i \) and column \( j \) of \( M\left( f\right) \) is the \( i \) -th coordinate of \( f\left( {u}_{j}\right) \) over the basis \( \left( {{v}_{1},\ldots ,{v}_{m}}\right) \) . By definition of \( {v}_{i}^{ * } \), we have \( \left\langle {{v}_{i}^{ * }, f\left( {u}_{... | Yes |
Proposition 11.15. Given an \( m \times n \) matrix \( A \) over a field \( K \), we have \( \operatorname{rk}\left( A\right) = \operatorname{rk}\left( {A}^{\top }\right) \) . | Proof. The matrix \( A \) corresponds to a linear map \( f : {K}^{n} \rightarrow {K}^{m} \), and by Theorem 11.12, \( \operatorname{rk}\left( f\right) = \operatorname{rk}\left( {f}^{\top }\right) \) . By Proposition 11.14, the linear map \( {f}^{\top } \) corresponds to \( {A}^{\top } \) . Since \( \operatorname{rk}\le... | Yes |
Proposition 11.16. Given any \( m \times n \) matrix \( A \) over a field \( K \) (typically \( K = \mathbb{R} \) or \( K = \mathbb{C} \) ), the rank of \( A \) is the maximum natural number \( r \) such that there is an invertible \( r \times r \) submatrix of \( A \) obtained by selecting \( r \) rows and \( r \) col... | For example, the \( 3 \times 2 \) matrix\n\n\[ A = \left( \begin{array}{ll} {a}_{11} & {a}_{12} \\ {a}_{21} & {a}_{22} \\ {a}_{31} & {a}_{32} \end{array}\right) \]\n\nhas rank 2 iff one of the three \( 2 \times 2 \) matrices\n\n\[ \left( \begin{array}{ll} {a}_{11} & {a}_{12} \\ {a}_{21} & {a}_{22} \end{array}\right) \;... | "No" |
Proposition 11.18. For any linear map \( f : E \rightarrow F \), we have\n\n\[ \n{f}^{\top \top } \circ {\operatorname{eval}}_{E} = {\operatorname{eval}}_{F} \circ f \n\]\n\nor equivalently the following diagram commutes:\n\n and every \( \varphi \in {F}^{ * } \), we have\n\n\[ \n\left( {{f}^{\top \top } \circ {\operatorname{eval}}_{E}}\right) \left( u\right) \left( \varphi \right) = \left\langle {{f}^{\top \top }\left( {{\operatorname{eval}}_{E}\left( u\right) }\right) ,\varphi }\right\rangle \n\]\n\n\[ \n= \... | Yes |
Proposition 11.19. If \( \dim \left( E\right) \) is finite, then we have\n\n\[ \operatorname{Ker}\left( {f}^{\top \top }\right) = {\operatorname{eval}}_{E}\left( {\operatorname{Ker}\left( f\right) }\right) \] | Proof. Indeed, if \( E \) is finite-dimensional, the map \( {\operatorname{eval}}_{E} : E \rightarrow {E}^{* * } \) is an isomorphism, so every \( \varphi \in {E}^{* * } \) is of the form \( \varphi = {\operatorname{eval}}_{E}\left( u\right) \) for some \( u \in E \), the map \( {\operatorname{eval}}_{F} : F \rightarro... | Yes |
Proposition 11.20. If \( \dim \left( E\right) \) is finite, then for any linear map \( f : E \rightarrow F \), we have\n\n\[ \operatorname{Im}{f}^{\top } = {\left( \operatorname{Ker}\left( f\right) \right) }^{0}. \]\n | Proof. From\n\n\[ \left\langle {{f}^{\top }\left( \varphi \right), u}\right\rangle = \langle \varphi, f\left( u\right) \rangle \]\n\nfor all \( \varphi \in {F}^{ * } \) and all \( u \in E \), we see that if \( u \in \operatorname{Ker}\left( f\right) \), then \( \left\langle {{f}^{\top }\left( \varphi \right), u}\right\... | Yes |
Problem 11.1. Prove the following properties of transposition: | \[ {\left( f + g\right) }^{\top } = {f}^{\top } + {g}^{\top } \] \[ {\left( g \circ f\right) }^{\top } = {f}^{\top } \circ {g}^{\top } \] \[ {\mathrm{{id}}}_{E}^{\top } = {\mathrm{{id}}}_{{E}^{ * }} \] | Yes |
Let \( \varphi : {\mathbb{R}}^{n} \times {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) be the map defined by\n\n\[ \varphi \left( {\left( {{x}_{1},\ldots ,{x}_{n}}\right) ,\left( {{y}_{1},\ldots ,{y}_{n}}\right) }\right) = {x}_{1}{y}_{1} + \cdots + {x}_{n}{y}_{n}. \]\n\nProve that \( \varphi \) is a bilinear nondegenerate... | Prove that \( \varphi \left( {x, x}\right) = 0 \) iff \( x = 0 \) . | No |
Let \( {U}_{1},\ldots ,{U}_{p} \) be some subspaces of a vector space \( E \), and assume that they form a direct sum \( U = {U}_{1} \oplus \cdots \oplus {U}_{p} \). Let \( {j}_{i} : {U}_{i} \rightarrow {U}_{1} \oplus \cdots \oplus {U}_{p} \) be the canonical injections, and let \( {\pi }_{i} : {U}_{1}^{ * } \times \cd... | \[ {\pi }_{i} \circ f = {j}_{i}^{\top },\;1 \leq i \leq p. \] | Yes |
Proposition 12.1. We have\n\n\[ \varphi \left( {u, v}\right) = \frac{1}{2}\left\lbrack {\Phi \left( {u + v}\right) - \Phi \left( u\right) - \Phi \left( v\right) }\right\rbrack \]\n\nfor all \( u, v \in E \) . We say that \( \varphi \) is the polar form of \( \Phi \) . | Proof. By bilinearity and symmetry, we have\n\n\[ \Phi \left( {u + v}\right) = \varphi \left( {u + v, u + v}\right) \]\n\n\[ = \varphi \left( {u, u + v}\right) + \varphi \left( {v, u + v}\right) \]\n\n\[ = \varphi \left( {u, u}\right) + {2\varphi }\left( {u, v}\right) + \varphi \left( {v, v}\right) \]\n\n\[ = \Phi \lef... | Yes |
Going back to Example 12.3 and to the inner product\n\n\\[ \n\\langle f, g\\rangle = {\\int }_{-\\pi }^{\\pi }f\\left( t\\right) g\\left( t\\right) {dt} \n\\]\n\non the vector space \\( \\mathcal{C}\\left\\lbrack {-\\pi ,\\pi }\\right\\rbrack \\), it is easily checked that\n\n\\[ \n\\langle \\sin {px},\\sin {qx}\\rangl... | As a consequence, the family \\( {\\left( \\sin px\\right) }_{p \\geq 1} \\cup {\\left( \\cos qx\\right) }_{q \\geq 0} \\) is orthogonal. It is not orthonormal, but becomes so if we divide every trigonometric function by \\( \\sqrt{\\pi } \\), and 1 by \\( \\sqrt{2\\pi } \\) . | Yes |
Proposition 12.4. Given a Euclidean space \( E \), for any family \( {\left( {u}_{i}\right) }_{i \in I} \) of nonnull vectors in \( E \), if \( {\left( {u}_{i}\right) }_{i \in I} \) is orthogonal, then it is linearly independent. | Proof. Assume there is a linear dependence\n\n\[ \mathop{\sum }\limits_{{j \in J}}{\lambda }_{j}{u}_{j} = 0 \]\n\nfor some \( {\lambda }_{j} \in \mathbb{R} \) and some finite subset \( J \) of \( I \) . By taking the inner product with \( {u}_{i} \) for any \( i \in J \), and using the the bilinearity of the inner prod... | Yes |
Theorem 12.6. Given a Euclidean space \( E \), the map \( b : E \rightarrow {E}^{ * } \) defined such that\n\n\[ b\left( u\right) = {\varphi }_{u} \]\n\nis linear and injective. When \( E \) is also of finite dimension, the map \( b : E \rightarrow {E}^{ * } \) is a canonical isomorphism. | Proof. That \( \flat : E \rightarrow {E}^{ * } \) is a linear map follows immediately from the fact that the inner product is bilinear. If \( {\varphi }_{u} = {\varphi }_{v} \), then \( {\varphi }_{u}\left( w\right) = {\varphi }_{v}\left( w\right) \) for all \( w \in E \), which by definition of \( {\varphi }_{u} \) me... | Yes |
Consider \( {\mathbb{R}}^{n} \) with its usual Euclidean inner product. Given any differentiable function \( f : U \rightarrow \mathbb{R} \), where \( U \) is some open subset of \( {\mathbb{R}}^{n} \), by definition, for any \( x \in U \) , the total derivative \( d{f}_{x} \) of \( f \) at \( x \) is the linear form d... | \[ d{f}_{x}\left( u\right) = \left( \begin{array}{lll} \frac{\partial f}{\partial {x}_{1}}\left( x\right) & \cdots & \frac{\partial f}{\partial {x}_{n}}\left( x\right) \end{array}\right) \left( \begin{matrix} {u}_{1} \\ \vdots \\ {u}_{n} \end{matrix}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}\frac{\partial f}{\partia... | Yes |
Given any two vectors \( u, v \in {\mathbb{R}}^{3} \), let \( c\left( {u, v}\right) \) be the linear form given by\n\n\[ c\left( {u, v}\right) \left( w\right) = \det \left( {u, v, w}\right) \;\text{ for all }w \in {\mathbb{R}}^{3}. \] | Since\n\n\[ \det \left( {u, v, w}\right) = \left| \begin{array}{lll} {u}_{1} & {v}_{1} & {w}_{1} \\ {u}_{2} & {v}_{2} & {w}_{2} \\ {u}_{3} & {v}_{3} & {w}_{3} \end{array}\right| = {w}_{1}\left| \begin{array}{ll} {u}_{2} & {v}_{2} \\ {u}_{3} & {v}_{3} \end{array}\right| - {w}_{2}\left| \begin{array}{ll} {u}_{1} & {v}_{1... | Yes |
Consider the vector space \( {\mathrm{M}}_{n}\left( \mathbb{R}\right) \) of real \( n \times n \) matrices with the inner product\n\n\[ \langle A, B\rangle = \operatorname{tr}\left( {{A}^{\top }B}\right) \]\n\nLet \( s : {\mathrm{M}}_{n}\left( \mathbb{R}\right) \rightarrow \mathbb{R} \) be the function given by\n\n\[ s... | is the matrix \( Z = \operatorname{\mathbf{o} \mathbf{n} \mathbf{e} \mathbf{s} }\left( {n, n}\right) \) whose entries are all equal to 1 . | Yes |
Given a Euclidean space \( E \) of finite dimension, for every linear map \( f : E \rightarrow E \), there is a unique linear map \( {f}^{ * } : E \rightarrow E \) such that\n\n\[ \n{f}^{ * }\left( u\right) \cdot v = u \cdot f\left( v\right) ,\;\text{ for all }u, v \in E.\n\] | Proof. Given \( {u}_{1},{u}_{2} \in E \), since the inner product is bilinear, we have\n\n\[ \n\left( {{u}_{1} + {u}_{2}}\right) \cdot f\left( v\right) = {u}_{1} \cdot f\left( v\right) + {u}_{2} \cdot f\left( v\right)\n\]\n\nfor all \( v \in E \), and\n\n\[ \n\left( {{f}^{ * }\left( {u}_{1}\right) + {f}^{ * }\left( {u}... | Yes |
For a specific example of this procedure, let \( E = {\mathbb{R}}^{3} \) with the standard Euclidean norm. Take the basis\n\n\[ \n{e}_{1} = \left( \begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right) \;{e}_{2} = \left( \begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right) \;{e}_{3} = \left( \begin{array}{l} 1 \\ 1 \\ 0 \end{array... | Then\n\n\[ \n{u}_{1} = 1/\sqrt{3}\left( \begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right)\n\]\n\nand\n\n\[ \n{u}_{2}^{\prime } = {e}_{2} - \left( {{e}_{2} \cdot {u}_{1}}\right) {u}_{1} = \left( \begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right) - 2/3\left( \begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right) = 1/3\left( \begin{m... | Yes |
Let us apply the modified Gram-Schmidt method to the \( \left( {{e}_{1},{e}_{2},{e}_{3}}\right) \) basis of Example 12.9. The only change is the computation of \( {u}_{3}^{\prime } \) . | For the modified Gram-Schmidt procedure, we first calculate\n\n\[ \n{u}_{1}^{3} = {e}_{3} - \left( {{e}_{3} \cdot {u}_{1}}\right) {u}_{1} = \left( \begin{array}{l} 1 \\ 1 \\ 0 \end{array}\right) - 2/3\left( \begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right) = 1/3\left( \begin{matrix} 1 \\ 1 \\ - 2 \end{matrix}\right) .\n\... | Yes |
If we consider polynomials and the inner product\n\n\\[ \n\\langle f, g\\rangle = {\\int }_{-1}^{1}f\\left( t\\right) g\\left( t\\right) {dt} \n\\]\n\napplying the Gram-Schmidt orthonormalization procedure to the polynomials\n\n\\[ \n1, x,{x}^{2},\\ldots ,{x}^{n},\\ldots ,\n\\]\n\nwhich form a basis of the polynomials ... | The Legendre polynomials \\( {P}_{n}\\left( x\\right) \\) have many nice properties. They are orthogonal, but their norm is not always 1 . The Legendre polynomials \\( {P}_{n}\\left( x\\right) \\) can be defined as follows. Letting \\( {f}_{n} \\) be the function\n\n\\[ \n{f}_{n}\\left( x\\right) = {\\left( {x}^{2} - 1... | Yes |
Consider polynomials over \( \left\lbrack {-1,1}\right\rbrack \), with the symmetric bilinear form\n\n\[ \langle f, g\rangle = {\int }_{-1}^{1}\frac{1}{\sqrt{1 - {t}^{2}}}f\left( t\right) g\left( t\right) {dt}. \] | We leave it as an exercise to prove that the above defines an inner product. | No |
Proposition 12.11. Given any nontrivial Euclidean space \( E \) of finite dimension \( n \geq 1 \), for any subspace \( F \) of dimension \( k \), the orthogonal complement \( {F}^{ \bot } \) of \( F \) has dimension \( n - k \) , and \( E = F \oplus {F}^{ \bot } \) . Furthermore, we have \( {F}^{ \bot \bot } = F \) . | Proof. From Proposition 12.9, the subspace \( F \) has some orthonormal basis \( \left( {{u}_{1},\ldots ,{u}_{k}}\right) \) . This linearly independent family \( \left( {{u}_{1},\ldots ,{u}_{k}}\right) \) can be extended to a basis \( \left( {{u}_{1},\ldots ,{u}_{k},{v}_{k + 1},\ldots ,{v}_{n}}\right) \) , and by Propo... | Yes |
Proposition 12.14. Let \( E \) be any Euclidean space of finite dimension \( n \), and let \( f : E \rightarrow E \) be any linear map. The following properties hold:\n\n(1) The linear map \( f : E \rightarrow E \) is an isometry iff\n\n\[ f \circ {f}^{ * } = {f}^{ * } \circ f = \mathrm{{id}}. \]\n\n(2) For every ortho... | Proof. (1) The linear map \( f : E \rightarrow E \) is an isometry iff\n\n\[ f\left( u\right) \cdot f\left( v\right) = u \cdot v \]\n\nfor all \( u, v \in E \), iff\n\n\[ {f}^{ * }\left( {f\left( u\right) }\right) \cdot v = f\left( u\right) \cdot f\left( v\right) = u \cdot v \]\n\nfor all \( u, v \in E \), which implie... | Yes |
Proposition 12.15. The exponential map \( \exp : \mathfrak{{so}}\left( 3\right) \rightarrow \mathbf{{SO}}\left( 3\right) \) is given by\n\n\[ {e}^{A} = \cos \theta {I}_{3} + \frac{\sin \theta }{\theta }A + \frac{\left( 1 - \cos \theta \right) }{{\theta }^{2}}B, \]\n\n or, equivalently, by\n\n\[ {e}^{A} = {I}_{3} + \fra... | Proof sketch. First observe that\n\n\[ {A}^{2} = - {\theta }^{2}{I}_{3} + B \]\n\n since\n\n\[ {A}^{2} = \left( \begin{matrix} 0 & - c & b \\ c & 0 & - a \\ - b & a & 0 \end{matrix}\right) \left( \begin{matrix} 0 & - c & b \\ c & 0 & - a \\ - b & a & 0 \end{matrix}\right) = \left( \begin{matrix} - {c}^{2} - {b}^{2} & {... | Yes |
Given any real \( n \times n \) matrix \( A \), if \( A \) is invertible, then there is an orthogonal matrix \( Q \) and an upper triangular matrix \( R \) with positive diagonal entries such that \( A = {QR} \) . | We can view the columns of \( A \) as vectors \( {A}^{1},\ldots ,{A}^{n} \) in \( {\mathbb{E}}^{n} \) . If \( A \) is invertible, then they are linearly independent, and we can apply Proposition 12.10 to produce an orthonormal basis using the Gram-Schmidt orthonormalization procedure. Recall that we construct vectors \... | Yes |
Consider the matrix\n\n\[ A = \left( \begin{array}{lll} 0 & 0 & 5 \\ 0 & 4 & 1 \\ 1 & 1 & 1 \end{array}\right) \]\n\nTo determine the \( {QR} \)-decomposition of \( A \), we first use the Gram-Schmidt orthonormalization procedure to calculate \( Q = \left( {{Q}^{1}{Q}^{2}{Q}^{3}}\right) \) . | By definition\n\n\[ {A}^{1} = {Q}^{\prime 1} = {Q}^{1} = \left( \begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right) \]\n\nand since \( {A}^{2} = \left( \begin{array}{l} 0 \\ 4 \\ 1 \end{array}\right) \), we discover that\n\n\[ {Q}^{\prime 2} = {A}^{2} - \left( {{A}^{2} \cdot {Q}^{1}}\right) {Q}^{1} = \left( \begin{array}{l... | Yes |
Another example of \( {QR} \) -decomposition is\n\n\[ A = \left( \begin{array}{lll} 1 & 1 & 2 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array}\right) \] | \[ A = \left( \begin{array}{lll} 1 & 1 & 2 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array}\right) = \left( \begin{matrix} 1/\sqrt{2} & 1/\sqrt{2} & 0 \\ 0 & 0 & 1 \\ 1/\sqrt{2} & - 1/\sqrt{2} & 0 \end{matrix}\right) \left( \begin{matrix} \sqrt{2} & 1/\sqrt{2} & \sqrt{2} \\ 0 & 1/\sqrt{2} & \sqrt{2} \\ 0 & 0 & 1 \end{matrix}\righ... | Yes |
If we apply the above Matlab function to the matrix\n\n\[ A = \left( \begin{array}{lllll} 4 & 1 & 0 & 0 & 0 \\ 1 & 4 & 1 & 0 & 0 \\ 0 & 1 & 4 & 1 & 0 \\ 0 & 0 & 1 & 4 & 1 \\ 0 & 0 & 0 & 1 & 4 \end{array}\right) \] | we obtain\n\n\[ Q = \left( \begin{matrix} {0.9701} & - {0.2339} & {0.0619} & - {0.0166} & {0.0046} \\ {0.2425} & {0.9354} & - {0.2477} & {0.0663} & - {0.0184} \\ 0 & {0.2650} & {0.9291} & - {0.2486} & {0.0691} \\ 0 & 0 & {0.2677} & {0.9283} & - {0.2581} \\ 0 & 0 & 0 & {0.2679} & {0.9634} \end{matrix}\right) \]\n\nand\n... | Yes |
Proposition 12.17. (Hadamard) For any real \( n \times n \) matrix \( A = \left( {a}_{ij}\right) \), we have\n\n\[ \n\left| {\det \left( A\right) }\right| \leq \mathop{\prod }\limits_{{i = 1}}^{n}{\left( \mathop{\sum }\limits_{{j = 1}}^{n}{a}_{ij}^{2}\right) }^{1/2}\;\text{ and }\;\left| {\det \left( A\right) }\right| ... | Proof. If \( \det \left( A\right) = 0 \), then the inequality is trivial. In addition, if the righthand side is also 0, then either some column or some row is zero. If \( \det \left( A\right) \neq 0 \), then we can factor \( A \) as \( A = {QR} \), with \( Q \) is orthogonal and \( R = \left( {r}_{ij}\right) \) upper t... | Yes |
Proposition 12.18. (Hadamard) For any real \( n \times n \) matrix \( A = \left( {a}_{ij}\right) \), if \( A \) is symmetric positive semidefinite, then we have\n\n\[ \det \left( A\right) \leq \mathop{\prod }\limits_{{i = 1}}^{n}{a}_{ii} \]\n\nMoreover, if \( A \) is positive definite, then equality holds iff \( A \) i... | Proof. If \( \det \left( A\right) = 0 \), the inequality is trivial. Otherwise, \( A \) is positive definite, and by Theorem 8.10 (the Cholesky Factorization), there is a unique upper triangular matrix \( B \) with positive diagonal entries such that\n\n\[ A = {B}^{\top }B\text{.} \]\n\nThus, \( \det \left( A\right) = ... | Yes |
Problem 12.5. Let \( E \) and \( F \) be two finite Euclidean spaces, let \( \left( {{u}_{1},\ldots ,{u}_{n}}\right) \) be a basis of \( E \), and let \( \left( {{v}_{1},\ldots ,{v}_{m}}\right) \) be a basis of \( F \) . For any linear map \( f : E \rightarrow F \), if \( A \) is the matrix of \( f \) w.r.t. the basis ... | \[ B = {G}_{1}^{-1}{A}^{\top }{G}_{2} \] | Yes |
Problem 12.9. Given \( p \) vectors \( \left( {{u}_{1},\ldots ,{u}_{p}}\right) \) in a Euclidean space \( E \) of dimension \( n \geq p \) , the Gram determinant (or Gramian) of the vectors \( \left( {{u}_{1},\ldots ,{u}_{p}}\right) \) is the determinant\n\n\[ \operatorname{Gram}\left( {{u}_{1},\ldots ,{u}_{p}}\right) ... | Hint. If \( \left( {{e}_{1},\ldots ,{e}_{n}}\right) \) is an orthonormal basis and \( A \) is the matrix of the vectors \( \left( {{u}_{1},\ldots ,{u}_{n}}\right) \) over this basis,\n\n\[ \det {\left( A\right) }^{2} = \det \left( {{A}^{\top }A}\right) = \det \left( {{A}^{i} \cdot {A}^{j}}\right) \]\n\nwhere \( {A}^{i}... | No |
For any two vectors \( x \) and \( y \), if \( X \) and \( Y \) denote the column vectors of coordinates of \( x \) and \( y \) w.r.t. the basis \( \left( {{e}_{1},\ldots ,{e}_{n}}\right) \), prove that | \[\varphi \left( {x, y}\right) = {X}^{\top }{AY}\] | No |
Prove that if \( \varphi \left( {x, x}\right) = 0 \) for all \( x \in E \), then \( \varphi \) is identically null on \( E \). | Otherwise, we can assume that there is some vector \( x \in E \) such that \( \varphi \left( {x, x}\right) \neq 0 \). | No |
Check that \( {K}^{\left( I\right) } \) is a vector space. | \[ {\left( {\lambda }_{i}\right) }_{i \in I} + {\left( {\mu }_{i}\right) }_{i \in I} = {\left( {\lambda }_{i} + {\mu }_{i}\right) }_{i \in I} \] and \[ \alpha \cdot {\left( {\mu }_{i}\right) }_{i \in I} = {\left( \alpha {\mu }_{i}\right) }_{i \in I} \] | No |
Proposition 13.2. Let \( E \) be any nontrivial Euclidean space. For any two vectors \( u, v \in E \) , if \( \parallel u\parallel = \parallel v\parallel \), then there is a hyperplane \( H \) such that the reflection \( s \) about \( H \) maps \( u \) to \( v \) , and if \( u \neq v \), then this reflection is unique.... | Proof. If \( u = v \), then any hyperplane containing \( u \) does the job. Otherwise, we must have \( H = \{ v - u{\} }^{ \bot } \), and by the above formula,\n\n\[ s\left( u\right) = u - 2\frac{\left( u \cdot \left( v - u\right) \right) }{\parallel \left( {v - u}\right) {\parallel }^{2}}\left( {v - u}\right) = u + \f... | Yes |
For every real \( n \times n \) matrix \( A \), there is a sequence \( {H}_{1},\ldots ,{H}_{n} \) of matrices, where each \( {H}_{i} \) is either a Householder matrix or the identity, and an upper triangular matrix \( R \) such that\n\n\[ R = {H}_{n}\cdots {H}_{2}{H}_{1}A \] | The \( j \) th column of \( A \) can be viewed as a vector \( {v}_{j} \) over the canonical basis \( \left( {{e}_{1},\ldots ,{e}_{n}}\right) \) of \( {\mathbb{E}}^{n} \) (where \( {\left( {e}_{j}\right) }_{i} = 1 \) if \( i = j \), and 0 otherwise, \( 1 \leq i, j \leq n \) ). Applying Proposition 13.3 to \( \left( {{v}... | Yes |
Consider the matrix\n\n\[ A = \left( \begin{array}{llll} 1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 5 \\ 3 & 4 & 5 & 6 \\ 4 & 5 & 6 & 7 \end{array}\right) \] | Running the function buildhouseQR, we get\n\n\[ Q = \left( \begin{matrix} {0.1826} & {0.8165} & {0.4001} & {0.3741} \\ {0.3651} & {0.4082} & - {0.2546} & - {0.7970} \\ {0.5477} & - {0.0000} & - {0.6910} & {0.4717} \\ {0.7303} & - {0.4082} & {0.5455} & - {0.0488} \end{matrix}\right) \]\n\nand\n\n\[ R = \left( \begin{mat... | No |
Problem 13.6. The purpose of this problem is to prove that given any self-adjoint linear map \( f : E \rightarrow E \) (i.e., such that \( {f}^{ * } = f \) ), where \( E \) is a Euclidean space of dimension \( n \geq 3 \) , given an orthonormal basis \( \left( {{e}_{1},\ldots ,{e}_{n}}\right) \), there are \( n - 2 \) ... | (1) Prove that for any isometry \( f : E \rightarrow E \), we have \( f = {f}^{ * } = {f}^{-1} \) iff \( f \circ f = \mathrm{{id}} \) .\n\nProve that if \( f \) and \( h \) are self-adjoint linear maps \( \left( {{f}^{ * } = f}\right. \) and \( \left. {{h}^{ * } = h}\right) \), then \( h \circ f \circ h \) is a self-ad... | Yes |
Proposition 14.1. Given a complex vector space \( E \), the following properties hold:\n\n(1) A sesquilinear form \( \varphi : E \times E \rightarrow \mathbb{C} \) is a Hermitian form iff \( \varphi \left( {u, u}\right) \in \mathbb{R} \) for all \( u \in E \) . | Proof. (1) If \( \varphi \) is a Hermitian form, then\n\n\[ \varphi \left( {v, u}\right) = \overline{\varphi \left( {u, v}\right) } \]\n\nimplies that\n\n\[ \varphi \left( {u, u}\right) = \overline{\varphi \left( {u, u}\right) } \]\n\nand thus \( \varphi \left( {u, u}\right) \in \mathbb{R} \) . If \( \varphi \) is sesq... | Yes |
Let \( {\ell }^{2} \) denote the set of all countably infinite sequences \( x = {\left( {x}_{i}\right) }_{i \in \mathbb{N}} \) of complex numbers such that \( \mathop{\sum }\limits_{{i = 0}}^{\infty }{\left| {x}_{i}\right| }^{2} \) is defined (i.e., the sequence \( \mathop{\sum }\limits_{{i = 0}}^{n}{\left| {x}_{i}\rig... | Actually, \( {\ell }^{2} \) is even a Hilbert space. | Yes |
Example 14.3. Let \( {\mathcal{C}}_{\text{piece }}\left\lbrack {a, b}\right\rbrack \) be the set of bounded piecewise continuous functions \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{C} \) under the Hermitian form \[ \langle f, g\rangle = {\int }_{a}^{b}f\left( x\right) \overline{g\left( x\right) }{dx} ... | It is easy to check that this Hermitian form is positive, but it is not definite. Thus, under this Hermitian form, \( {\mathcal{C}}_{\text{piece }}\left\lbrack {a, b}\right\rbrack \) is only a pre-Hilbert space. | Yes |
Let \( \mathcal{C}\left\lbrack {a, b}\right\rbrack \) be the set of complex-valued continuous functions \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{C} \) under the Hermitian form\n\n\[ \langle f, g\rangle = {\int }_{a}^{b}f\left( x\right) \overline{g\left( x\right) }{dx} \] | It is easy to check that this Hermitian form is positive definite. Thus, \( \mathcal{C}\left\lbrack {a, b}\right\rbrack \) is a Hermitian space. | No |
Proposition 14.3. Given any Hermitian space \( E \) with Hermitian product \( \langle - , - \rangle \), for any linear map \( f : E \rightarrow E \), if \( \langle f\left( x\right), x\rangle = 0 \) for all \( x \in E \), then \( f = 0 \) . | Proof. Compute \( \langle f\left( {x + y}\right), x + y\rangle \) and \( \langle f\left( {x - y}\right), x - y\rangle \) :\n\n\[ \langle f\left( {x + y}\right), x + y\rangle = \langle f\left( x\right), x\rangle + \langle f\left( x\right), y\rangle + \langle f\left( y\right), x\rangle + \langle y, y\rangle \]\n\n\[ \lan... | Yes |
Proposition 14.4. Let \( \langle E,\varphi \rangle \) be a pre-Hilbert space with associated quadratic form \( \Phi \) . For all \( u, v \in E \), we have the Cauchy-Schwarz inequality\n\n\[ \left| {\varphi \left( {u, v}\right) }\right| \leq \sqrt{\Phi \left( u\right) }\sqrt{\Phi \left( v\right) }.\] | Proof. For all \( u, v \in E \) and all \( \mu \in \mathbb{C} \), we have observed that\n\n\[ \varphi \left( {u + {\mu v}, u + {\mu v}}\right) = \varphi \left( {u, u}\right) + 2\Re \left( {\bar{\mu }\varphi \left( {u, v}\right) }\right) + {\left| \mu \right| }^{2}\varphi \left( {v, v}\right) . \]\n\nLet \( \varphi \lef... | Yes |
Proposition 14.5. The equations \( {\varphi }_{u}^{l} = {\varphi }_{u}^{r} \) and \( {b}^{l} = {b}^{r} \) hold. | Proof. Indeed, for all \( u, v \in E \), we have\n\n\[ \n{b}^{l}\left( u\right) \left( v\right) = {\varphi }_{u}^{l}\left( v\right) \n\]\n\n\[ \n= \overline{u \cdot v} \n\]\n\n\[ \n= v \cdot u \n\]\n\n\[ \n= {\varphi }_{u}^{r}\left( v\right) \n\]\n\n\[ \n= {b}^{r}\left( u\right) \left( v\right) \text{.} \n\]\n\nTherefo... | Yes |
Theorem 14.6. Let \( E \) be a Hermitian space \( E \) . The map \( b : E \rightarrow {E}^{ * } \) defined such that\n\n\[ b\left( u\right) = {\varphi }_{u}^{l} = {\varphi }_{u}^{r}\;\text{ for all }u \in E \]\n\nis semilinear and injective. When \( E \) is also of finite dimension, the map \( b : \bar{E} \rightarrow {... | Proof. That \( \mathfrak{b} : E \rightarrow {E}^{ * } \) is a semilinear map follows immediately from the fact that \( \mathfrak{b} = {\mathfrak{b}}^{r} \) , and that the Hermitian product is semilinear in its second argument. If \( {\varphi }_{u} = {\varphi }_{v} \), then \( {\varphi }_{u}\left( w\right) = {\varphi }_... | Yes |
Given a Hermitian space \( E \) of finite dimension, for every linear map \( f : E \rightarrow E \) there is a unique linear map \( {f}^{ * } : E \rightarrow E \) such that\n\n\[ \n{f}^{ * }\left( u\right) \cdot v = u \cdot f\left( v\right) ,\;\text{ for all }u, v \in E.\n\] | Proof. Careful inspection of the proof of Proposition 12.8 reveals that it applies unchanged. The only potential problem is in proving that \( {f}^{ * }\left( {\lambda u}\right) = \lambda {f}^{ * }\left( u\right) \), but everything takes place in the first argument of the Hermitian product, and there, we have linearity... | No |
Proposition 14.10. Given any finite-dimensional Hermitian space \( E \) with Hermitian product \( \langle - , - \rangle \), for any linear map \( f : E \rightarrow E \), if \( \langle f\left( x\right), x\rangle \in \mathbb{R} \) for all \( x \in E \), then \( f \) is self-adjoint. In particular, any positive semidefini... | Proof. Since \( \langle f\left( x\right), x\rangle \in \mathbb{R} \) for all \( x \in E \), we have\n\n\[ \langle f\left( x\right), x\rangle = \overline{\langle f\left( x\right), x\rangle } \]\n\n\[ = \langle x, f\left( x\right) \rangle \]\n\n\[ = \left\langle {{f}^{ * }\left( x\right), x}\right\rangle \]\n\nso we have... | Yes |
Proposition 14.14. Given any two nontrivial Hermitian spaces \( E \) and \( F \) of the same finite dimension \( n \), for every function \( f : E \rightarrow F \), the following properties are equivalent:\n\n(1) \( f \) is a linear map and \( \parallel f\left( u\right) \parallel = \parallel u\parallel \), for all \( u... | Proof. The proof that (2) implies (3) given in Proposition 12.12 needs to be revised as follows. We use the polarization identity\n\n\[ \n{2\varphi }\left( {u, v}\right) = \left( {1 + i}\right) \left( {\parallel u{\parallel }^{2} + \parallel v{\parallel }^{2}}\right) - \parallel u - v{\parallel }^{2} - i\parallel u - {... | No |
Proposition 14.15. Let \( E \) be any Hermitian space of finite dimension \( n \), and let \( f : E \rightarrow E \) be any linear map. The following properties hold:\n\n(1) The linear map \( f : E \rightarrow E \) is an isometry iff\n\n\[ f \circ {f}^{ * } = {f}^{ * } \circ f = \mathrm{{id}}. \]\n\n(2) For every ortho... | Proof. (1) The proof is identical to that of Proposition 12.14 (1).\n\n(2) If \( \left( {{e}_{1},\ldots ,{e}_{n}}\right) \) is an orthonormal basis for \( E \), let \( A = \left( {a}_{ij}\right) \) be the matrix of \( f \), and let \( B = \left( {b}_{ij}\right) \) be the matrix of \( {f}^{ * } \) . Since \( {f}^{ * } \... | Yes |
Proposition 14.16. Given any \( n \times n \) complex matrix \( A \), if \( A \) is invertible, then there is a unitary matrix \( U \) and an upper triangular matrix \( R \) with positive diagonal entries such that \( A = {UR} \) . | The proof is absolutely the same as in the real case! | No |
Proposition 14.17. (Hadamard) For any complex \( n \times n \) matrix \( A = \left( {a}_{ij}\right) \), we have\n\n\[ \left| {\det \left( A\right) }\right| \leq \mathop{\prod }\limits_{{i = 1}}^{n}{\left( \mathop{\sum }\limits_{{j = 1}}^{n}{\left| {a}_{ij}\right| }^{2}\right) }^{1/2}\;\text{ and }\;\left| {\det \left( ... | Moreover, equality holds iff either A has a zero row in the left inequality or a zero column in the right inequality, or \( A \) is unitary. | Yes |
Proposition 14.19. Let \( E \) be any nontrivial Hermitian space.\n\n(1) For any two vectors \( u, v \in E \) such that \( u \neq v \) and \( \parallel u\parallel = \parallel v\parallel \), if \( u \cdot v = {e}^{i\theta }\left| {u \cdot v}\right| \), then the (usual) reflection \( s \) about the hyperplane orthogonal ... | Proof. (1) Consider the (usual) reflection about the hyperplane orthogonal to \( w = v - {e}^{-{i\theta }}u \) .\n\nWe have\n\[ s\left( u\right) = u - 2\frac{\left( u \cdot \left( v - {e}^{-{i\theta }}u\right) \right) }{{\begin{Vmatrix}v - {e}^{-{i\theta }}u\end{Vmatrix}}^{2}}\left( {v - {e}^{-{i\theta }}u}\right) .\n\... | Yes |
Let \( E \) be a nontrivial Hermitian space of dimension \( n \) . Given any orthonormal basis \( \left( {{e}_{1},\ldots ,{e}_{n}}\right) \), for any \( n \) -tuple of vectors \( \left( {{v}_{1},\ldots ,{v}_{n}}\right) \), there is a sequence of \( n - 1 \) isometries \( {h}_{1},\ldots ,{h}_{n - 1} \), such that \( {h}... | The proof is very similar to the proof of Proposition 13.3, but it needs to be modified a little bit since Proposition 14.19 is weaker than Proposition 13.2. We explain how to modify the induction step, leaving the base case and the rest of the proof as an exercise.\n\nAs in the proof of Proposition 13.3, the vectors \... | No |
For every complex \( n \times n \) -matrix \( A \), there is a sequence \( {H}_{1},\ldots ,{H}_{n - 1} \) of matrices, where each \( {H}_{i} \) is either a Householder matrix or the identity, and an upper triangular matrix \( R \), such that\n\n\[ R = {H}_{n - 1}\cdots {H}_{2}{H}_{1}A. \] | Proof. It is essentially identical to the proof of Proposition 13.4, and we leave the details as an exercise. For the last statement, observe that \( {h}_{n} \circ \cdots \circ {h}_{1} \) is also an isometry. | No |
Proposition 14.22. For any linear map \( f : E \rightarrow E \), we have \( {f}^{2} = \operatorname{id} \) iff \( \frac{1}{2}\left( {\mathrm{{id}} - f}\right) \) is a projection iff \( \frac{1}{2}\left( {\mathrm{{id}} + f}\right) \) is a projection; in this case, \( f \) is equal to the difference of the two projection... | Proof. We have\n\n\[ \n{\left( \frac{1}{2}\left( \mathrm{{id}} - f\right) \right) }^{2} = \frac{1}{4}\left( {\mathrm{{id}} - {2f} + {f}^{2}}\right) \n\]\n\nso\n\[ \n{\left( \frac{1}{2}\left( \mathrm{{id}} - f\right) \right) }^{2} = \frac{1}{2}\left( {\mathrm{{id}} - f}\right) \;\text{ iff }\;{f}^{2} = \mathrm{{id}}. \n... | Yes |
Proposition 14.23. For any linear map \( f : E \rightarrow E \), let \( {U}^{ + } = \operatorname{Ker}\left( {\frac{1}{2}\left( {\mathrm{{id}} - f}\right) }\right) \) and let \( {U}^{ - } = \operatorname{Im}\left( {\frac{1}{2}\left( {\mathrm{{id}} - f}\right) }\right) \) . If \( {f}^{2} = \mathrm{{id}} \), then\n\n\[ \... | Proof. If \( {f}^{2} = \mathrm{{id}} \), then\n\n\[ \n\left( {\mathrm{{id}} + f}\right) \circ \left( {\mathrm{{id}} - f}\right) = \mathrm{{id}} - {f}^{2} = \mathrm{{id}} - \mathrm{{id}} = 0,\n\]\n\nwhich implies that\n\n\[ \n\operatorname{Im}\left( {\frac{1}{2}\left( {\mathrm{{id}} + f}\right) }\right) \subseteq \opera... | Yes |
Proposition 14.24. Let \( f \in \mathbf{{GL}}\left( E\right) \) be an involution. The following properties are equivalent:\n\n(a) The map \( f \) is unitary; that is, \( f \in \mathbf{U}\left( E\right) \).\n\n(b) The subspaces \( {U}^{ - } = \operatorname{Im}\left( {\frac{1}{2}\left( {\mathrm{{id}} - f}\right) }\right)... | Proof. If \( f \) is unitary, then from \( \langle f\left( u\right), f\left( v\right) \rangle = \langle u, v\rangle \) for all \( u, v \in E \), we see that if \( u \in {U}^{ + } \) and \( v \in {U}^{ - } \), we get\n\n\[ \langle u, v\rangle = \langle f\left( u\right), f\left( v\right) \rangle = \langle u, - v\rangle =... | Yes |
Proposition 14.25. If \( f : E \rightarrow E \) is a projection \( \left( {{f}^{2} = f}\right) \), then \( \operatorname{Ker}\left( f\right) \) and \( \operatorname{Im}\left( f\right) \) are orthogonal iff \( {f}^{ * } = f \) . | Proof. Apply Proposition 14.24 to \( g = \mathrm{{id}} - {2f} \) . Since \( \mathrm{{id}} - g = {2f} \) we have\n\n\[
{U}^{ + } = \operatorname{Ker}\left( {\frac{1}{2}\left( {\mathrm{{id}} - g}\right) }\right) = \operatorname{Ker}\left( f\right)
\]\n\nand\n\n\[
{U}^{ - } = \operatorname{Im}\left( {\frac{1}{2}\left( {\m... | No |
Proposition 14.26. For any function \( p : E \rightarrow \mathbb{R} \), if we define \( {p}^{D} \) by\n\n\[ \n{p}^{D}\left( x\right) = \mathop{\sup }\limits_{{p\left( z\right) = 1}}\left| {\langle z, x\rangle }\right| \n\]\n\nthen we have\n\n\[ \n{p}^{D}\left( {x + y}\right) \leq {p}^{D}\left( x\right) + {p}^{D}\left( ... | Proof. We have\n\n\[ \n{p}^{D}\left( {x + y}\right) = \mathop{\sup }\limits_{{p\left( z\right) = 1}}\left| {\langle z, x + y\rangle }\right| \n\]\n\n\[ \n= \mathop{\sup }\limits_{{p\left( z\right) = 1}}\left( \left| {\langle z, x\rangle +\langle z, y\rangle }\right| \right) \n\]\n\n\[ \n\leq \mathop{\sup }\limits_{{p\l... | Yes |
Proposition 14.28. For all \( y \in E \), we have\n\n\[ \parallel y{\parallel }^{D} = \mathop{\sup }\limits_{\substack{{x \in E} \\ {\parallel x\parallel = 1} }}\left| {\langle x, y\rangle }\right| = \mathop{\sup }\limits_{\substack{{x \in E} \\ {\parallel x\parallel = 1} }}\Re \langle x, y\rangle . \] | Proof. Since \( E \) is finite dimensional, the unit sphere \( {S}^{n - 1} = \{ x \in E \mid \parallel x\parallel = 1\} \) is compact, so there is some \( {x}_{0} \in {S}^{n - 1} \) such that\n\n\[ \parallel y{\parallel }^{D} = \left| \left\langle {{x}_{0}, y}\right\rangle \right| \]\n\nIf \( \left\langle {{x}_{0}, y}\... | Yes |
Proposition 14.29. For all \( x, y \in E \), we have\n\n\[ \left| {\langle x, y\rangle }\right| \leq \parallel x\parallel \parallel y{\parallel }^{D} \]\n\n\[ \left| {\langle x, y\rangle }\right| \leq \parallel x{\parallel }^{D}\parallel y\parallel \] | Proof. If \( x = 0 \), then \( \langle x, y\rangle = 0 \) and these inequalities are trivial. If \( x \neq 0 \), since \( \parallel x/\parallel x\parallel \parallel = 1 \) , by definition of \( \parallel y{\parallel }^{D} \), we have\n\n\[ \left| {\langle x/\parallel x\parallel, y\rangle }\right| \leq \mathop{\sup }\li... | Yes |
Proposition 14.30. If \( p, q \geq 1 \) and \( 1/p + 1/q = 1 \), then for all \( y \in {\mathbb{C}}^{n} \), we have\n\n\[ \parallel y{\parallel }_{p}^{D} = \parallel y{\parallel }_{q} \] | Proof. By Hölder’s inequality (Corollary 9.2), for all \( x, y \in {\mathbb{C}}^{n} \), we have\n\n\[ \left| {\langle x, y\rangle }\right| \leq \parallel x{\parallel }_{p}\parallel y{\parallel }_{q} \]\n\nso\n\n\[ \parallel y{\parallel }_{p}^{D} = \mathop{\sup }\limits_{\substack{{x \in {\mathbb{C}}^{n}} \\ {\parallel ... | Yes |
Proposition 14.31. The dual of the spectral norm is given by\n\n\[ \parallel A{\parallel }_{2}^{D} = {\sigma }_{1} + \cdots + {\sigma }_{r} \]\n\nwhere \( {\sigma }_{1} > \cdots > {\sigma }_{r} > 0 \) are the singular values of \( A \in {\mathrm{M}}_{n}\left( \mathbb{C}\right) \) (which has rank \( r \) ). | Proof. In this case the inner product on \( {\mathrm{M}}_{n}\left( \mathbb{C}\right) \) is the Frobenius inner product \( \langle A, B\rangle = \) \( \operatorname{tr}\left( {{B}^{ * }A}\right) \), and the dual norm of the spectral norm is given by\n\n\[ \parallel A{\parallel }_{2}^{D} = \sup \left\{ {\left| {\operator... | Yes |
Theorem 14.32. If \( E \) is a finite-dimensional Hermitian space, then for any norm \( \parallel \parallel \) on \( E \), we have\n\n\[ \parallel y{\parallel }^{DD} = \parallel y\parallel \]\n\nfor all \( y \in E \) . | Proof. By Proposition 14.29, we have\n\n\[ \left| {\langle x, y\rangle }\right| \leq \parallel x{\parallel }^{D}\parallel y\parallel \]\n\nso we get\n\n\[ \parallel y{\parallel }^{DD} = \mathop{\sup }\limits_{{\parallel x{\parallel }^{D} = 1}}\left| {\langle x, y\rangle }\right| \leq \parallel y\parallel ,\;\text{ for ... | No |
Problem 14.10. Prove that all \( y \in {\mathbb{C}}^{n} \) , | \[ \parallel y{\parallel }_{1}^{D} = \parallel y{\parallel }_{\infty } \] \[ \parallel y{\parallel }_{\infty }^{D} = \parallel y{\parallel }_{1} \] \[ \parallel y{\parallel }_{2}^{D} = \parallel y{\parallel }_{2} \] | No |
Prove that the squares \( {\sigma }_{1}^{2} \) and \( {\sigma }_{2}^{2} \) of the singular values of\n\n\[ A = \left( \begin{array}{ll} 1 & 2 \\ c & d \end{array}\right) \]\n\nare the zeros of the equation\n\n\[ {\lambda }^{2} - \left( {5 + {c}^{2} + {d}^{2}}\right) \lambda + {\left( 2c - d\right) }^{2} = 0. \] | Using the fact that\n\n\[ \parallel A{\parallel }_{N} = {\sigma }_{1} + {\sigma }_{2} = \sqrt{{\sigma }_{1}^{2} + {\sigma }_{2}^{2} + 2{\sigma }_{1}{\sigma }_{2}}, \]\n\nprove that\n\n\[ \parallel A{\parallel }_{N}^{2} = 5 + {c}^{2} + {d}^{2} + 2\left| {{2c} - d}\right| \]\n\nConsider the cases where \( {2c} - d \geq 0... | No |
Proposition 15.1. Let \( E \) be any vector space of finite dimension \( n \) and let \( f \) be any linear map \( f : E \rightarrow E \). The eigenvalues of \( f \) are the roots (in \( K \)) of the polynomial\n\n\[ \det \left( {\lambda \mathrm{{id}} - f}\right) \] | Proof. A scalar \( \lambda \in K \) is an eigenvalue of \( f \) iff there is some vector \( u \neq 0 \) in \( E \) such that\n\n\[ f\left( u\right) = {\lambda u} \]\n\niff\n\n\[ \left( {\lambda \mathrm{{id}} - f}\right) \left( u\right) = 0 \]\n\niff \( \left( {\lambda \mathrm{{id}} - f}\right) \) is not invertible iff,... | Yes |
Proposition 15.2. Let \( A \) be an \( n \times n \) matrix over a field \( K \) and assume that all the roots of the characteristic polynomial \( {\chi }_{A}\left( X\right) = \det \left( {{XI} - A}\right) \) of \( A \) belong to \( K \) . For every eigenvalue \( {\lambda }_{i} \) of \( A \), the geometric multiplicity... | Proof. To see this, if \( {n}_{i} \) is the dimension of the eigenspace \( {E}_{{\lambda }_{i}} \) associated with the eigenvalue \( {\lambda }_{i} \), we can form a basis of \( {K}^{n} \) obtained by picking a basis of \( {E}_{{\lambda }_{i}} \) and completing this linearly independent family to a basis of \( {K}^{n} ... | Yes |
Proposition 15.3. Let \( E \) be any vector space of finite dimension \( n \) and let \( f \) be any linear map. If \( {u}_{1},\ldots ,{u}_{m} \) are eigenvectors associated with pairwise distinct eigenvalues \( {\lambda }_{1},\ldots ,{\lambda }_{m} \) , then the family \( \left( {{u}_{1},\ldots ,{u}_{m}}\right) \) is ... | Proof. Assume that \( \left( {{u}_{1},\ldots ,{u}_{m}}\right) \) is linearly dependent. Then there exists \( {\mu }_{1},\ldots ,{\mu }_{k} \in K \) such that\n\n\[{\mu }_{1}{u}_{{i}_{1}} + \cdots + {\mu }_{k}{u}_{{i}_{k}} = 0\]\n\nwhere \( 1 \leq k \leq m,{\mu }_{i} \neq 0 \) for all \( i,1 \leq i \leq k,\left\{ {{i}_{... | Yes |
Corollary 15.4. If \( {\lambda }_{1},\ldots ,{\lambda }_{m} \) are all the pairwise distinct eigenvalues of \( f \) (where \( m \leq n \) ), we have a direct sum\n\n\[ \n{E}_{{\lambda }_{1}} \oplus \cdots \oplus {E}_{{\lambda }_{m}} \n\]\n\nof the eigenspaces \( {E}_{{\lambda }_{i}} \) . | Unfortunately, it is not always the case that\n\n\[ \nE = {E}_{{\lambda }_{1}} \oplus \cdots \oplus {E}_{{\lambda }_{m}} \n\]\n\nDefinition 15.4. When\n\n\[ \nE = {E}_{{\lambda }_{1}} \oplus \cdots \oplus {E}_{{\lambda }_{m}} \n\]\n\nwe say that \( f \) is diagonalizable (and similarly for any matrix associated with \(... | No |
Theorem 15.6. (Schur decomposition) Given any linear map \( f : E \rightarrow E \) over a complex Hermitian space \( E \), there is an orthonormal basis \( \left( {{u}_{1},\ldots ,{u}_{n}}\right) \) with respect to which \( f \) is represented by an upper triangular matrix. Equivalently, for every \( n \times n \) matr... | Proof. During the induction, we choose \( F \) to be the orthogonal complement of \( \mathbb{C}{u}_{1} \) and we pick orthonormal bases (use Propositions 14.13 and 14.12). If \( E \) is a real Euclidean space and if the eigenvalues of \( f \) are all real, the proof also goes through with real matrices (use Proposition... | No |
Proposition 15.7. Given any \( n \times n \) matrix \( A \in {\mathrm{M}}_{n}\left( K\right) \) with coefficients in a field \( K \) , if all eigenvalues \( {\lambda }_{1},\ldots ,{\lambda }_{n} \) of \( A \) are in \( K \), then for every polynomial \( q\left( X\right) \in K\left\lbrack X\right\rbrack \), the eigenval... | Proof. By Theorem 15.5, there is an upper triangular matrix \( T \) and an invertible matrix \( P \) (both in \( {\mathrm{M}}_{n}\left( K\right) \) ) such that\n\n\[ A = {PT}{P}^{-1} \]\n\nSince \( A \) and \( T \) are similar, they have the same eigenvalues (with the same multiplicities), so the diagonal entries of \(... | Yes |
If \( A \) is a Hermitian matrix (i.e. \( {A}^{ * } = A \) ), then its eigenvalues are real and \( A \) can be diagonalized with respect to an orthonormal basis of eigenvectors. In matrix terms, there is a unitary matrix \( U \) and a real diagonal matrix \( D \) such that \( A = {UD}{U}^{ * } \) . If \( A \) is a real... | By Theorem 15.6, we can write \( A = {UT}{U}^{ * } \) where \( T = \left( {t}_{ij}\right) \) is upper triangular and \( U \) is a unitary matrix. If \( {A}^{ * } = A \), we get\n\n\[ \n{UT}{U}^{ * } = U{T}^{ * }{U}^{ * }\n\]\n\nand this implies that \( T = {T}^{ * } \) . Since \( T \) is an upper triangular matrix, \( ... | Yes |
Theorem 15.9. (Gershgorin’s disc theorem) For any complex \( n \times n \) matrix \( A \), all the eigenvalues of \( A \) belong to the Gershgorin domain \( G\left( A\right) \) . Furthermore the following properties hold:\n\n(1) If \( A \) is strictly row diagonally dominant, that is\n\n\[ \left| {a}_{ii}\right| > \mat... | Proof. Let \( \lambda \) be any eigenvalue of \( A \) and let \( u \) be a corresponding eigenvector (recall that we must have \( u \neq 0 \) ). Let \( k \) be an index such that\n\n\[ \left| {u}_{k}\right| = \mathop{\max }\limits_{{1 \leq i \leq n}}\left| {u}_{i}\right| \]\n\nSince \( {Au} = {\lambda u} \), we have\n\... | Yes |
Proposition 15.11. Let \( A \in {\mathrm{M}}_{n}\left( \mathbb{C}\right) \) be a diagonalizable matrix, \( P \) be an invertible matrix, and \( D \) be a diagonal matrix \( D = \operatorname{diag}\left( {{\lambda }_{1},\ldots ,{\lambda }_{n}}\right) \) such that\n\n\[ A = {PD}{P}^{-1} \]\n\nand let \( \parallel \parall... | Proof. Let \( \lambda \) be any eigenvalue of the matrix \( A + {\Delta A} \) . If \( \lambda = {\lambda }_{j} \) for some \( j \), then the result is trivial. Thus assume that \( \lambda \neq {\lambda }_{j} \) for \( j = 1,\ldots, n \) . In this case the matrix \( D - {\lambda I} \) is invertible (since its eigenvalue... | Yes |
Proposition 15.12. Let \( A \) and \( U \) be (real or complex) matrices and assume that \( U \) is invertible. Then\n\n\[ \n{e}^{{UA}{U}^{-1}} = U{e}^{A}{U}^{-1} \n\] | Proof. A trivial induction shows that\n\n\[ \nU{A}^{p}{U}^{-1} = {\left( UA{U}^{-1}\right) }^{p}, \n\]\n\nand thus\n\n\[ \n{e}^{{UA}{U}^{-1}} = \mathop{\sum }\limits_{{p \geq 0}}\frac{{\left( UA{U}^{-1}\right) }^{p}}{p!} = \mathop{\sum }\limits_{{p \geq 0}}\frac{U{A}^{p}{U}^{-1}}{p!} \n\]\n\n\[ \n= U\left( {\mathop{\su... | Yes |
Given any complex \( n \times n \) matrix \( A \), if \( {\lambda }_{1},\ldots ,{\lambda }_{n} \) are the eigenvalues of \( A \), then \( {e}^{{\lambda }_{1}},\ldots ,{e}^{{\lambda }_{n}} \) are the eigenvalues of \( {e}^{A} \). Furthermore, if \( u \) is an eigenvector of \( A \) for \( {\lambda }_{i} \), then \( u \)... | By Theorem 15.5, there is an invertible matrix \( P \) and an upper triangular matrix \( T \) such that\n\n\[ A = {PT}{P}^{-1}. \]\n\nBy Proposition 15.12,\n\n\[ {e}^{{PT}{P}^{-1}} = P{e}^{T}{P}^{-1}. \]\n\nNote that \( {e}^{T} = \mathop{\sum }\limits_{{p > 0}}\frac{{T}^{p}}{p!} \) is upper triangular since \( {T}^{p} ... | Yes |
Proposition 15.14. For every complex (or real) square matrix \( A \), we have\n\n\[ \det \left( {e}^{A}\right) = {e}^{\operatorname{tr}\left( A\right) } \]\n\nwhere \( \operatorname{tr}\left( A\right) \) is the trace of \( A \), i.e., the sum \( {a}_{11} + \cdots + {a}_{nn} \) of its diagonal entries. | Proof. The trace of a matrix \( A \) is equal to the sum of the eigenvalues of \( A \) . The determinant of a matrix is equal to the product of its eigenvalues, and if \( {\lambda }_{1},\ldots ,{\lambda }_{n} \) are the eigenvalues of \( A \), then by Proposition 15.13, \( {e}^{{\lambda }_{1}},\ldots ,{e}^{{\lambda }_{... | Yes |
Proposition 15.15. For every skew symmetric matrix \( B \in \mathfrak{{so}}\left( n\right) \), we have \( {e}^{B} \in \mathbf{{SO}}\left( n\right) \) , that is, \( {e}^{B} \) is a rotation. | Proof. By Proposition 9.23, \( {e}^{B} \) is an orthogonal matrix. Since \( \operatorname{tr}\left( B\right) = 0 \), we deduce that \( \det \left( {e}^{B}\right) = {e}^{0} = 1 \) . Therefore, \( {e}^{B} \in \mathbf{{SO}}\left( n\right) \) . | Yes |
Problem 15.4. Let \( A \) be the real symmetric \( 2 \times 2 \) matrix\n\n\[ A = \left( \begin{array}{ll} a & b \\ b & c \end{array}\right) \]\n\n(1) Prove that the eigenvalues of \( A \) are real and given by\n\n\[ {\lambda }_{1} = \frac{a + c + \sqrt{4{b}^{2} + {\left( a - c\right) }^{2}}}{2},\;{\lambda }_{2} = \fra... | \[ {\lambda }_{1} = \frac{a + c + \sqrt{4{b}^{2} + {\left( a - c\right) }^{2}}}{2},\;{\lambda }_{2} = \frac{a + c - \sqrt{4{b}^{2} + {\left( a - c\right) }^{2}}}{2}. \] | No |
Let \( A \) be a real symmetric \( n \times n \) matrix and \( B \) be a real symmetric positive definite \( n \times n \) matrix. We would like to solve the generalized eigenvalue problem: find \( \lambda \in \mathbb{R} \) and \( u \neq 0 \) such that\n\n\[ \n{Au} = {\lambda Bu}.\n\]\n\n\( \left( *\right) \) | (1) Use the Cholseky decomposition \( B = C{C}^{\top } \) to show that \( \lambda \) and \( u \) are solutions of the generalized eigenvalue problem (*) iff \( \lambda \) and \( v \) are solutions the (ordinary) eigenvalue problem\n\n\[ \n{C}^{-1}A{\left( {C}^{\top }\right) }^{-1}v = {\lambda v},\;\text{ with }v = {C}^... | Yes |
Consider the following tridiagonal \( n \times n \) matrices\n\n\[ A = \left( \begin{matrix} 2 & - 1 & 0 & & \\ - 1 & 2 & - 1 & & \\ & \ddots & \ddots & \ddots & \\ & & - 1 & 2 & - 1 \\ & & 0 & - 1 & 2 \end{matrix}\right) ,\;S = \left( \begin{matrix} 0 & 1 & 0 & & \\ 1 & 0 & 1 & & \\ & \ddots & \ddots & \ddots & \\ & &... | (2) Using Problem 10.6, prove that the eigenvalues of the matrix \( A \) are given by\n\n\[ {\lambda }_{k} = 4{\sin }^{2}\left( \frac{k\pi }{2\left( {n + 1}\right) }\right) ,\;k = 1,\ldots, n. \] | No |
Consider the following real tridiagonal symmetric \( n \times n \) matrix\n\n\[ A = \left( \begin{matrix} c & 1 & 0 & & \\ 1 & c & 1 & & \\ & \ddots & \ddots & \ddots & \\ & & 1 & c & 1 \\ & & 0 & 1 & c \end{matrix}\right) \]\n\n(1) Using Problem 10.6, prove that the eigenvalues of the matrix \( A \) are given by\n\n\[... | Using Problem 10.6, prove that the eigenvalues of the matrix \( A \) are given by\n\n\[ {\lambda }_{k} = c + 2\cos \left( \frac{k\pi }{n + 1}\right) ,\;k = 1,\ldots, n. \] | No |
Let \( A \) be an \( m \times n \) matrix and \( B \) be an \( n \times m \) matrix (over \( \mathbb{C} \) ).\n\n(1) Prove that\n\n\[ \det \left( {{I}_{m} - {AB}}\right) = \det \left( {{I}_{n} - {BA}}\right) . \] | Hint. Consider the matrices\n\n\[ X = \left( \begin{matrix} {I}_{m} & A \\ B & {I}_{n} \end{matrix}\right) \;\text{ and }\;Y = \left( \begin{matrix} {I}_{m} & 0 \\ - B & {I}_{n} \end{matrix}\right) . \] | No |
The purpose of this problem is to prove that the characteristic polynomial of the matrix\n\n\[ A = \left( \begin{matrix} 1 & 2 & 3 & 4 & \cdots & n \\ 2 & 3 & 4 & 5 & \cdots & n + 1 \\ 3 & 4 & 5 & 6 & \cdots & n + 2 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \\ n & n + 1 & n + 2 & n + 3 & \cdots & {2n} - 1 \end{ma... | (1) Prove that the characteristic polynomial \( {P}_{A}\left( \lambda \right) \) is given by\n\n\[ {P}_{A}\left( \lambda \right) = {\lambda }^{n - 2}P\left( \lambda \right) \]\n\nwith\n\n\[ P\left( \lambda \right) = \left| \begin{matrix} \lambda - 1 & - 2 & - 3 & - 4 & \cdots & - n + 3 & - n + 2 & - n + 1 & - n \\ - \l... | Yes |
Proposition 16.1. For every unit quaternion \( q \in \mathbf{SU}\left( 2\right) \), the linear map \( {r}_{q} \) is orthogonal, that is, \( {r}_{q} \in \mathbf{O}\left( 3\right) \) . | Proof. Since\n\n\[ \n- \parallel \left( {x, y, z}\right) {\parallel }^{2} = - \left( {{x}^{2} + {y}^{2} + {z}^{2}}\right) = \det \left( {x{\sigma }^{3} + y{\sigma }^{2} + z{\sigma }_{1}}\right) = \det \left( {\varphi \left( {x, y, z}\right) }\right) , \n\]\n\nwe have\n\n\[ \n- {\begin{Vmatrix}{r}_{q}\left( x, y, z\righ... | Yes |
Theorem 16.3. The map \( r : \mathbf{SU}\left( 2\right) \rightarrow \mathbf{SO}\left( 3\right) \) is homomorphism whose kernel is \( \{ I, - I\} \) . | Proof. This map is a homomorphism, because if \( {q}_{1},{q}_{2} \in \mathbf{SU}\left( 2\right) \), then\n\n\[ \n{r}_{{q}_{2}}\left( {{r}_{{q}_{1}}\left( {x, y, z}\right) }\right) = {\varphi }^{-1}\left( {{q}_{2}\varphi \left( {{r}_{{q}_{1}}\left( {x, y, z}\right) }\right) {q}_{2}^{ * }}\right) \n\]\n\n\[ \n= {\varphi ... | Yes |
Proposition 16.4. The matrix representing \( {r}_{q} \) is\n\n\[ \n{R}_{q} = \left( \begin{matrix} {a}^{2} + {b}^{2} - {c}^{2} - {d}^{2} & {2bc} - {2ad} & {2ac} + {2bd} \\ {2bc} + {2ad} & {a}^{2} - {b}^{2} + {c}^{2} - {d}^{2} & - {2ab} + {2cd} \\ - {2ac} + {2bd} & {2ab} + {2cd} & {a}^{2} - {b}^{2} - {c}^{2} + {d}^{2} \... | Since \( {a}^{2} + {b}^{2} + {c}^{2} + {d}^{2} = 1 \), this matrix can also be written as\n\n\[ \n{R}_{q} = \left( \begin{matrix} 2{a}^{2} + 2{b}^{2} - 1 & {2bc} - {2ad} & {2ac} + {2bd} \\ {2bc} + {2ad} & 2{a}^{2} + 2{c}^{2} - 1 & - {2ab} + {2cd} \\ - {2ac} + {2bd} & {2ab} + {2cd} & 2{a}^{2} + 2{d}^{2} - 1 \end{matrix}... | Yes |
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