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Proposition 5.2. (Sylvester,1867) If \( H \) is a Hadamard matrix of dimension \( n \), then the block matrix of dimension \( {2n} \) , | \[ \left( \begin{matrix} H & H \\ H & - H \end{matrix}\right) \] | Yes |
Show that given any vector \( c = \left( {{c}_{1},{c}_{2},{c}_{3},{c}_{4},{c}_{5},{c}_{6},{c}_{7},{c}_{8}}\right) \), the result \( {W}_{3,3}c \) of applying \( {W}_{3,3} \) to \( c \) is\n\n\[ {W}_{3,3}c = \left( {{c}_{1} + {c}_{5},{c}_{1} - {c}_{5},{c}_{2} + {c}_{6},{c}_{2} - {c}_{6},{c}_{3} + {c}_{7},{c}_{3} - {c}_{... | \[ {W}_{3,3} = \left( \begin{matrix} 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & - 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & - 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & - 1 \end{matrix}\r... | Yes |
Given any two vector spaces, \( {E}_{1} \) and \( {E}_{2} \), the set \( {E}_{1} \coprod {E}_{2} \) is a vector space. For every pair of linear maps, \( f : {E}_{1} \rightarrow G \) and \( g : {E}_{2} \rightarrow G \), there is a unique linear map, \( f + g : {E}_{1} \coprod {E}_{2} \rightarrow G \), such that \( \left... | Proof. Define\n\n\[ \left( {f + g}\right) \left( {\{ \langle 1, u\rangle ,\langle 2, v\rangle \} }\right) = f\left( u\right) + g\left( v\right) ,\]\n\nfor every \( u \in {E}_{1} \) and \( v \in {E}_{2} \) . It is immediately verified that \( f + g \) is the unique linear map with the required properties. | Yes |
Proposition 6.2. Given any two vector spaces, \( {E}_{1} \) and \( {E}_{2} \), for every pair of linear maps, \( f : D \rightarrow {E}_{1} \) and \( g : D \rightarrow {E}_{2} \), there is a unique linear map, \( f \times g : D \rightarrow {E}_{1} \coprod {E}_{2} \), such that \( {\pi }_{1} \circ \left( {f \times g}\rig... | Proof. Define\n\n\[ \left( {f \times g}\right) \left( w\right) = \{ \langle 1, f\left( w\right) \rangle ,\langle 2, g\left( w\right) \rangle \} \]\n\nfor every \( w \in D \) . It is immediately verified that \( f \times g \) is the unique linear map with the required properties. | Yes |
Proposition 6.3. If the map \( a : {U}_{1} \times \cdots \times {U}_{p} \rightarrow E \) is injective, then every \( u \in {U}_{1} + \cdots + {U}_{p} \) has a unique expression as a sum\n\n\[ u = {u}_{1} + \cdots + {u}_{p} \]\n\nwith \( {u}_{i} \in {U}_{i} \), for \( i = 1,\ldots, p \) . | Proof. If\n\n\[ u = {v}_{1} + \cdots + {v}_{p} = {w}_{1} + \cdots + {w}_{p}, \]\n\nwith \( {v}_{i},{w}_{i} \in {U}_{i} \), for \( i = 1,\ldots, p \), then we have\n\n\[ {w}_{1} - {v}_{1} + \cdots + {w}_{p} - {v}_{p} = 0 \]\n\nand since \( {v}_{i},{w}_{i} \in {U}_{i} \) and each \( {U}_{i} \) is a subspace, \( {w}_{i} -... | Yes |
Proposition 6.4. If the map \( a : {U}_{1} \times \cdots \times {U}_{p} \rightarrow E \) is injective, then any \( p \) nonzero vectors \( {u}_{1},\ldots ,{u}_{p} \) with \( {u}_{i} \in {U}_{i} \) are linearly independent. | Proof. To see this, assume that\n\n\[ \n{\lambda }_{1}{u}_{1} + \cdots + {\lambda }_{p}{u}_{p} = 0 \n\] \n\nfor some \( {\lambda }_{i} \in \mathbb{R} \) . Since \( {u}_{i} \in {U}_{i} \) and \( {U}_{i} \) is a subspace, \( {\lambda }_{i}{u}_{i} \in {U}_{i} \), and the injectivity of \( a \) implies that \( {\lambda }_{... | Yes |
Proposition 6.7. If \( E \) is any vector space, for any (finite-dimensional) subspaces \( {U}_{1},\ldots \) , \( {U}_{p} \) of \( E \), we have \[ \dim \left( {{U}_{1} \oplus \cdots \oplus {U}_{p}}\right) = \dim \left( {U}_{1}\right) + \cdots + \dim \left( {U}_{p}\right) . \] | If \( E \) is a direct sum \[ E = {U}_{1} \oplus \cdots \oplus {U}_{p} \] since every \( u \in E \) can be written in a unique way as \[ u = {u}_{1} + \cdots + {u}_{p} \] with \( {u}_{i} \in {U}_{i} \) for \( i = 1\ldots, p \), we can define the maps \( {\pi }_{i} : E \rightarrow {U}_{i} \), called projections, by \[ {... | Yes |
Let \( E \) and \( F \) be vector spaces, and let \( f : E \rightarrow F \) be a linear map. If \( f : E \rightarrow F \) is injective, then there is a surjective linear map \( r : F \rightarrow E \) called a retraction, such that \( r \circ f = {\mathrm{{id}}}_{E} \). | Let \( {\left( {u}_{i}\right) }_{i \in I} \) be a basis of \( E \) . Since \( f : E \rightarrow F \) is an injective linear map, by Proposition \( {3.15},{\left( f\left( {u}_{i}\right) \right) }_{i \in I} \) is linearly independent in \( F \) . By Theorem 3.7, there is a basis \( {\left( {v}_{j}\right) }_{j \in J} \) o... | Yes |
Proposition 6.12. Let \( E, F \) and \( G \), be three vector spaces, \( f : E \rightarrow F \) an injective linear map, \( g : F \rightarrow G \) a surjective linear map, and assume that \( \operatorname{Im}f = \operatorname{Ker}g \) . Then, the following properties hold. (a) For any section \( s : G \rightarrow F \) ... | Proof. (a) Since \( s : G \rightarrow F \) is a section of \( g \), we have \( g \circ s = {\operatorname{id}}_{G} \), and for every \( u \in F \) ,\n\n\[ g\left( {u - s\left( {g\left( u\right) }\right) }\right) = g\left( u\right) - g\left( {s\left( {g\left( u\right) }\right) }\right) = g\left( u\right) - g\left( u\rig... | Yes |
Theorem 6.13. (Rank-nullity theorem) Let \( E \) and \( F \) be vector spaces, and let \( f : E \rightarrow F \) be a linear map. Then, \( E \) is isomorphic to \( \operatorname{Ker}f \oplus \operatorname{Im}f \), and thus,\n\n\[ \dim \left( E\right) = \dim \left( {\operatorname{Ker}f}\right) + \dim \left( {\operatorna... | Proof. Consider\n\n\[ \operatorname{Ker}f\overset{i}{ \rightarrow }E\overset{{f}^{\prime }}{ \rightarrow }\operatorname{Im}f \]\n\nwhere \( \operatorname{Ker}f\overset{i}{ \rightarrow }E \) is the inclusion map, and \( E\overset{{f}^{\prime }}{ \rightarrow }\operatorname{Im}f \) is the surjection associated with \( E\o... | Yes |
Given a vector space \( E \), if \( U \) and \( V \) are any two subspaces of \( E \), then\n\n\[ \dim \left( U\right) + \dim \left( V\right) = \dim \left( {U + V}\right) + \dim \left( {U \cap V}\right) \]\n\nan equation known as Grassmann's relation. | Proof. Recall that \( U + V \) is the image of the linear map\n\n\[ a : U \times V \rightarrow E \]\n\ngiven by\n\n\[ a\left( {u, v}\right) = u + v \]\n\nand that we proved earlier that the kernel Ker \( a \) of \( a \) is isomorphic to \( U \cap V \) . By Theorem 6.13,\n\n\[ \dim \left( {U \times V}\right) = \dim \lef... | Yes |
Proposition 6.15. If \( {U}_{1},\ldots ,{U}_{p} \) are any subspaces of a finite dimensional vector space \( E \), then \[ \dim \left( {{U}_{1} + \cdots + {U}_{p}}\right) \leq \dim \left( {U}_{1}\right) + \cdots + \dim \left( {U}_{p}\right) \] and \[ \dim \left( {{U}_{1} + \cdots + {U}_{p}}\right) = \dim \left( {U}_{1}... | Proof. If we apply Theorem 6.13 to the linear map \[ a : {U}_{1} \times \cdots \times {U}_{p} \rightarrow {U}_{1} + \cdots + {U}_{p} \] given by \( a\left( {{u}_{1},\ldots ,{u}_{p}}\right) = {u}_{1} + \cdots + {u}_{p} \), we get \[ \dim \left( {{U}_{1} + \cdots + {U}_{p}}\right) = \dim \left( {{U}_{1} \times \cdots \ti... | Yes |
Proposition 6.16. Let \( E \) and \( F \) be two vector spaces with the same finite dimension \( \dim \left( E\right) = \dim \left( F\right) = n \) . For every linear map \( f : E \rightarrow F \), the following properties are equivalent:\n\n(a) \( f \) is bijective.\n\n(b) \( f \) is surjective.\n\n(c) \( f \) is inje... | Proof. Obviously, (a) implies (b).\n\nIf \( f \) is surjective, then \( \operatorname{Im}f = F \), and so \( \dim \left( {\operatorname{Im}f}\right) = n \) . By Theorem 6.13,\n\n\[ \dim \left( E\right) = \dim \left( {\operatorname{Ker}f}\right) + \dim \left( {\operatorname{Im}f}\right) \]\n\nand since \( \dim \left( E\... | Yes |
Let \( E \) be a vector space. If \( E = U \oplus V \) and \( E = U \oplus W \), then there is an isomorphism \( f : V \rightarrow W \) between \( V \) and \( W \) . | Let \( R \) be the relation between \( V \) and \( W \), defined such that\n\n\[ \langle v, w\rangle \in R\;\text{ iff }\;w - v \in U. \]\n\nWe claim that \( R \) is a functional relation that defines a linear isomorphism \( f : V \rightarrow W \) between \( V \) and \( W \), where \( f\left( v\right) = w \) iff \( \la... | Yes |
Proposition 6.18. Given a linear map \( f : E \rightarrow F \), the following properties hold:\n\n(i) \( \operatorname{rk}\left( f\right) = \operatorname{codim}\left( {\operatorname{Ker}f}\right) \) .\n\n(ii) \( \operatorname{rk}\left( f\right) + \dim \left( {\operatorname{Ker}f}\right) = \dim \left( E\right) \) .\n\n(... | Proof. Since by Proposition 6.13, \( \dim \left( E\right) = \dim \left( {\operatorname{Ker}f}\right) + \dim \left( {\operatorname{Im}f}\right) \), and by definition, \( \operatorname{rk}\left( f\right) = \dim \left( {\operatorname{Im}f}\right) \), we have \( \operatorname{rk}\left( f\right) = \operatorname{codim}\left(... | Yes |
If we let \( {Z}_{i} \subseteq {U}_{1} \times \cdots \times {U}_{p} \) be given by\n\n\[ \n{Z}_{i} = \left\{ {\left( {{u}_{1},\ldots ,{u}_{i - 1}, - \mathop{\sum }\limits_{{j = 1, j \neq i}}^{p}{u}_{j},{u}_{i + 1},\ldots ,{u}_{p}}\right) \mid \mathop{\sum }\limits_{{j = 1, j \neq i}}^{p}{u}_{j} \in {U}_{i} \cap \left( ... | In general, for any given \( i \), the condition \( {U}_{i} \cap \left( {\mathop{\sum }\limits_{{j = 1, j \neq i}}^{p}{U}_{j}}\right) = \left( 0\right) \) does not necessarily imply that \( {Z}_{i} = \left( 0\right) \) . Thus, let\n\n\[ \nZ = \left\{ {\left( {{u}_{1},\ldots ,{u}_{i - 1},{u}_{i},{u}_{i + 1},\ldots ,{u}_... | Yes |
For every \( n \geq 2 \), for every permutation \( \pi : \left\lbrack n\right\rbrack \rightarrow \left\lbrack n\right\rbrack \), there is a partition of \( \left\lbrack n\right\rbrack \) into \( r \) subsets called the orbits of \( \pi \), with \( 1 \leq r \leq n \), where each set \( J \) in this partition is either a... | Consider the relation \( {R}_{\pi } \) defined on \( \left\lbrack n\right\rbrack \) as follows: \( i{R}_{\pi }j \) iff there is some \( k \geq 1 \) such that \( j = {\pi }^{k}\left( i\right) \) . We claim that \( {R}_{\pi } \) is an equivalence relation. Transitivity is obvious. We claim that for every \( i \in \left\l... | Yes |
Proposition 7.2. For every \( n \geq 2 \), for every permutation \( \pi : \left\lbrack n\right\rbrack \rightarrow \left\lbrack n\right\rbrack \), for every transposition \( \tau \), we have\n\n\[ \epsilon \left( {\tau \circ \pi }\right) = - \epsilon \left( \pi \right) \]\n\nConsequently, for every product of transposit... | Proof. Assume that \( \tau \left( i\right) = j \) and \( \tau \left( j\right) = i \), where \( i < j \) . There are two cases, depending whether \( i \) and \( j \) are in the same equivalence class \( {J}_{l} \) of \( {R}_{\pi } \), or if they are in distinct equivalence classes. If \( i \) and \( j \) are in the same... | Yes |
Proposition 7.3. Let \( f : E \times \ldots \times E \rightarrow F \) be an \( n \) -linear alternating map, with \( n \geq 2 \) . The following properties hold:\n\n(1)\n\n\[ f\left( {\ldots ,{x}_{i},{x}_{i + 1},\ldots }\right) = - f\left( {\ldots ,{x}_{i + 1},{x}_{i},\ldots }\right) \]\n\n(2)\n\n\[ f\left( {\ldots ,{x... | Proof. (1) By multilinearity applied twice, we have\n\n\[ f\left( {\ldots ,{x}_{i} + {x}_{i + 1},{x}_{i} + {x}_{i + 1},\ldots }\right) = f\left( {\ldots ,{x}_{i},{x}_{i},\ldots }\right) + f\left( {\ldots ,{x}_{i},{x}_{i + 1},\ldots }\right) \]\n\n\[ + f\left( {\ldots ,{x}_{i + 1},{x}_{i},\ldots }\right) + f\left( {\ldo... | Yes |
Lemma 7.4. Let \( f : E \times \ldots \times E \rightarrow F \) be an \( n \) -linear alternating map. Let \( \left( {{u}_{1},\ldots ,{u}_{n}}\right) \) and \( \left( {{v}_{1},\ldots ,{v}_{n}}\right) \) be two families of \( n \) vectors, such that,\n\n\[ \n{v}_{1} = {a}_{11}{u}_{1} + \cdots + {a}_{n1}{u}_{n} \n\]\n\n\... | Proof. Expanding \( f\left( {{v}_{1},\ldots ,{v}_{n}}\right) \) by multilinearity, we get a sum of terms of the form\n\n\[ \n{a}_{\pi \left( 1\right) 1}\cdots {a}_{\pi \left( n\right) n}f\left( {{u}_{\pi \left( 1\right) },\ldots ,{u}_{\pi \left( n\right) }}\right) , \n\]\n\nfor all possible functions \( \pi : \{ 1,\ldo... | Yes |
When \( n = 2 \), if\n\n\[ A = \left( \begin{array}{ll} a & b \\ c & d \end{array}\right) \]\n\nthen by expanding according to any row, we have\n\n\[ D\left( A\right) = {ad} - {bc}. \] | \[ D\left( A\right) = {ad} - {bc}. \] | No |
Lemma 7.5. For every \( n \geq 1 \), for every \( D \in {\mathcal{D}}_{n} \) as defined in Definition 7.6, \( D \) is an alternating multilinear map such that \( D\left( {I}_{n}\right) = 1 \) . | Proof. By induction on \( n \), it is obvious that \( D\left( {I}_{n}\right) = 1 \) . Let us now prove that \( D \) is multilinear. Let us show that \( D \) is linear in each column. Consider any Column \( k \) . Since\n\n\[ D\left( A\right) = {\left( -1\right) }^{i + 1}{a}_{i1}D\left( {A}_{i1}\right) + \cdots + {\left... | Yes |
Theorem 7.6. For every \( n \geq 1 \), for every \( D \in {\mathcal{D}}_{n} \), for every matrix \( A \in {\mathrm{M}}_{n}\left( K\right) \), we have\n\n\[ D\left( A\right) = \mathop{\sum }\limits_{{\pi \in {\mathfrak{S}}_{n}}}\epsilon \left( \pi \right) {a}_{\pi \left( 1\right) 1}\cdots {a}_{\pi \left( n\right) n}, \]... | Proof. Consider the standard basis \( \left( {{e}_{1},\ldots ,{e}_{n}}\right) \) of \( {K}^{n} \), where \( {\left( {e}_{i}\right) }_{i} = 1 \) and \( {\left( {e}_{i}\right) }_{j} = 0 \), for \( j \neq i \) . Then each column \( {A}^{j} \) of \( A \) corresponds to a vector \( {v}_{j} \) whose coordinates over the basi... | Yes |
For every matrix \( A \in {\mathrm{M}}_{n}\left( K\right) \), we have \( \det \left( A\right) = \det \left( {A}^{\top }\right) \). | By Theorem 7.6, we have\n\n\[ \det \left( A\right) = \mathop{\sum }\limits_{{\pi \in {\mathfrak{S}}_{n}}}\epsilon \left( \pi \right) {a}_{\pi \left( 1\right) 1}\cdots {a}_{\pi \left( n\right) n}, \]\n\nwhere the sum ranges over all permutations \( \pi \) on \( \{ 1,\ldots, n\} \). Since a permutation is invertible, eve... | Yes |
Consider the so-called Vandermonde determinant\n\n\\[ \nV\\left( {{x}_{1},\\ldots ,{x}_{n}}\\right) = \\left| \\begin{matrix} 1 & 1 & \\ldots & 1 \\\\ {x}_{1} & {x}_{2} & \\ldots & {x}_{n} \\\\ {x}_{1}^{2} & {x}_{2}^{2} & \\ldots & {x}_{n}^{2} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ {x}_{1}^{n - 1} & {x}_{2}^{n... | We prove it by induction on \\( n \\geq 1 \\) . The case \\( n = 1 \\) is obvious. Assume \\( n \\geq 2 \\) . We proceed as follows: multiply Row \\( n - 1 \\) by \\( {x}_{1} \\) and subtract it from Row \\( n \\) (the last row), then multiply Row \\( n - 2 \\) by \\( {x}_{1} \\) and subtract it from Row \\( n - 1 \\) ... | Yes |
Proposition 7.8. Let \( f : E \times \ldots \times E \rightarrow F \) be an \( n \) -linear alternating map. Let \( \left( {{u}_{1},\ldots ,{u}_{n}}\right) \) and \( \left( {{v}_{1},\ldots ,{v}_{n}}\right) \) be two families of \( n \) vectors, such that\n\n\[ \n{v}_{1} = {a}_{11}{u}_{1} + \cdots + {a}_{1n}{u}_{n} \n\]... | Proof. The only difference with Lemma 7.4 is that here we are using \( {A}^{\top } \) instead of \( A \) . Thus, by Lemma 7.4 and Corollary 7.7, we get the desired result. | Yes |
Proposition 7.9. For any two \( n \times n \) -matrices \( A \) and \( B \), we have \( \det \left( {AB}\right) = \det \left( A\right) \det \left( B\right) \) . | Proof. We use Proposition 7.8 as follows: let \( \left( {{e}_{1},\ldots ,{e}_{n}}\right) \) be the standard basis of \( {K}^{n} \), and let\n\n\[ \left( \begin{matrix} {w}_{1} \\ {w}_{2} \\ \vdots \\ {w}_{n} \end{matrix}\right) = {AB}\left( \begin{matrix} {e}_{1} \\ {e}_{2} \\ \vdots \\ {e}_{n} \end{matrix}\right) \]\n... | Yes |
Proposition 7.11. Given an \( n \times n \) -matrix \( A \) over a field \( K \), the columns \( {A}^{1},\ldots ,{A}^{n} \) of \( A \) are linearly dependent iff \( \det \left( A\right) = \det \left( {{A}^{1},\ldots ,{A}^{n}}\right) = 0 \) . Equivalently, \( A \) has rank \( n \) iff \( \det \left( A\right) \neq 0 \) . | Proof. First assume that the columns \( {A}^{1},\ldots ,{A}^{n} \) of \( A \) are linearly dependent. Then there are \( {x}_{1},\ldots ,{x}_{n} \in K \), such that\n\n\[ \n{x}_{1}{A}^{1} + \cdots + {x}_{j}{A}^{j} + \cdots + {x}_{n}{A}^{n} = 0, \n\]\n\nwhere \( {x}_{j} \neq 0 \) for some \( j \) . If we compute\n\n\[ \n... | Yes |
Given an \( n \times n \) -matrix \( A \) over a field \( K \), the following properties hold:\n\n(1) For every column vector \( b \), there is a unique column vector \( x \) such that \( {Ax} = b \) iff the only solution to \( {Ax} = 0 \) is the trivial vector \( x = 0 \), iff \( \det \left( A\right) \neq 0 \) . | Proof. (1) Assume that \( {Ax} = b \) has a single solution \( {x}_{0} \), and assume that \( {Ay} = 0 \) with \( y \neq 0 \) . Then,\n\n\[ A\left( {{x}_{0} + y}\right) = A{x}_{0} + {Ay} = A{x}_{0} + 0 = b,\]\n\nand \( {x}_{0} + y \neq {x}_{0} \) is another solution of \( {Ax} = b \), contradicting the hypothesis that ... | Yes |
Proposition 7.13. Given any vector space \( E \) of finite dimension \( n \), a linear map \( f : E \rightarrow E \) is invertible iff \( \det \left( f\right) \neq 0 \) . | Proof. The linear map \( f : E \rightarrow E \) is invertible iff its matrix \( M\left( f\right) \) in any basis is invertible (by Proposition 4.2), iff \( \det \left( {M\left( f\right) }\right) \neq 0 \), by Proposition 7.10. | Yes |
Theorem 7.15. (Cayley-Hamilton) For every finite-dimensional vector space over a field \( K \), for every linear map \( f : E \rightarrow E \), for every basis \( \left( {{e}_{1},\ldots ,{e}_{n}}\right) \), if \( A \) is the matrix over \( f \) over the basis \( \left( {{e}_{1},\ldots ,{e}_{n}}\right) \) and if\n\n\[ \... | Proof. Since the columns of \( A \) consist of the vector \( f\left( {e}_{j}\right) \) expressed over the basis \( \left( {{e}_{1},\ldots ,{e}_{n}}\right) \), we have \n\n\[ \nf\left( {e}_{j}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{ij}{e}_{i},\;1 \leq j \leq n. \n\] \n\nUsing our action of \( K\left\lbrack X\r... | Yes |
Let \( A \) be the \( \left( {m + n}\right) \times \left( {m + n}\right) \) block matrix (over any field \( K \) ) given by\n\n\[ A = \left( \begin{matrix} {A}_{1} & {A}_{2} \\ 0 & {A}_{4} \end{matrix}\right) \]\n\nwhere \( {A}_{1} \) is an \( m \times m \) matrix, \( {A}_{2} \) is an \( m \times n \) matrix, and \( {A... | Use the above result to prove that if \( A \) is an upper triangular \( n \times n \) matrix, then \( \det \left( A\right) = \) \( {a}_{11}{a}_{22}\cdots {a}_{nn}. \) | No |
Compute the product of the following determinants\n\n\[ \left| \begin{matrix} a & - b & - c & - d \\ b & a & - d & c \\ c & d & a & - b \\ d & - c & b & a \end{matrix}\right| \left| \begin{matrix} x & - y & - z & - t \\ y & x & - t & z \\ z & t & x & - y \\ t & - z & y & x \end{matrix}\right| \] | to prove the following identity (due to Euler):\n\n\[ \left( {{a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}}\right) \left( {{x}^{2} + {y}^{2} + {z}^{2} + {t}^{2}}\right) = {\left( ax + by + cz + dt\right) }^{2} + {\left( ay - bx + ct - dz\right) }^{2} \]\n\n\[ + {\left( az - bt - cx + dy\right) }^{2} + {\left( at + bz - cy + d... | No |
Theorem 8.1. (Gaussian elimination) Let \( A \) be an \( n \times n \) matrix (invertible or not). Then there is some invertible matrix \( M \) so that \( U = {MA} \) is upper-triangular. The pivots are all nonzero iff \( A \) is invertible. | Proof. We already proved the theorem when \( A \) is invertible, as well as the last assertion. Now \( A \) is singular iff some pivot is zero, say at Stage \( k \) of the elimination. If so, we must have \( {a}_{ik}^{\left( k\right) } = 0 \) for \( i = k,\ldots, n \) ; but in this case, \( {A}_{k + 1} = {A}_{k} \) and... | No |
Corollary 8.3. (LU-Factorization) Let \( A \) be an invertible \( n \times n \) -matrix. If every matrix \( A\left( {1 : k,1 : k}\right) \) is invertible for \( k = 1,\ldots, n \), then Gaussian elimination requires no pivoting and yields an LU-factorization \( A = {LU} \) . | Proof. We proved in Proposition 8.2 that in this case Gaussian elimination requires no pivoting. Then since every elementary matrix \( {E}_{i, k;\beta } \) is lower-triangular (since we always arrange that the pivot \( {\pi }_{k} \) occurs above the rows that it operates on), since \( {E}_{i, k;\beta }^{-1} = {E}_{i, k... | Yes |
Given\n\n\[ A = {A}_{1} = \\left( \\begin{matrix} 1 & 1 & 1 & 0 \\\\ 1 & - 1 & 0 & 1 \\\\ 1 & 1 & - 1 & 0 \\\\ 1 & - 1 & 0 & - 1 \\end{matrix}\\right) \]\n\nwe have the following sequence of steps: The first pivot is \( {\\pi }_{1} = 1 \) in row 1, and we substract row 1 from rows 2, 3, and 4 . We get\n\n\[ {A}_{2} = \... | We now show how to extend the above method to deal with pivoting efficiently. This is the \( {PA} = {LU} \) factorization. | Yes |
Proposition 8.4. Let \( A \) be an invertible \( n \times n \) -matrix. There is some permutation matrix \( P \) so that \( \left( {PA}\right) \left( {1 : k,1 : k}\right) \) is invertible for \( k = 1,\ldots, n \) . | Proof. The case \( n = 1 \) is trivial, and so is the case \( n = 2 \) (we swap the rows if necessary). If \( n \geq 3 \), we proceed by induction. Since \( A \) is invertible, its columns are linearly independent; in particular, its first \( n - 1 \) columns are also linearly independent. Delete the last column of \( ... | Yes |
For every invertible \( n \times n \) -matrix \( A \), the following hold:\n\n(1) There is some permutation matrix \( P \), some upper-triangular matrix \( U \), and some unit lower-triangular matrix \( L \), so that \( {PA} = {LU} \) (recall, \( {L}_{ii} = 1 \) for \( i = 1,\ldots, n \) ). Furthermore, if \( P = I \),... | (2) If \( {E}_{n - 1}\ldots {E}_{1}A = U \) is the result of Gaussian elimination without pivoting, write as usual \( {A}_{k} = {E}_{k - 1}\ldots {E}_{1}A \) (with \( {A}_{k} = \left( {a}_{ij}^{\left( k\right) }\right) \) ), and let \( {\ell }_{ik} = {a}_{ik}^{\left( k\right) }/{a}_{kk}^{\left( k\right) } \), with \( 1... | Yes |
Proposition 8.6. If an invertible real symmetric matrix \( A \) has an LU-decomposition, then A has a factorization of the form\n\n\[ A = {LD}{L}^{\top } \]\n\nwhere \( L \) is a lower-triangular matrix whose diagonal entries are equal to 1, and where \( D \) consists of the pivots. Furthermore, such a decomposition is... | Proof. If \( A \) has an \( {LU} \) -factorization, then it has an \( {LDU} \) factorization\n\n\[ A = {LDU} \]\n\nwhere \( L \) is lower-triangular, \( U \) is upper-triangular, and the diagonal entries of both \( L \) and \( U \) are equal to 1 . Since \( A \) is symmetric, we have\n\n\[ {LDU} = A = {A}^{\top } = {U}... | Yes |
Proposition 8.7. If \( A \) is the tridiagonal matrix above, then \( {\delta }_{k} = \det \left( {A\left( {1 : k,1 : k}\right) }\right) \) for \( k = 1,\ldots, n \) . | Proof. By expanding \( \det \left( {A\left( {1 : k,1 : k}\right) }\right) \) with respect to its last row, the proposition follows by induction on \( k \) . | No |
Proposition 8.9. If \( A \) is a real symmetric positive definite matrix, then \( A\left( {1 : k,1 : k}\right) \) is symmetric positive definite and thus invertible for \( k = 1,\ldots, n \) . | Proof. Since \( A \) is symmetric, each \( A\left( {1 : k,1 : k}\right) \) is also symmetric. If \( w \in {\mathbb{R}}^{k} \), with \( 1 \leq k \leq n \), we let \( x \in {\mathbb{R}}^{n} \) be the vector with \( {x}_{i} = {w}_{i} \) for \( i = 1,\ldots, k \) and \( {x}_{i} = 0 \) for \( i = k + 1,\ldots, n \) . Now si... | Yes |
Theorem 8.10. (Cholesky factorization) Let \( A \) be a real symmetric positive definite matrix. Then there is some real lower-triangular matrix \( B \) so that \( A = B{B}^{\top } \) . Furthermore, \( B \) can be chosen so that its diagonal elements are strictly positive, in which case \( B \) is unique. | Proof. We proceed by induction on the dimension \( n \) of \( A \) . For \( n = 1 \), we must have \( {a}_{11} > 0 \) , and if we let \( \alpha = \sqrt{{a}_{11}} \) and \( B = \left( \alpha \right) \), the theorem holds trivially. If \( n \geq 2 \), as we explained above, again we must have \( {a}_{11} > 0 \), and we c... | Yes |
Proposition 8.11. Let \( A \) be any \( n \times n \) real symmetric matrix. The following conditions are equivalent:\n\n(a) \( A \) is positive definite.\n\n(b) All principal minors of \( A \) are positive; that is: \( \det \left( {A\left( {1 : k,1 : k}\right) }\right) > 0 \) for \( k = 1,\ldots, n \) (Sylvester's cri... | Proof. By Proposition 8.9, if \( A \) is symmetric positive definite, then each matrix \( A\left( {1 : k,1 : k}\right) \) is symmetric positive definite for \( k = 1,\ldots, n \) . By the Cholsesky decomposition, \( A(1 : k,1 \) : \( k) = {Q}^{\top }Q \) for some invertible matrix \( Q \), so \( \det \left( {A\left( {1... | Yes |
Proposition 8.12. Given any \( m \times n \) matrix \( A \) and any vector \( b \in {\mathbb{R}}^{m} \), for any sequence of elementary row operations \( {E}_{1},\ldots ,{E}_{k} \), if \( P = {E}_{k}\cdots {E}_{1} \) and \( \left( {{A}^{\prime },{b}^{\prime }}\right) = P\left( {A, b}\right) \), then the solutions of \(... | Proof. Since each elementary row operation \( {E}_{i} \) is invertible, so is \( P \), and since \( \left( {{A}^{\prime },{b}^{\prime }}\right) = \) \( P\left( {A, b}\right) \), then \( {A}^{\prime } = {PA} \) and \( {b}^{\prime } = {Pb} \) . If \( x \) is a solution of the original system \( {Ax} = b \), then multiply... | Yes |
Proposition 8.13. Given an \( m \times n \) matrix \( A \), for any sequence of row operations \( {E}_{1},\ldots ,{E}_{k} \) , if \( P = {E}_{k}\cdots {E}_{1} \) and \( B = {PA} \), then the subspaces spanned by the rows of \( A \) and the rows of \( B \) are identical. Therefore, \( A \) and \( B \) have the same row ... | Proof. Since \( B = {PA} \), from a previous observation, the rows of \( B \) are linear combinations of the rows of \( A \), so the span of the rows of \( B \) is a subspace of the span of the rows of \( A \) . Since \( P \) is invertible, \( A = {P}^{-1}B \), so by the same reasoning the span of the rows of \( A \) i... | Yes |
Proposition 8.14. Given any \( m \times n \) matrix \( A \), there is a sequence of row operations \( {E}_{1},\ldots ,{E}_{k} \) such that if \( P = {E}_{k}\cdots {E}_{1} \), then \( U = {PA} \) is a reduced row echelon matrix. | Proof. We proceed by induction on \( m \) . If \( m = 1 \), then either all entries on this row are zero, so \( A = 0 \), or if \( {a}_{j} \) is the first nonzero entry in \( A \), let \( P = \left( {a}_{j}^{-1}\right) \) (a \( 1 \times 1 \) matrix); clearly, \( {PA} \) is a reduced row echelon matrix.\n\nLet us now as... | Yes |
Proposition 8.17. Given any homogeneous system \( {Ax} = 0 \) of \( m \) equations in \( n \) variables, if \( m < n \), then there is a nonzero vector \( x \in {\mathbb{R}}^{n} \) such that \( {Ax} = 0 \) . | Proof. Convert the matrix \( A \) to a reduced row echelon matrix \( {A}^{\prime } \) . We know that \( {Ax} = 0 \) iff \( {A}^{\prime }x = 0 \) . If \( r \) is the number of pivots of \( {A}^{\prime } \), we must have \( r \leq m \), so by Theorem 8.16 we may assign arbitrary values to \( n - r > 0 \) nonpivot variabl... | Yes |
Proposition 8.18. Let \( A \) be a square matrix of dimension \( n \) . The following conditions are equivalent:\n\n(a) The matrix \( A \) can be reduced to the identity by a sequence of elementary row operations.\n\n(b) The matrix \( A \) is a product of elementary matrices.\n\n(c) The matrix \( A \) is invertible.\n\... | Proof. First we prove that (a) implies (b). If (a) can be reduced to the identity by a sequence of row operations \( {E}_{1},\ldots ,{E}_{p} \), this means that \( {E}_{p}\cdots {E}_{1}A = I \) . Since each \( {E}_{i} \) is invertible, we get\n\n\[ A = {E}_{1}^{-1}\cdots {E}_{p}^{-1} \]\n\nwhere each \( {E}_{i}^{-1} \)... | Yes |
Proposition 8.20. Let \( A \) be any \( m \times n \) matrix and let \( b \in {\mathbb{R}}^{m} \) be any vector. If the system \( {Ax} = b \) has a solution, then the set \( Z \) of all solutions of this system is the set \[ Z = {x}_{0} + \operatorname{Ker}\left( A\right) = \left\{ {{x}_{0} + x \mid {Ax} = 0}\right\} ,... | Proof. Assume that the system \( {Ax} = b \) is solvable and let \( {x}_{0} \) and \( {x}_{1} \) be any two solutions so that \( A{x}_{0} = b \) and \( A{x}_{1} = b \) . Subtracting the first equation from the second, we get \[ A\left( {{x}_{1} - {x}_{0}}\right) = 0 \] which means that \( {x}_{1} - {x}_{0} \in \operato... | Yes |
Proposition 8.22. Let \( f : E \rightarrow E \) be a bijective linear map and assume that \( f \neq \mathrm{{id}} \) and that \( f\left( x\right) = x \) for all \( x \in H \), where \( H \) is some hyperplane in \( E \). If there is some nonzero vector \( u \in E \) such that \( u \notin H \) and \( f\left( u\right) - ... | Proof. Using the notation as above, for some \( v \notin H \), we have \( f\left( v\right) = h + {\alpha v} \) with \( \alpha \neq 0 \), and write \( u = y + {tv} \) with \( y \in H \) and \( t \neq 0 \) since \( u \notin H \). If \( f\left( u\right) - u \in H \), from \[ f\left( u\right) - u = t\left( {h + \left( {\al... | Yes |
Proposition 8.25. If the field \( K \) is not of characteristic 2, then every transvection \( f \) of hyperplane \( H \) can be written as \( f = {d}_{2} \circ {d}_{1} \), where \( {d}_{1},{d}_{2} \) are dilatations of hyperplane \( H \) , where the direction of \( {d}_{1} \) can be chosen arbitrarily. | Proof. Pick some dilatation \( {d}_{1} \) of hyperplane \( H \) and scale factor \( \alpha \neq 0,1 \) . Then, \( {d}_{2} = f \circ {d}_{1}^{-1} \) leaves every vector in \( H \) fixed, and \( \det \left( {d}_{2}\right) = {\alpha }^{-1} \neq 1 \) . By Proposition 8.23, the linear map \( {d}_{2} \) is a dilatation of hy... | Yes |
Let \( E \) be any finite-dimensional vector space over a field \( K \) of characteristic not equal to 2. Then the group \( \mathbf{{SL}}\left( E\right) \) is generated by the transvections, and the group \( \mathbf{{GL}}\left( E\right) \) is generated by the dilatations. | Consider any \( f \in \mathbf{{SL}}\left( E\right) \), and let \( A \) be its matrix in any basis. By Proposition 8.24, there is a matrix \( S \) such that\n\n\[ \n{SA} = \left( \begin{matrix} {I}_{n - 1} & 0 \\ 0 & \alpha \end{matrix}\right) = {E}_{n,\alpha },\n\]\n\nwith \( \alpha = \det \left( A\right) \), and where... | Yes |
Proposition 8.27. Let \( E \) be any finite-dimensional vector space. For every transvection \( {\tau }_{\varphi, u}\left( {u \neq 0}\right) \) and every linear map \( g \in \mathbf{{GL}}\left( E\right) \), the map \( g \circ {\tau }_{\varphi, u} \circ {g}^{-1} \) is the transvection of hyperplane \( g\left( H\right) \... | Proof. We just need to prove that if \( n \geq 3 \), then for any two transvections \( {\tau }_{\varphi, u} \) and \( {\tau }_{\psi ,{u}^{\prime }} \) \( \left( {u,{u}^{\prime } \neq 0}\right) \), there is some \( g \in \mathbf{{SL}}\left( E\right) \) such that \( {\tau }_{\psi ,{u}^{\prime }} = g \circ {\tau }_{\varph... | Yes |
Problem 8.1. Solve the following linear systems by Gaussian elimination: | \[ \left( \begin{matrix} 2 & 3 & 1 \\ 1 & 2 & - 1 \\ - 3 & - 5 & 1 \end{matrix}\right) \left( \begin{array}{l} x \\ y \\ z \end{array}\right) = \left( \begin{matrix} 6 \\ 2 \\ - 7 \end{matrix}\right) ,\;\left( \begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 2 \\ 1 & 2 & 3 \end{array}\right) \left( \begin{array}{l} x \\ y \\ z ... | No |
Problem 8.2. Solve the following linear system by Gaussian elimination: | \[ \left( \begin{matrix} 1 & 2 & 1 & 1 \\ 2 & 3 & 2 & 3 \\ - 1 & 0 & 1 & - 1 \\ - 2 & - 1 & 4 & 0 \end{matrix}\right) \left( \begin{array}{l} {x}_{1} \\ {x}_{2} \\ {x}_{3} \\ {x}_{4} \end{array}\right) = \left( \begin{matrix} 7 \\ {14} \\ - 1 \\ 2 \end{matrix}\right) \] | No |
A permutation can be viewed as an operation permuting the rows of a matrix. For example, the permutation\n\n\\[ \n\\left( \\begin{array}{llll} 1 & 2 & 3 & 4 \\\\ 3 & 4 & 2 & 1 \\end{array}\\right)\n\\]\n\ncorresponds to the matrix\n\n\\[ \n{P}_{\\pi } = \\left( \\begin{array}{llll} 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 \\\\... | For example,\n\n\\[ \n{P}_{\\pi }A = \\left( \\begin{array}{llll} 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 \\\\ 1 & 0 & 0 & 0 \\\\ 0 & 1 & 0 & 0 \\end{array}\\right) \\left( \\begin{array}{llll} {a}_{11} & {a}_{12} & {a}_{13} & {a}_{14} \\\\ {a}_{21} & {a}_{22} & {a}_{23} & {a}_{24} \\\\ {a}_{31} & {a}_{32} & {a}_{33} & {a}_{3... | Yes |
Proposition 9.1. If \( E = {\mathbb{C}}^{n} \) or \( E = {\mathbb{R}}^{n} \), for every real number \( p \geq 1 \), the \( {\ell }^{p} \) -norm is indeed a norm. | Proof. The cases \( p = 1 \) and \( p = \infty \) are easy and left to the reader. If \( p > 1 \), then let \( q > 1 \) such that \[ \frac{1}{p} + \frac{1}{q} = 1 \] We will make use of the following fact: for all \( \alpha ,\beta \in \mathbb{R} \), if \( \alpha ,\beta \geq 0 \), then \[ {\alpha \beta } \leq \frac{{\al... | Yes |
Corollary 9.4. For any norm \( u \mapsto \parallel u\parallel \) on a finite-dimensional (complex or real) vector space \( E \), the map \( u \mapsto \parallel u\parallel \) is continuous with respect to the norm \( \parallel {\parallel }_{1} \) . | Let \( {S}_{1}^{n - 1} \) be the unit sphere with respect to the norm \( \parallel {\parallel }_{1} \), namely\n\n\[ \n{S}_{1}^{n - 1} = \left\{ {x \in E \mid \parallel x{\parallel }_{1} = 1}\right\} .\n\]\n\nNow \( {S}_{1}^{n - 1} \) is a closed and bounded subset of a finite-dimensional vector space, so by Heine-Bore... | Yes |
Theorem 9.5. If \( E \) is any real or complex vector space of finite dimension, then any two norms on \( E \) are equivalent. | Proof. It is enough to prove that any norm \( \parallel \parallel \) is equivalent to the 1-norm. We already proved that the function \( x \mapsto \parallel x\parallel \) is continuous with respect to the norm \( \parallel {\parallel }_{1} \), and we observed that the unit sphere \( {S}_{1}^{n - 1} \) is compact. Now w... | Yes |
Proposition 9.6. For any matrix norm \( \parallel \parallel \) on \( {\mathrm{M}}_{n}\left( \mathbb{C}\right) \) and for any square \( n \times n \) matrix \( A \in {\mathrm{M}}_{n}\left( \mathbb{C}\right) \), we have\n\n\[ \rho \left( A\right) \leq \parallel A\parallel \] | Proof. Let \( \lambda \) be some eigenvalue of \( A \) for which \( \left| \lambda \right| \) is maximum, that is, such that \( \left| \lambda \right| = \rho \left( A\right) \) . If \( u\left( { \neq 0}\right) \) is any eigenvector associated with \( \lambda \) and if \( U \) is the \( n \times n \) matrix whose column... | Yes |
Proposition 9.7. The Frobenius norm \( \parallel {\parallel }_{F} \) on \( {\mathrm{M}}_{n}\left( \mathbb{C}\right) \) satisfies the following properties:\n\n(1) It is a matrix norm; that is, \( \parallel {AB}{\parallel }_{F} \leq \parallel A{\parallel }_{F}\parallel B{\parallel }_{F} \), for all \( A, B \in {\mathrm{M... | Proof. (1) The only property that requires a proof is the fact \( \parallel {AB}{\parallel }_{F} \leq \parallel A{\parallel }_{F}\parallel B{\parallel }_{F} \) . This follows from the Cauchy-Schwarz inequality:\n\n\[ \parallel {AB}{\parallel }_{F}^{2} = \mathop{\sum }\limits_{{i, j = 1}}^{n}{\left| \mathop{\sum }\limit... | Yes |
Proposition 9.8. For every norm \( \parallel \parallel \) on \( {\mathbb{C}}^{n} \) (or \( {\mathbb{R}}^{n} \) ), for every matrix \( A \in {\mathrm{M}}_{n}\left( \mathbb{C}\right) \) (or \( A \in {\mathrm{M}}_{n}\left( \mathbb{R}\right) ) \), there is a real constant \( {C}_{A} \geq 0 \), such that\n\n\[ \parallel {Au... | Proof. For every basis \( \left( {{e}_{1},\ldots ,{e}_{n}}\right) \) of \( {\mathbb{C}}^{n} \) (or \( {\mathbb{R}}^{n} \) ), for every vector \( u = {u}_{1}{e}_{1} + \cdots + {u}_{n}{e}_{n} \), we have\n\n\[ \parallel {Au}\parallel = \begin{Vmatrix}{{u}_{1}A\left( {e}_{1}\right) + \cdots + {u}_{n}A\left( {e}_{n}\right)... | Yes |
Proposition 9.9. For any matrix norm \( \parallel \parallel \) on \( {\mathrm{M}}_{n}\left( \mathbb{R}\right) \) and for any square \( n \times n \) matrix \( A \in {\mathrm{M}}_{n}\left( \mathbb{R}\right) \), we have\n\n\[ \rho \left( A\right) \leq \parallel A\parallel \] | Proof. We follow the proof in Denis Serre’s book [154]. If \( A \) is a real matrix, the problem is that the eigenvectors associated with the eigenvalue of maximum modulus may be complex. We use a trick based on the fact that for every matrix \( A \) (real or complex),\n\n\[ \rho \left( {A}^{k}\right) = {\left( \rho \l... | No |
Proposition 9.10. For every square matrix \( A = \left( {a}_{ij}\right) \in {\mathrm{M}}_{n}\left( \mathbb{C}\right) \), we have\n\n\[ \parallel A{\parallel }_{1} = \mathop{\sup }\limits_{\substack{{x \in {\mathbb{C}}^{n}} \\ {\parallel x{\parallel }_{1} = 1} }}\parallel {Ax}{\parallel }_{1} = \mathop{\max }\limits_{j}... | Proof. For every vector \( u \), we have\n\n\[ \parallel {Au}{\parallel }_{1} = \mathop{\sum }\limits_{i}\left| {\mathop{\sum }\limits_{j}{a}_{ij}{u}_{j}}\right| \leq \mathop{\sum }\limits_{j}\left| {u}_{j}\right| \mathop{\sum }\limits_{i}\left| {a}_{ij}\right| \leq \left( {\mathop{\max }\limits_{j}\mathop{\sum }\limit... | Yes |
Proposition 9.11. Let \( \parallel \parallel \) be any matrix norm, and let \( B \in {\mathrm{M}}_{n}\left( \mathbb{C}\right) \) such that \( \parallel B\parallel < 1 \). (1) If \( \parallel \parallel \) is a subordinate matrix norm, then the matrix \( I + B \) is invertible and \[ \begin{Vmatrix}{\left( I + B\right) }... | Proof. (1) Observe that \( \left( {I + B}\right) u = 0 \) implies \( {Bu} = - u \), so \[ \parallel u\parallel = \parallel {Bu}\parallel \] Recall that \[ \parallel {Bu}\parallel \leq \parallel B\parallel \parallel u\parallel \] for every subordinate norm. Since \( \parallel B\parallel < 1 \), if \( u \neq 0 \), then \... | Yes |
Proposition 9.12. For every matrix \( A \in {\mathrm{M}}_{n}\left( \mathbb{C}\right) \) and for every \( \epsilon > 0 \), there is some subordinate matrix norm \( \parallel \parallel \) such that\n\n\[ \parallel A\parallel \leq \rho \left( A\right) + \epsilon \] | Proof. By Theorem 15.5, there exists some invertible matrix \( U \) and some upper triangular matrix \( T \) such that\n\n\[ A = {UT}{U}^{-1} \]\n\nand say that\n\n\[ T = \left( \begin{matrix} {\lambda }_{1} & {t}_{12} & {t}_{13} & \cdots & {t}_{1n} \\ 0 & {\lambda }_{2} & {t}_{23} & \cdots & {t}_{2n} \\ \vdots & \vdot... | Yes |
Proposition 9.13. Let \( A \) be an invertible matrix and let \( x \) and \( x + {\Delta x} \) be the solutions of the linear systems\n\n\[ \n{Ax} = b \n\]\n\n\[ \nA\left( {x + {\Delta x}}\right) = b + {\Delta b}. \n\]\n\n\nIf \( b \neq 0 \), then the inequality\n\n\[ \n\frac{\parallel {\Delta x}\parallel }{\parallel x... | Proof. We already proved the inequality. Now, because \( \parallel \parallel \) is a subordinate matrix norm, there exist some vectors \( x \neq 0 \) and \( {\Delta b} \neq 0 \) for which\n\n\[ \n\begin{Vmatrix}{{A}^{-1}{\Delta b}}\end{Vmatrix} = \begin{Vmatrix}{A}^{-1}\end{Vmatrix}\parallel {\Delta b}\parallel \;\text... | Yes |
Proposition 9.14. Let \( A \) be an invertible matrix and let \( x \) and \( x + {\Delta x} \) be the solutions of the two systems\n\n\[ \n{Ax} = b \n\]\n\n\[ \n\left( {A + {\Delta A}}\right) \left( {x + {\Delta x}}\right) = b. \n\]\n\nIf \( b \neq 0 \), then the inequality\n\n\[ \n\frac{\parallel {\Delta x}\parallel }... | Proof. The first inequality has already been proven. To show that equality can be achieved, let \( w \) be any vector such that \( w \neq 0 \) and\n\n\[ \n\begin{Vmatrix}{{A}^{-1}w}\end{Vmatrix} = \begin{Vmatrix}{A}^{-1}\end{Vmatrix}\parallel w\parallel \n\]\n\nand let \( \beta \neq 0 \) be any real number. Now the vec... | Yes |
Corollary 9.15. The spectral norm \( \parallel A{\parallel }_{2} \) of a matrix \( A \) is equal to the largest singular value of \( A \) . Equivalently, the spectral norm \( \parallel A{\parallel }_{2} \) of a matrix \( A \) is equal to the \( {\ell }^{\infty } \) -norm of its vector of singular values, | \[ \parallel A{\parallel }_{2} = \mathop{\max }\limits_{{1 \leq i \leq n}}{\sigma }_{i} = {\begin{Vmatrix}\left( {\sigma }_{1},\ldots ,{\sigma }_{n}\right) \end{Vmatrix}}_{\infty }. \] | Yes |
Proposition 9.17. For every invertible matrix \( A \in {\mathrm{M}}_{n}\left( \mathbb{C}\right) \), the following properties hold: (1)\n\n\[ \operatorname{cond}\left( A\right) \geq 1 \]\n\n\[ \operatorname{cond}\left( A\right) = \operatorname{cond}\left( {A}^{-1}\right) \]\n\n\[ \operatorname{cond}\left( {\alpha A}\rig... | Proof. The properties in (1) are immediate consequences of the properties of subordinate matrix norms. In particular, \( A{A}^{-1} = I \) implies\n\n\[ 1 = \parallel I\parallel \leq \parallel A\parallel \begin{Vmatrix}{A}^{-1}\end{Vmatrix} = \operatorname{cond}\left( A\right) \]\n\n(2) We showed earlier that \( \parall... | Yes |
Proposition 9.18. Assume \( \left( {E,\parallel \parallel }\right) \) is a complete normed vector space. If a series \( \mathop{\sum }\limits_{{k = 0}}^{\infty }{u}_{k} \) is absolutely convergent, then it is convergent. | Proof. If \( \mathop{\sum }\limits_{{k = 0}}^{\infty }{u}_{k} \) is absolutely convergent, then we prove that the sequence \( \left( {S}_{m}\right) \) is a Cauchy sequence; that is, for every \( \epsilon > 0 \), there is some \( p > 0 \) such that for all \( n \geq m \geq p \) ,\n\n\[\n\begin{Vmatrix}{{S}_{n} - {S}_{m}... | Yes |
Proposition 9.19. Assume \( \\left( {E,\\parallel \\parallel }\\right) \) is a normed vector space. If a series \( \\mathop{\\sum }\\limits_{{k = 0}}^{\\infty }{u}_{k} \) is convergent as well as absolutely convergent, then for every permutation \( \\sigma \) of \( \\mathbb{N} \), the series \( \\mathop{\\sum }\\limits... | \[ \\mathop{\\sum }\\limits_{{k = 0}}^{\\infty }{u}_{\\sigma \\left( k\\right) } = \\mathop{\\sum }\\limits_{{k = 0}}^{\\infty }{u}_{k} \] | Yes |
Proposition 9.20. For any \( n \times n \) real or complex matrix \( A \), the series\n\n\[ \mathop{\sum }\limits_{{k = 0}}^{\infty }\frac{{A}^{k}}{k!} \]\n\nconverges absolutely for any operator norm on \( {\mathrm{M}}_{n}\left( \mathbb{C}\right) \) (or \( {\mathrm{M}}_{n}\left( \mathbb{R}\right) \) ). | Proof. Pick any norm on \( {\mathbb{C}}^{n} \) (or \( {\mathbb{R}}^{n} \) ) and let \( \parallel \parallel \) be the corresponding operator norm on \( {\mathrm{M}}_{n}\left( \mathbb{C}\right) \) . Since \( {\mathrm{M}}_{n}\left( \mathbb{C}\right) \) has dimension \( {n}^{2} \), it is complete. By Proposition 9.18, it s... | Yes |
For any two \( n \times n \) complex matrices \( A \) and \( B \), if \( A \) and \( B \) commute, that is, \( {AB} = {BA} \), then\n\n\[ \n{e}^{A + B} = {e}^{A}{e}^{B} \n\] | A proof of Proposition 9.21 can be found in Gallier [73]. | No |
Proposition 9.22. If \( A = {\theta J} \), then\n\n\[ \n{e}^{A} = \cos {\theta I} + \sin {\theta J} = \left( \begin{matrix} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{matrix}\right) .\n\] | Proof. We have\n\n\[ \n{A}^{4n} = {\theta }^{4n}{I}_{2}\n\]\n\n\[ \n{A}^{{4n} + 1} = {\theta }^{{4n} + 1}J\n\]\n\n\[ \n{A}^{{4n} + 2} = - {\theta }^{{4n} + 2}{I}_{2}\n\]\n\n\[ \n{A}^{{4n} + 3} = - {\theta }^{{4n} + 3}J\n\]\n\nand so\n\[ \n{e}^{A} = {I}_{2} + \frac{\theta }{1!}J - \frac{{\theta }^{2}}{2!}{I}_{2} - \frac... | Yes |
Proposition 9.23. If \( B \) is an \( n \times n \) (real) skew symmetric matrix, that is, \( {B}^{\top } = - B \), then \( Q = {e}^{B} \) is an orthogonal matrix, that is\n\n\[ \n{Q}^{\top }Q = Q{Q}^{\top } = I \n\] | Proof. Since \( {B}^{\top } = - B \), we have\n\n\[ \n{Q}^{\top } = {\left( {e}^{B}\right) }^{\top } = {e}^{{B}^{\top }} = {e}^{-B}. \n\]\n\nSince \( B \) and \( - B \) commute, we have\n\n\[ \n{Q}^{\top }Q = {e}^{-B}{e}^{B} = {e}^{-B + B} = {e}^{0} = I. \n\]\n\nSimilarly,\n\n\[ \nQ{Q}^{\top } = {e}^{B}{e}^{-B} = {e}^{... | Yes |
For any \( p \geq 1 \), prove that for all \( x \in {\mathbb{R}}^{n} \) , | \[ \mathop{\lim }\limits_{{p \mapsto \infty }}\parallel x{\parallel }_{p} = \parallel x{\parallel }_{\infty } \] | Yes |
For any nonzero vector \( v \), prove that\n\n\[ \parallel {Av}{\parallel }_{\infty } \geq \parallel v{\parallel }_{\infty }\delta \] | Use the above to prove that \( A \) is invertible. | No |
Consider a real \( 2 \times 2 \) matrix with zero trace of the form\n\n\[ A = \left( \begin{matrix} a & b \\ c & - a \end{matrix}\right) \]\n\n(1) Prove that\n\n\[ {A}^{2} = \left( {{a}^{2} + {bc}}\right) {I}_{2} = - \det \left( A\right) {I}_{2} \] | If \( {a}^{2} + {bc} = 0 \), prove that\n\n\[ {e}^{A} = {I}_{2} + A \] | No |
Theorem 10.1. For any square matrix \( B \), the following conditions are equivalent:\n\n(1) \( \mathop{\lim }\limits_{{k \mapsto \infty }}{B}^{k} = 0 \) ,\n\n(2) \( \mathop{\lim }\limits_{{k \mapsto \infty }}{B}^{k}v = 0 \), for all vectors \( v \) ,\n\n(3) \( \rho \left( B\right) < 1 \) ,\n\n(4) \( \parallel B\parall... | Proof. Assume (1) and let \( \parallel \parallel \) be a vector norm on \( E \) and \( \parallel \parallel \) be the corresponding matrix norm. For every vector \( v \in E \), because \( \parallel \parallel \) is a matrix norm, we have\n\n\[ \begin{Vmatrix}{{B}^{k}v}\end{Vmatrix} \leq \begin{Vmatrix}{B}^{k}\end{Vmatrix... | Yes |
Proposition 10.2. For every square matrix \( B \in {\mathrm{M}}_{n}\left( \mathbb{C}\right) \) and every matrix norm \( \parallel \parallel \), we have\n\n\[ \mathop{\lim }\limits_{{k \mapsto \infty }}{\begin{Vmatrix}{B}^{k}\end{Vmatrix}}^{1/k} = \rho \left( B\right) \] | Proof. We know from Proposition 9.6 that \( \rho \left( B\right) \leq \parallel B\parallel \), and since \( \rho \left( B\right) = {\left( \rho \left( {B}^{k}\right) \right) }^{1/k} \), we deduce that\n\n\[ \rho \left( B\right) \leq {\begin{Vmatrix}{B}^{k}\end{Vmatrix}}^{1/k}\;\text{ for all }k \geq 1 \]\n\nand so\n\n\... | Yes |
Theorem 10.3. Given a system \( u = {Bu} + c \) as above, where \( I - B \) is invertible, the following statements are equivalent:\n\n(1) The iterative method is convergent.\n\n(2) \( \rho \left( B\right) < 1 \) .\n\n(3) \( \parallel B\parallel < 1 \), for some subordinate matrix norm \( \parallel \parallel \) . | Proof. Define the vector \( {e}_{k} \) (error vector) by\n\n\[ {e}_{k} = {u}_{k} - \widetilde{u} \]\n\nwhere \( \widetilde{u} \) is the unique solution of the system \( u = {Bu} + c \) . Clearly, the iterative method is convergent iff\n\n\[ \mathop{\lim }\limits_{{k \mapsto \infty }}{e}_{k} = 0 \]\n\nWe claim that\n\n\... | Yes |
Proposition 10.4. Let \( \parallel \parallel \) be any vector norm, let \( B \in {\mathrm{M}}_{n}\left( \mathbb{C}\right) \) be a matrix such that \( I - B \) is invertible, and let \( \widetilde{u} \) be the unique solution of \( u = {Bu} + c \) . (1) If \( \left( {u}_{k}\right) \) is any sequence defined iteratively ... | Proof. Let \( \parallel \parallel \) be the subordinate matrix norm. Recall that \[ {u}_{k} - \widetilde{u} = {B}^{k}{e}_{0} \] with \( {e}_{0} = {u}_{0} - \widetilde{u} \) . For every \( k \in \mathbb{N} \), we have \[ {\left( \rho \left( {B}_{1}\right) \right) }^{k} = \rho \left( {B}_{1}^{k}\right) \leq \begin{Vmatri... | Yes |
Consider the linear system\n\n\[ \n\\left( \\begin{matrix} 2 & - 1 & 0 & 0 \\ - 1 & 2 & - 1 & 0 \\ 0 & - 1 & 2 & - 1 \\ 0 & 0 & - 1 & 2 \\end{matrix}\\right) \\left( \\begin{array}{l} {x}_{1} \\ {x}_{2} \\ {x}_{3} \\ {x}_{4} \\end{array}\\right) = \\left( \\begin{matrix} {25} \\ - {24} \\ {21} \\ - {15} \\end{matrix}\\... | We check immediately that the solution is\n\n\[ \n{x}_{1} = {11},{x}_{2} = - 3,{x}_{3} = 7,{x}_{4} = - 4\\text{.} \n\] | Yes |
Consider the same linear system\n\n\\[ \n\\left( \\begin{matrix} 2 & - 1 & 0 & 0 \\\\ - 1 & 2 & - 1 & 0 \\\\ 0 & - 1 & 2 & - 1 \\\\ 0 & 0 & - 1 & 2 \\end{matrix}\\right) \\left( \\begin{array}{l} {x}_{1} \\\\ {x}_{2} \\\\ {x}_{3} \\\\ {x}_{4} \\end{array}\\right) = \\left( \\begin{matrix} {25} \\\\ - {24} \\\\ {21} \\\... | After 10 Gauss-Seidel iterations, we find the approximate solution\n\n\\[ \n{x}_{1} = {10.9966},{x}_{2} = - {3.0044},{x}_{3} = {6.9964},{x}_{4} = - {4.0018}\\text{.}\n\\]\n\nAfter 20 iterations, we find the approximate solution\n\n\\[ \n{x}_{1} = {11.0000},{x}_{2} = - {3.0001},{x}_{3} = {6.9999},{x}_{4} = - {4.0000}\\t... | Yes |
Consider the same linear system as in Examples 10.1 and 10.2, whose solution is\n\n\[ \n{x}_{1} = {11},{x}_{2} = - 3,{x}_{3} = 7,{x}_{4} = - 4\text{.} \n\] | After 10 relaxation iterations with \( \omega = {1.1} \), we find the approximate solution\n\n\[ \n{x}_{1} = {11.0026},{x}_{2} = - {2.9968},{x}_{3} = {7.0024},{x}_{4} = - {3.9989}. \n\]\n\nAfter 10 iterations with \( \omega = {1.2} \), we find the approximate solution\n\n\[ \n{x}_{1} = {11.0014},{x}_{2} = - {2.9985},{x... | Yes |
Proposition 10.5. Let \( A \) be any Hermitian positive definite matrix, written as\n\n\[ A = M - N \]\n\nwith \( M \) invertible. Then \( {M}^{ * } + N \) is Hermitian, and if it is positive definite, then\n\n\[ \rho \left( {{M}^{-1}N}\right) < 1 \]\n\nso that the iterative method converges. | Proof. Since \( M = A + N \) and \( A \) is Hermitian, \( {A}^{ * } = A \), so we get\n\n\[ {M}^{ * } + N = {A}^{ * } + {N}^{ * } + N = A + N + {N}^{ * } = M + {N}^{ * } = {\left( {M}^{ * } + N\right) }^{ * }, \]\n\nwhich shows that \( {M}^{ * } + N \) is indeed Hermitian.\n\nBecause \( A \) is Hermitian positive defin... | Yes |
Theorem 10.6. If \( A = D - E - F \) is Hermitian positive definite, and if \( 0 < \omega < 2 \), then the relaxation method converges. This also holds for a block decomposition of \( A \) . | Proof. Recall that for the relaxation method, \( A = M - N \) with\n\n\[ M = \frac{D}{\omega } - E \]\n\n\[ N = \frac{1 - \omega }{\omega }D + F \]\n\nand because \( {D}^{ * } = D,{E}^{ * } = F \) (since \( A \) is Hermitian) and \( \omega \neq 0 \) is real, we have\n\n\[ {M}^{ * } + N = \frac{{D}^{ * }}{\omega } - {E}... | Yes |
Given any matrix \( A = D - E - F \), with \( A \) and \( D \) invertible, for any \( \omega \neq 0 \), we have\n\n\[ \rho \left( {\mathcal{L}}_{\omega }\right) \geq \left| {\omega - 1}\right| \]\n\nwhere \( {\mathcal{L}}_{\omega } = {\left( \frac{D}{\omega } - E\right) }^{-1}\left( {\frac{1 - \omega }{\omega }D + F}\r... | Proof. Observe that the product \( {\lambda }_{1}\cdots {\lambda }_{n} \) of the eigenvalues of \( {\mathcal{L}}_{\omega } \), which is equal to \( \det \left( {\mathcal{L}}_{\omega }\right) \), is given by\n\n\[ {\lambda }_{1}\cdots {\lambda }_{n} = \det \left( {\mathcal{L}}_{\omega }\right) = \frac{\det \left( {\frac... | Yes |
Proposition 10.10. Let \( A \) be a tridiagonal matrix (possibly by blocks) which is Hermitian positive definite. Then the methods of Jacobi, Gauss-Seidel, and relaxation, all converge for \( \omega \in \left( {0,2}\right) \) . There is a unique optimal relaxation parameter \[ {\omega }_{0} = \frac{2}{1 + \sqrt{1 - {\l... | Proof. In order to apply Proposition 10.9, we have to check that \( J = {D}^{-1}\left( {E + F}\right) \) has real eigenvalues. However, if \( \alpha \) is any eigenvalue of \( J \) and if \( u \) is any corresponding eigenvector, then \[ {D}^{-1}\left( {E + F}\right) u = {\alpha u} \] implies that \[ \left( {E + F}\rig... | Yes |
Prove that the eigenvalues of the Jacobi matrix \( J \) are given by\n\n\[ {\lambda }_{k} = \cos \left( \frac{k\pi }{n + 1}\right) ,\;k = 1,\ldots, n. \] | Hint. First show that the Jacobi matrix is\n\n\[ J = \frac{1}{2}\left( \begin{matrix} 0 & 1 & 0 & & \\ 1 & 0 & 1 & & \\ & \ddots & \ddots & \ddots & \\ & & 1 & 0 & 1 \\ & & 0 & 1 & 0 \end{matrix}\right) . \]\n\nThen the eigenvalues and the eigenvectors of \( J \) are solutions of the system of equations\n\n\[ {y}_{0} =... | No |
For example, consider the columns of the Bézier matrix\n\n\[ \n{B}_{4} = \left( \begin{matrix} 1 & - 3 & 3 & - 1 \\ 0 & 3 & - 6 & 3 \\ 0 & 0 & 3 & - 3 \\ 0 & 0 & 0 & 1 \end{matrix}\right) \n\] | In other words, we have the basis\n\n\[ \n{u}_{1} = \left( \begin{array}{l} 1 \\ 0 \\ 0 \\ 0 \end{array}\right) \;{u}_{2} = \left( \begin{matrix} - 3 \\ 3 \\ 0 \\ 0 \end{matrix}\right) \;{u}_{3} = \left( \begin{matrix} 3 \\ - 6 \\ 3 \\ 0 \end{matrix}\right) \;{u}_{4} = \left( \begin{matrix} - 1 \\ 3 \\ - 3 \\ 1 \end{ma... | Yes |
Proposition 11.1. Let \( \left( {{u}_{1},\ldots ,{u}_{n}}\right) \) and \( \left( {{v}_{1},\ldots ,{v}_{n}}\right) \) be two bases of \( E \), and let \( P = \left( {a}_{ij}\right) \) be the change of basis matrix from \( \left( {{u}_{1},\ldots ,{u}_{n}}\right) \) to \( \left( {{v}_{1},\ldots ,{v}_{n}}\right) \), so th... | To best understand the preceding paragraph, recall Example 3.1, in which \( E = {\mathbb{R}}^{2} \) , \( {u}_{1} = \left( {1,0}\right) ,{u}_{2} = \left( {0,1}\right) \), and \( {v}_{1} = \left( {1,1}\right) ,{v}_{2} = \left( {-1,1}\right) \) . Then \( P \), the change of basis matrix from \( \left( {{u}_{1},{u}_{2}}\ri... | Yes |
Proposition 11.2. If \( {V}_{1} \subseteq {V}_{2} \subseteq E \), then \( {V}_{2}^{0} \subseteq {V}_{1}^{0} \subseteq {E}^{ * } \), and if \( {U}_{1} \subseteq {U}_{2} \subseteq {E}^{ * } \), then \( {U}_{2}^{0} \subseteq {U}_{1}^{0} \subseteq E \) . | Proof. Indeed, if \( {V}_{1} \subseteq {V}_{2} \subseteq E \), then for any \( {f}^{ * } \in {V}_{2}^{0} \) we have \( {f}^{ * }\left( v\right) = 0 \) for all \( v \in {V}_{2} \), and thus \( {f}^{ * }\left( v\right) = 0 \) for all \( v \in {V}_{1} \), so \( {f}^{ * } \in {V}_{1}^{0} \) . Similarly, if \( {U}_{1} \subs... | Yes |
Let \( E = {\mathrm{M}}_{2}\left( \mathbb{R}\right) \), the space of real \( 2 \times 2 \) matrices, and let \( V \) be the subspace of \( {\mathrm{M}}_{2}\left( \mathbb{R}\right) \) spanned by the matrices\n\n\[ \left( \begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right) ,\;\left( \begin{array}{ll} 1 & 0 \\ 0 & 0 \end{... | By the duality theorem (Theorem 11.4) we have\n\n\[ \dim \left( {V}^{0}\right) = \dim \left( E\right) - \dim \left( V\right) = 4 - 3 = 1. \] | Yes |
The above example generalizes to \( E = {\mathrm{M}}_{n}\left( \mathbb{R}\right) \) for any \( n \geq 1 \), but this time, consider the space \( U \) of linear forms asserting that a matrix \( A \) is symmetric; these are the linear forms spanned by the \( n\left( {n - 1}\right) /2 \) equations\n\n\[ {a}_{ij} - {a}_{ji... | Note there are no constraints on diagonal entries, and half of the equations\n\n\[ {a}_{ij} - {a}_{ji} = 0,\;1 \leq i \neq j \leq n \]\n\nare redundant. It is easy to check that the equations (linear forms) for which \( i < j \) are linearly independent. To be more precise, let \( U \) be the space of linear forms in \... | Yes |
If \( E = {\mathrm{M}}_{n}\left( \mathbb{R}\right) \), consider the subspace \( U \) of linear forms in \( {E}^{ * } \) spanned by the linear forms\n\n\[ \n{u}_{ij}^{ * }\left( {{a}_{11},\ldots ,{a}_{1n},{a}_{21},\ldots ,{a}_{2n},\ldots ,{a}_{n1},\ldots ,{a}_{nn}}\right) = {a}_{ij} + {a}_{ji},\;1 \leq i < j \leq n \n\]... | \[ \n\frac{n\left( {n - 1}\right) }{2} = {n}^{2} - \frac{n\left( {n + 1}\right) }{2}. \n\] | Yes |
Proposition 11.3. We have \( V \subseteq {V}^{00} \) for every subspace \( V \) of \( E \), and \( U \subseteq {U}^{00} \) for every subspace \( U \) of \( {E}^{ * } \) . | Proof. Indeed, for any \( v \in V \), to show that \( v \in {V}^{00} \) we need to prove that \( {u}^{ * }\left( v\right) = 0 \) for all \( {u}^{ * } \in {V}^{0} \) . However, \( {V}^{0} \) consists of all linear forms \( {u}^{ * } \) such that \( {u}^{ * }\left( y\right) = 0 \) for all \( y \in V \) ; in particular, f... | Yes |
(a) For every basis \( {\left( {u}_{i}\right) }_{i \in I} \) of \( E \), the family \( {\left( {u}_{i}^{ * }\right) }_{i \in I} \) of coordinate forms is linearly independent. | Assume that\n\n\[ \mathop{\sum }\limits_{{i \in I}}{\lambda }_{i}{u}_{i}^{ * } = 0 \]\n\nfor a family \( {\left( {\lambda }_{i}\right) }_{i \in I} \) (of scalars in \( K \) ). Since \( {\left( {\lambda }_{i}\right) }_{i \in I} \) has finite support, there is a finite subset \( J \) of \( I \) such that \( {\lambda }_{i... | Yes |
Problem 1 . Suppose that \( V \) is a subspace of \( {\mathbb{R}}^{n} \) of dimension \( m \) and that \( \left( {{v}_{1},\ldots ,{v}_{m}}\right) \) is a basis of \( V \) . The problem is to find a basis of \( {V}^{0} \) . | We first extend \( \left( {{v}_{1},\ldots ,{v}_{m}}\right) \) to a basis \( \left( {{v}_{1},\ldots ,{v}_{n}}\right) \) of \( {\mathbb{R}}^{n} \), and then by part (c) of Theorem 11.4, we know that \( \left( {{v}_{m + 1}^{ * },\ldots ,{v}_{n}^{ * }}\right) \) is a basis of \( {V}^{0} \) . | No |
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