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Proposition 3.2. Let \( E, F, D \) be modules, and let \( \mathfrak{B},{\mathfrak{B}}^{\prime },{\mathfrak{B}}^{\prime \prime } \) be finite bases of \( E, F, D \), respectively. Let\n\n\[ E\overset{f}{ \rightarrow }F\overset{g}{ \rightarrow }D \]\n\nbe linear maps. Then\n\n\[ {M}_{{\Phi }^{\prime \prime }}^{\Phi }\lef... | Proof. Let \( A \) and \( B \) be the matrices associated with the maps \( f, g \) respectively, with respect to our given bases. If \( X \) is the column vector associated with \( x \in E \), the vector associated with \( g\left( {f\left( x\right) }\right) \) is \( B\left( {AX}\right) = \left( {BA}\right) X \) . Hence... | Yes |
Corollary 3.3. Let \( E = F \) . Then\n\n\[ \n{M}_{{\varphi }^{\prime }}^{\varphi }\left( \mathrm{{id}}\right) {M}_{\varphi }^{{\varphi }^{\prime }}\left( \mathrm{{id}}\right) = {M}_{{\varphi }^{\prime }}^{{\varphi }^{\prime }}\left( \mathrm{{id}}\right) = I.\n\]\n\nEach matrix \( {M}_{\mathfrak{G}}^{\mathfrak{B}} \) (... | Proof. Obvious. | No |
Corollary 3.4. Let \( N = {M}_{{\alpha }^{\prime }}^{\alpha } \) (id). Then\n\n\[ \n{M}_{{\mathbb{G}}^{\prime }}^{{\mathbb{G}}^{\prime }}\left( f\right) = {M}_{{\mathbb{G}}^{\prime }}^{\mathbb{G}}\left( \mathrm{{id}}\right) {M}_{\mathbb{G}}^{\mathbb{G}}\left( f\right) {M}_{\mathbb{G}}^{{\mathbb{G}}^{\prime }}\left( \ma... | Proof. Obvious | No |
Proposition 4.1. Let \( f \) be an n-multilinear alternating map on \( E \) . Let \( {x}_{1},\ldots ,{x}_{n} \in E \) . Then\n\n\[ f\left( {\ldots ,{x}_{i},{x}_{i + 1},\ldots }\right) = - f\left( {\ldots ,{x}_{i + 1},{x}_{i},\ldots }\right) . \]\n\nIn other words, when we interchange two adjacent arguments of \( f \), ... | Proof. Restricting our attention to the factors in the \( i \) -th and \( j \) -th place, with \( j = i + 1 \), we may assume \( f \) is bilinear for the first statement. Then for all \( x \) , \( y \in E \) we have\n\n\[ 0 = f\left( {x + y, x + y}\right) = f\left( {x, y}\right) + f\left( {y, x}\right) . \]\n\nThis pro... | Yes |
Corollary 4.2. Let \( f \) be an n-multilinear alternating map on \( E \) . Let \( {x}_{1},\ldots ,{x}_{n} \in E \) . Let \( i \neq j \) and let \( a \in R \) . Then the value of \( f \) on \( \left( {{x}_{1},\ldots ,{x}_{n}}\right) \) does not change if we replace \( {x}_{i} \) by \( {x}_{i} + a{x}_{j} \) and leave al... | Proof. Obvious. | No |
Theorem 4.4. (Cramer’s Rule). Let \( {A}^{1},\ldots ,{A}^{n} \) be column vectors of dimension \( n \) . Let \( {x}_{1},\ldots ,{x}_{n} \in R \) be such that\n\n\[ \n{x}_{1}{A}^{1} + \cdots + {x}_{n}{A}^{n} = B \n\]\n\nfor some column vector \( B \) . Then for each \( i \) we have\n\n\[ \n{x}_{i}D\left( {{A}^{1},\ldots... | Proof. Say \( i = 1 \) . We expand\n\n\[ \nD\left( {B,{A}^{2},\ldots ,{A}^{n}}\right) = \mathop{\sum }\limits_{{j = 1}}^{n}{x}_{j}D\left( {{A}^{j},{A}^{2},\ldots ,{A}^{n}}\right) , \n\]\n\nand use Proposition 4.1 to get what we want (all terms on the right are equal to 0 except the one having \( {x}_{1} \) in it). | Yes |
Corollary 4.5. Assume that \( R \) is a field. Then \( {A}^{1},\ldots ,{A}^{n} \) are linearly dependent if and only if \( D\left( {{A}^{1},\ldots ,{A}^{n}}\right) = 0 \) . | Proof. Assume we have a relation\n\n\[ \n{x}_{1}{A}^{1} + \cdots + {x}_{n}{A}^{n} = 0 \n\]\n\nwith \( {x}_{i} \in R \) . Then \( {x}_{i}D\left( A\right) = 0 \) for all \( i \) . If some \( {x}_{i} \neq 0 \) then \( D\left( A\right) = 0 \) . Conversely, assume that \( {A}^{1},\ldots ,{A}^{n} \) are linearly independent.... | Yes |
Proposition 4.6. If determinants exist, they are unique. If \( {A}^{1},\ldots ,{A}^{n} \) are the column vectors of dimension \( n \), of the matrix \( A = \left( {a}_{ij}\right) \), then\n\n\[ D\left( {{A}^{1},\ldots ,{A}^{n}}\right) = \mathop{\sum }\limits_{\sigma }\epsilon \left( \sigma \right) {a}_{\sigma \left( 1\... | Proof. Let \( {e}^{1},\ldots ,{e}^{n} \) be the unit vectors as usual. We can write\n\n\[ {A}^{1} = {a}_{11}{e}^{1} + \cdots + {a}_{n1}{e}^{n} \]\n\n\[ {A}^{n} = {a}_{1n}{e}^{n} + \cdots + {a}_{nn}{e}^{n} \]\n\nTherefore\n\n\[ D\left( {{A}^{1},\ldots ,{A}^{n}}\right) = \mathop{\sum }\limits_{\sigma }\epsilon \left( \si... | Yes |
Corollary 4.7. Let \( \varphi : R \rightarrow {R}^{\prime } \) be a ring-homomorphism into a commutative ring. If \( A \) is a square matrix in \( R \), define \( {\varphi A} \) to be the matrix obtained by applying \( \varphi \) to each component of \( A \) . Then\n\n\[ \varphi \left( {D\left( A\right) }\right) = D\le... | Proof. Apply \( \varphi \) to the expression of Proposition 4.6. | No |
Proposition 4.8. If \( A \) is a square matrix in \( R \) then\n\n\[ D\left( A\right) = D\left( {{}^{t}A}\right) \] | Proof. In a product\n\n\[ {a}_{\sigma \left( 1\right) ,1}\cdots {a}_{\sigma \left( n\right), n} \]\n\neach integer \( k \) from 1 to \( n \) occurs precisely once among the integers \( \sigma \left( 1\right) ,\ldots ,\sigma \left( n\right) \) . Hence we can rewrite this product in the form\n\n\[ {a}_{1,{\sigma }^{-1}\l... | Yes |
Corollary 4.9. The determinant is multilinear and alternating with respect to the rows of a matrix. | We shall now prove existence, and prove simultaneously one additional important property of determinants.\n\nWhen \( n = 1 \), we define \( D\left( a\right) = a \) for any \( a \in R \) .\n\nAssume that we have proved the existence of determinants for all integers \( < n\left( {n \geqq 2}\right) \) . Let \( A \) be an ... | Yes |
Theorem 4.11. Let \( E \) be a module over \( R \), and let \( {v}_{1},\ldots ,{v}_{n} \) be elements of \( E \) . Let \( A = \left( {a}_{ij}\right) \) be a matrix in \( R \), and let\n\n\[ A\left( \begin{matrix} {v}_{1} \\ \vdots \\ {v}_{n} \end{matrix}\right) = \left( \begin{matrix} {w}_{1} \\ \vdots \\ {w}_{n} \end{... | Proof. We expand\n\n\[ \Delta \left( {{a}_{11}{v}_{1} + \cdots + {a}_{1n}{v}_{n},\ldots ,{a}_{n1}{v}_{1} + \cdots + {a}_{nn}{v}_{n}}\right) ,\]\n\nand find precisely what we want, taking into account \( D\left( A\right) = D\left( {{}^{t}A}\right) \) . | No |
Corollary 4.12. Let \( E \) be a free module over \( R \), and let \( \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) be a basis.\n\nLet \( F \) be any module, and let \( w \in F \) . There exists a unique n-multilinear alternating map\n\n\[ \n{\Delta }_{w} : E \times \cdots \times E \rightarrow F \n\]\n\nsuch that \( {\D... | Proof. Without loss of generality, we may assume that \( E = {R}^{\left( n\right) } \), and then, if \( {A}^{1},\ldots ,{A}^{n} \) are column vectors, we define\n\n\[ \n{\Delta }_{w}\left( {{A}^{1},\ldots ,{A}^{n}}\right) = D\left( A\right) w.\n\]\n\nThen \( {\Delta }_{w} \) obviously has the required properties. | No |
Corollary 4.13. If \( E \) is free over \( R \), and has a basis consisting of \( n \) elements, then \( {L}_{a}^{n}\left( E\right) \) is free over \( R \), and has a basis consisting of 1 element. | Proof. We let \( {\Delta }_{1} \) be the multilinear alternating map taking the value 1 on a basis \( \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) . Any element \( \varphi \in {L}_{a}^{n}\left( E\right) \) can then be written in a unique way as \( c{\Delta }_{1} \), with some \( c \in R \), namely \( c = \varphi \left(... | Yes |
Corollary 4.14. Let \( R \) be a field. Let \( E \) be a vector space of dimension \( n \) . Let \( \Delta \) be any determinant on \( E \) . Let \( {v}_{1},\ldots ,{v}_{n} \in E \) . In order that \( \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) be a basis of \( E \) it is necessary and sufficient that | \[ \Delta \left( {{v}_{1},\ldots ,{v}_{n}}\right) \neq 0. \] | Yes |
Proposition 4.15. Let \( A, B \) be \( n \times n \) matrices in \( R \) . Then\n\n\[ D\left( {AB}\right) = D\left( A\right) D\left( B\right) \] | Proof. This is actually a corollary of Theorem 4.11. We take \( {v}_{1},\ldots ,{v}_{n} \)\nto be the unit vectors \( {e}^{1},\ldots ,{e}^{n} \), and consider\n\n\[ {AB}\left( \begin{matrix} {e}^{1} \\ \vdots \\ {e}^{n} \end{matrix}\right) = \left( \begin{matrix} {w}_{1} \\ \vdots \\ {w}_{n} \end{matrix}\right) \]\n\nW... | Yes |
Proposition 4.16. Let \( d = D\left( A\right) \) . Then \( A\widetilde{A} = \widetilde{A}A = {dI} \) . The determinant \( D\left( A\right) \) is invertible in \( R \) if and only if \( A \) is invertible, and then \[ {A}^{-1} = \frac{1}{d}\widetilde{A} \] | Proof. For any pair of indices \( i, k \) the \( {ik} \) -component of \( A\widetilde{A} \) is \[ {a}_{i1}{b}_{1k} + {a}_{i2}{b}_{2k} + \cdots + {a}_{in}{b}_{nk} = {a}_{i1}{\left( -1\right) }^{k + 1}D\left( {A}_{k1}\right) + \cdots + {a}_{in}{\left( -1\right) }^{k + n}D\left( {A}_{kn}\right) . \] If \( i = k \), then t... | Yes |
Corollary 4.17. Let \( F \) be any \( R \) -module, and let \( {w}_{1},\ldots ,{w}_{n} \) be elements of \( F \) . Let \( A = \left( {a}_{ij}\right) \) be an \( n \times n \) matrix in \( R \) . Let\n\n\[ \n{a}_{11}{w}_{1} + \cdots + {a}_{1n}{w}_{n} = {v}_{1} \]\n\n\[ \n{a}_{n1}{w}_{1} + \cdots + {a}_{nn}{w}_{n} = {v}_... | Proof. This is immediate from the relation \( \widetilde{A}A = D\left( A\right) I \), using the remarks in \( §3 \) about applying matrices to column vectors whose components lie in the module. | Yes |
Proposition 4.18. Let \( E, F \) be free modules of dimension \( n \) over \( R \) . Let \( f : E \rightarrow F \) be a linear map. Let \( \mathfrak{G},{\mathfrak{R}}^{\prime } \) be bases of \( E, F \) respectively over \( R \) . Then \( f \) is an isomorphism if and only if the determinant of its associated matrix \(... | Proof. Let \( A = {M}_{{\alpha }^{\prime }}^{\alpha }\left( f\right) \) . By definition, \( f \) is an isomorphism if and only if there exists a linear map \( g : F \rightarrow E \) such that \( g \circ f = \mathrm{{id}} \) and \( f \circ g = \mathrm{{id}} \) . If \( f \) is an isomorphism, and \( B = {M}_{\mathfrak{G}... | Yes |
Proposition 4.19. Let \( E \) be a free module over \( R \), of dimension \( n \) . Let \( \{ \Delta \} \) be a basis of \( {L}_{a}^{n}\left( E\right) \) . Let \( f : E \rightarrow E \) be an endomorphism of \( E \) . Then \[ {f}^{ * }\Delta = D\left( f\right) \Delta \] | Proof. This is an immediate consequence of Theorem 4.11. Namely, we let \( \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) be a basis of \( E \), and then take \( A \) (or ’ \( A \) ) to be a matrix of \( f \) relative to this basis. By definition, \[ {f}^{ * }\Delta \left( {{v}_{1},\ldots ,{v}_{n}}\right) = \Delta \left(... | Yes |
Proposition 4.20. Let \( R \) be a principal entire ring. Let \( F \) be a free module over \( R \) and let \( M \) be a finitely generated submodule. Let \( \left\{ {{e}_{1},\ldots ,{e}_{m},\ldots }\right\} \) be a basis of \( F \) such that there exist non-zero elements \( {a}_{1},\ldots ,{a}_{m} \in R \) such that:\... | Proof. We first show that \( {J}_{s} \subset \left( {{a}_{1}\cdots {a}_{s}}\right) \) . Indeed, an element \( y \in M \) can be written in the form\n\n\[ \ny = {c}_{1}{a}_{1}{e}_{1} + \cdots + {c}_{r}{a}_{r}{e}_{r} \n\]\n\nHence if \( {y}_{1},\ldots ,{y}_{s} \in M \), and \( f \) is multilinear alternating on \( F \), ... | Yes |
Proposition 5.1. Let \( f : E \times E \rightarrow R \) be a non-singular bilinear form. Let \( A : E \rightarrow E \) be a linear map. Then \( A \) is an automorphism of \( f \) if and only if \( {}^{t}{AA} = \mathrm{{id}} \), and \( A \) is invertible. | Proof. From the equality\n\n\[ \langle x, y\rangle = \langle {Ax},{Ay}\rangle = \left\langle {x,{}^{t}{AAy}}\right\rangle \]\n\nholding for all \( x, y \in E \), we conclude that \( {}^{t}{AA} = \mathrm{{id}} \) if \( A \) is an automorphism of \( f \) . The converse is equally clear. | Yes |
Proposition 6.1. Let \( E, F \) be free modules of dimension \( n \) over \( R \) and let \( f : E \times F \rightarrow R \) be a bilinear form. Then the following conditions are equivalent:\n\n\( f \) is non-singular on the left.\n\n\( f \) is non-singular on the right.\n\n\( f \) is non-singular.\n\nThe determinant o... | Proof. Assume that \( f \) is non-singular on the left. Fix bases of \( E \) and \( F \) relative to which we write elements of these modules as column vectors, and giving rise to the matrix \( G \) for \( f \) . Then our form is given by\n\n\[ \left( {X, Y}\right) \mapsto {}^{t}{XGY} \]\n\nwhere \( X, Y \) are column ... | Yes |
Corollary 6.4. Let the notation be as in Proposition 6.2, and let \( E = F \) , \( \mathcal{B} = {\mathcal{B}}^{\prime } \) . An \( n \times n \) matrix \( M \) is the matrix of an automorphism of the form \( f \) (relative to our basis) if and only if\n\n\[ \n{}^{t}{MGM} = G\text{.} \n\]\n\nIf this condition is satisf... | Proof. We use the definitions, together with the formula given in Proposition 6.2. We note that \( M \) is invertible, for instance because its determinant is a unit in \( R \) . | Yes |
Proposition 6.5. Let \( E \) be a free module of dimension \( n \) over \( R \), and let \( \mathbb{G} \) be a fixed basis. The map\n\n\[ f \mapsto {M}_{\mathfrak{B}}^{\mathfrak{B}}\left( f\right) \]\n\ninduces an isomorphism between the module of symmetric bilinear forms on \( E \times E \) (resp. the module of altern... | Proof. Consider first the symmetric case. Assume that \( f \) is symmetric. In terms of coordinates, let \( G = {M}_{\mathfrak{G}}\left( f\right) \) . Our form is given by \( {}^{t}{XGY} \) which must be equal to \( {}^{t}{YGX} \) by symmetry. However, \( {}^{t}{XGY} \) may be viewed as a \( 1 \times 1 \) matrix, and i... | Yes |
Proposition 7.1. Let \( f : E \times E \rightarrow R \) be a non-singular sesquilinear form. Let \( A : E \rightarrow E \) be a linear map. Then \( A \) is an automorphism of \( f \) if and only if \( {A}^{ * }A = \mathrm{{id}} \), and \( A \) is invertible. | The proof, and also the proofs of subsequent propositions, which are completely similar to those of the bilinear case, will be omitted. | No |
Proposition 7.3. Let \( E, F \) be free over \( R \), of dimension \( n \) . Let \( f : E \times F \rightarrow R \) be a non-singular sesquilinear form. Let \( \mathbb{G},{\mathbb{R}}^{\prime } \) be bases of \( E \) and \( F \) respectively over \( R \), and let \( G \) be the matrix of frelative to these bases. Let \... | \[ {\left( {\bar{G}}^{-1}\right) }^{t}\bar{M}\bar{G} \] | Yes |
Lemma 8.1. The matrices\n\n\\[ \nX\\left( b\\right) = \\left( \\begin{array}{ll} 1 & b \\\\ 0 & 1 \\end{array}\\right) \\text{ and }\\;Y\\left( c\\right) = \\left( \\begin{array}{ll} 1 & 0 \\\\ c & 1 \\end{array}\\right) \n\\]\n\ngenerate \\( S{L}_{2}\\left( F\\right) \\) . | Proof. Multiplying an arbitrary element of \\( S{L}_{2}\\left( F\\right) \\) by matrices of the\n\nabove type on the right and on the left corresponds to elementary row and column operations, that is adding a scalar multiple of a row to the other, etc. Thus a given matrix can always be brought into a form\n\n\\[ \n\\le... | Yes |
Proposition 8.2. The Borel subgroup \( B \) is a maximal proper subgroup. | Proof. By the Bruhat decomposition, any element not in \( B \) lies in \( {BwB} \) , so the assertion follows since \( B,{BwB} \) cover \( G \) . | Yes |
Theorem 8.3. If \( F \) has at least four elements, then \( S{L}_{2}\left( F\right) \) is equal to its own commutator group. | Proof. We have the commutator relation (by matrix multiplication)\n\n\[ s\left( a\right) u\left( b\right) s{\left( a\right) }^{-1}u{\left( b\right) }^{-1} = u\left( {b{a}^{2} - b}\right) = u\left( {b\left( {{a}^{2} - 1}\right) }\right) .\n\]\n\nLet \( G = S{L}_{2}\left( F\right) \) for this proof. We let \( {G}^{\prime... | Yes |
Theorem 8.4. If \( F \) has at least four elements, then \( S{L}_{2}\left( F\right) /Z \) is simple. | The proof will result from two lemmas.\n\nLemma 8.5. The intersection of all conjugates of \( B \) in \( G \) is equal to \( Z \).\n\nProof. We leave this to the reader, as a simple fact using conjugation with \( w \).\n\nLemma 8.6. Let \( G = S{L}_{2}\left( F\right) \). If \( H \) is normal in \( G \), then either \( ... | No |
Lemma 8.5. The intersection of all conjugates of \( B \) in \( G \) is equal to \( Z \) . | Proof. We leave this to the reader, as a simple fact using conjugation with \( w \) . | No |
Lemma 8.6. Let \( G = S{L}_{2}\left( F\right) \) . If \( H \) is normal in \( G \), then either \( H \subset Z \) or \( H \supset {G}^{\prime } \) . | Proof. By the maximality of \( B \) we must have \[ {HB} = B\text{ or }{HB} = G. \] If \( {HB} = B \) then \( H \subset B \) . Since \( H \) is normal, we conclude that \( H \) is contained in every conjugate of \( B \), whence in the center by Lemma 8.5. On the other hand, suppose that \( {HB} = G \) . Write \[ w = {h... | Yes |
Proposition 9.1. The group \( S{L}_{n}\left( F\right) \) is generated by the elementary matrices. If \( A \in G{L}_{n}\left( F\right) \), then \( A \) can be written in the form\n\n\[ A = {SD}, \]\n\nwhere \( S \in S{L}_{n}\left( F\right) \) and \( D \) is a diagonal matrix of the form\n\n\[ D = \left( \begin{matrix} 1... | Proof. Let \( A \in G{L}_{n}\left( F\right) \) . Since \( A \) is non-singular, the first component of some row is not zero, and by an elementary row operation, we can make \( {a}_{11} \neq 0 \) . Adding a suitable multiple of the first row to the second row, we make \( {a}_{21} \neq 0 \), and then adding a suitable mu... | Yes |
Theorem 9.2. For \( n \geqq 3, S{L}_{n}\left( F\right) \) is equal to its own commutator group. | Proof. It suffices to prove that \( {E}_{ij}\left( c\right) \) is a commutator. Using \( n \geqq 3 \), let \( k \neq i, j \) . Then by direct computation,\n\n\[ \n{E}_{ij}\left( c\right) = {E}_{ik}\left( c\right) {E}_{kj}\left( 1\right) {E}_{ik}\left( {-c}\right) {E}_{kj}\left( {-1}\right) \n\]\n\nexpresses \( {E}_{ij}... | Yes |
Lemma 9.4. For \( n \geqq 3 \), the transvections \( \neq I \) form a single conjugacy class in \( S{L}_{n}\left( F\right) \) . | Proof. First, by picking a basis of a hyperplane \( H = {H}_{\lambda } \) and using one more element to form a basis of \( {F}^{\left( n\right) } \), one sees from the matrix of a transvection \( T \) that det \( T = 1 \), i.e. transvections are in \( S{L}_{n}\left( F\right) \) .\n\nLet \( {T}^{\prime } \) be another t... | Yes |
Lemma 9.5. Let \( n \geqq 3 \) . Let \( G \) be \( S{L}_{n} \) -invariant, and suppose that \( G \) contains a transvection \( T \neq I \) . Then \( S{L}_{n}\left( F\right) \subset G \) . | Proof. By Lemma 9.4, all transvections are conjugate, and the set of transvections contains the elementary matrices which generate \( S{L}_{n}\left( F\right) \) by Proposition 9.1, so the lemma follows. | Yes |
Theorem 9.6. Let \( n \geqq 3 \) . If \( G \) is a subgroup of \( G{L}_{n}\left( F\right) \) which is \( S{L}_{n} \) -invariant and which is not contained in the center of \( G{L}_{n}\left( F\right) \), then \( S{L}_{n}\left( F\right) \subset G \) . | Proof. By the preceding lemma, it suffices to prove that \( G \) contains a transvection, and this is the key step in the proof of Theorem 9.3.\n\nWe start with an element \( A \in G \) which moves some line. This is possible since \( G \) is not contained in the center. So there exists a vector \( u \neq 0 \) such tha... | Yes |
Theorem 2.1. Let \( E \) be a non-zero finite-dimensional space over the field \( k \) , and let \( A \in {\operatorname{End}}_{k}\left( E\right) \) . Then \( E \) admits a direct sum decomposition\n\n\[ E = {E}_{1} \oplus \cdots \oplus {E}_{r} \]\n\nwhere each \( {E}_{i} \) is a principal \( k\left\lbrack A\right\rbra... | Proof. The first statement is simply a rephrasing in the present language for the structure theorem for modules over principal rings. Furthermore, it is clear that \( {q}_{r}\left( A\right) = 0 \) since \( {q}_{i} \mid {q}_{r} \) for each \( i \) . No polynomial of lower degree than \( {q}_{r} \) can annihilate \( E \)... | Yes |
Corollary 2.2. Let \( {k}^{\prime } \) be an extension field of \( k \) and let \( A \) be an \( n \times n \) matrix in \( k \) . The invariants of \( A \) over \( k \) are the same as its invariants over \( {k}^{\prime } \) . | Proof. Let \( \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) be a basis of \( {k}^{\left( n\right) } \) over \( k \) . Then we may view it also as a basis of \( {k}^{\prime \left( n\right) } \) over \( {k}^{\prime } \) . (The unit vectors are in the \( k \) -space generated by \( {v}_{1},\ldots ,{v}_{n} \) ; hence \( {v}... | Yes |
Corollary 2.3. Let \( A, B \) be \( n \times n \) matrices over a field \( k \) and let \( {k}^{\prime } \) be an extension field of \( k \) . Assume that there is an invertible matrix \( {C}^{\prime } \) in \( {k}^{\prime } \) such that \( B = {C}^{\prime }A{C}^{\prime - 1} \) . Then there is an invertible matrix \( C... | Proof. Exercise. | No |
Theorem 2.4. Let \( {q}_{A}\left( t\right) = {\left( t - \alpha \right) }^{e} \) for some \( \alpha \in k, e \geqq 1 \) . Assume that \( E \) is isomorphic to \( k\left\lbrack t\right\rbrack /\left( q\right) \) . Then \( E \) has a basis over \( k \) such that the matrix of \( A \) relative to this basis is of type\n\n... | Proof. Since \( E \) is isomorphic to \( k\left\lbrack t\right\rbrack /\left( q\right) \), there exists an element \( v \in E \) such that \( k\left\lbrack t\right\rbrack v = E \) . This element corresponds to the unit element of \( k\left\lbrack t\right\rbrack \) in the isomorphism\n\n\[ k\left\lbrack t\right\rbrack /... | Yes |
Theorem 2.6. Two pairs \( \left( {E, A}\right) \) and \( \left( {F, B}\right) \) are isomorphic if and only if they have the same invariants. | You can prove this as Exercise 19. The Jordan basis gives a normalized form for the matrix associated with such a pair and an appropriate basis. | No |
Corollary 2.7. Let \( k \) be a field and let \( K \) be a finite separable extension of degree \( n \) . Let \( V \) be a finite dimensional vector space of dimension \( n \) over \( k \), and let \( \rho ,{\rho }^{\prime } : K \rightarrow {\operatorname{End}}_{k}\left( V\right) \) be two representations of \( K \) on... | Proof. By the primitive element theorem of field theory, there exists an element \( \alpha \in K \) such that \( K = k\left\lbrack \alpha \right\rbrack \) . Let \( p\left( t\right) \) be the irreducible polynomial of \( \alpha \) over \( k \) . Then \( \left( {V,\rho \left( \alpha \right) }\right) \) and \( \left( {V,{... | Yes |
Theorem 3.1. (Cayley-Hamilton). We have \( {P}_{A}\left( A\right) = 0 \) . | Proof. Let \( \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) be a basis of \( E \) over \( k \) . Then\n\n\[ t{v}_{j} = \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{ij}{v}_{i} \]\n\nwhere \( \left( {a}_{ij}\right) = M \) is the matrix of \( A \) with respect to the basis. Let \( \widetilde{B}\left( t\right) \) be the matrix w... | Yes |
Theorem 3.2. The eigenvalues of \( A \) are precisely the roots of the characteristic polynomial of \( A \) . | Proof. Let \( \lambda \) be an eigenvalue. Then \( A - {\lambda I} \) is not invertible in \( {\operatorname{End}}_{k}\left( E\right) \) , and hence \( \det \left( {A - {\lambda I}}\right) = 0 \) . Hence \( \lambda \) is a root of \( {P}_{A} \) . The arguments are reversible, so we also get the converse. | Yes |
Theorem 3.3. Let \( {w}_{1},\ldots ,{w}_{m} \) be non-zero eigenvectors of \( A \), having distinct eigenvalues. Then they are linearly independent. | Proof. Suppose that we have\n\n\[ \n{a}_{1}{w}_{1} + \cdots + {a}_{m}{w}_{m} = 0 \n\]\n\nwith \( {a}_{i} \in k \), and let this be a shortest relation with not all \( {a}_{i} = 0 \) (assuming such exists). Then \( {a}_{i} \neq 0 \) for all \( i \) . Let \( {\lambda }_{1},\ldots ,{\lambda }_{m} \) be the eigenvalues of ... | Yes |
Corollary 3.4. If \( A \) has \( n \) distinct eigenvalues \( {\lambda }_{1},\ldots ,{\lambda }_{n} \) belonging to eigenvectors \( {v}_{1},\ldots ,{v}_{n} \), and \( \dim E = n \), then \( \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) is a basis for \( E \) . The matrix of \( A \) with respect to this basis is the diag... | \[ \left( \begin{array}{llll} {\lambda }_{1} & & & 0 \\ & {\lambda }_{2} & & \\ & & \ddots & \\ 0 & & & {\lambda }_{n} \end{array}\right) \] | Yes |
Theorem 3.5. Let \( E \) be a finite-dimensional vector space over a field \( k \), let \( A \in {\operatorname{End}}_{k}\left( E\right) \), and let \( {q}_{1},\ldots ,{q}_{r} \) be the invariants of \( \left( {E, A}\right) \) . Then\n\n\[ \n{P}_{A}\left( t\right) = {q}_{1}\left( t\right) \cdots {q}_{r}\left( t\right) ... | Proof. We assume that \( E = {k}^{\left( n\right) } \) and that \( A \) is represented by a matrix \( M \) . We have seen that the invariants do not change when we extend \( k \) to a larger field, and neither does the characteristic polynomial. Hence we may assume that \( k \) is algebraically closed. In view of Theor... | Yes |
Corollary 3.6. The minimal polynomial of \( A \) and its characteristic polynomial have the same irreducible factors. | Proof. Because \( {q}_{r} \) is the minimal polynomial, by Theorem 2.1. | No |
Theorem 3.7. Let \( k \) be a commutative ring, and in the following diagram,\n\n\n\nlet the rows be exact sequences of free modules over \( k \), of finite dimension, and let the vertical maps be \( k \) -linear map... | Proof. We may assume that \( {E}^{\prime } \) is a submodule of \( E \) . We select a basis \( \left\{ {{v}_{1},\ldots ,{v}_{m}}\right\} \) for \( {E}^{\prime } \) . Let \( \left\{ {{\widetilde{v}}_{m + 1},\ldots ,\widetilde{v}}\right\} \) be a basis for \( {E}^{\prime \prime } \), and let \( {v}_{m + 1},\ldots ,{v}_{n... | Yes |
Theorem 3.8. Let \( k \) be a commutative ring, and \( E \) a free module of dimension \( n \) over \( k \) . Let \( A \in {\operatorname{End}}_{k}\left( E\right) \) . Let\n\n\[ \n{P}_{A}\left( t\right) = {t}^{n} + {c}_{n - 1}{t}^{n - 1} + \cdots + {c}_{0}.\n\]\n\nThen\n\n\[ \n\operatorname{tr}\left( A\right) = - {c}_{... | Proof. For the determinant, we observe that \( {P}_{A}\left( 0\right) = {c}_{0} \) . Substituting \( t = 0 \) in the definition of the characteristic polynomial by the determinant shows that \( {c}_{0} = {\left( -1\right) }^{n}\det \left( A\right) \) .\n\nFor the trace, let \( M \) be the matrix representing \( A \) wi... | Yes |
Corollary 3.9. Let the notation be as in Theorem 3.7. Then\n\n\\[ \n\\operatorname{tr}\\left( A\\right) = \\operatorname{tr}\\left( {A}^{\\prime }\\right) + \\operatorname{tr}\\left( {A}^{\\prime \\prime }\\right) \\;\\text{ and }\\;\\det \\left( A\\right) = \\det \\left( {A}^{\\prime }\\right) \\det \\left( {A}^{\\pri... | Proof. Clear. | No |
Theorem 3.10. Let \( k \) be a commutative ring, \( M \) an \( n \times n \) matrix in \( k \), and \( f \) a polynomial in \( k\left\lbrack t\right\rbrack \) . Assume that \( {P}_{M}\left( t\right) \) has a factorization,\n\n\[ \n{P}_{M}\left( t\right) = \mathop{\prod }\limits_{{i = 1}}^{n}\left( {t - {\alpha }_{i}}\r... | Proof. Assume first that \( k \) is a field. Then using the canonical decomposition in terms of matrices given in Theorem 2.4, we find that our assertion is immediately obvious. When \( k \) is a ring, we use a substitution argument. It is however necessary to know that if \( X = \left( {x}_{ij}\right) \) is a matrix w... | Yes |
Proposition 1.1. Let \( E \) be a vector space over the field \( k \), and let \( g \) be a form of one of the three above types. Suppose that \( E \) is expressed as an orthogonal sum,\n\n\[ E = {E}_{1} \bot \cdots \bot {E}_{m} \]\n\nThen \( g \) is non-degenerate on \( E \) if and only if it is non-degenerate on each... | Proof. Elements \( v \) , \( w \) of \( E \) can be written uniquely\n\n\[ v = \mathop{\sum }\limits_{{i = 1}}^{m}{v}_{i},\;w = \mathop{\sum }\limits_{{i = 1}}^{m}{w}_{i} \]\n\nwith \( {v}_{i},{w}_{i} \in {E}_{i} \) . Then\n\n\[ v \cdot w = \mathop{\sum }\limits_{{i = 1}}^{m}{v}_{i} \cdot {w}_{i} \]\n\nand \( v \cdot w... | Yes |
Proposition 1.2. Let \( E \) be a finite-dimensional space over the field \( k \), and let \( g \) be a form of the preceding type on \( E \) . Assume that \( g \) is non-degenerate. Let \( F \) be a subspace of \( E \) . The form is non-degenerate on \( F \) if and only if \( F + {F}^{ \bot } = E \), and also if and o... | Proof. We have (as a trivial consequence of Chapter III, §5)\n\n\[ \n\dim F + \dim {F}^{ \bot } = \dim E = \dim \left( {F + {F}^{ \bot }}\right) + \dim \left( {F \cap {F}^{ \bot }}\right) .\n\]\n\nHence \( F + {F}^{ \bot } = E \) if and only if \( \dim \left( {F \cap {F}^{ \bot }}\right) = 0 \) . Our first assertion fo... | Yes |
Proposition 2.1. Assume that \( F \) is without 2-torsion. Let \( f : E \rightarrow F \) be quadratic, expressed as above in terms of a symmetric bilinear map and a linear map. Then \( g \) , \( h \) are uniquely determined by \( f \) . For all \( x, y \in E \) we have\n\n\[ \n{2g}\left( {x, y}\right) = f\left( {x + y}... | Proof. If we compute \( f\left( {x + y}\right) - f\left( x\right) - f\left( y\right) \), then we obtain \( {2g}\left( {x, y}\right) \) . If \( {g}_{1} \) is symmetric bilinear, \( {h}_{1} \) is linear, and \( f\left( x\right) = {g}_{1}\left( {x, x}\right) + {h}_{1}\left( x\right) \), then \( {2g}\left( {x, y}\right) = ... | Yes |
Proposition 2.2. Let \( f : E \rightarrow F \) be a map such that \( {\Delta f} \) is bilinear. Assume that \( F \) is uniquely divisible by 2 . Then the map \( x \mapsto f\left( x\right) - \frac{1}{2}{\Delta f}\left( {x, x}\right) \) is Z-linear. If \( f \) satisfies the condition \( f\left( {2x}\right) = {4f}\left( x... | Proof. Obvious. | No |
Theorem 3.1. Let \( E \) be \( \neq 0 \) and finite dimensional over \( k \) . Let \( g \) be a symmetric form on \( E \) . Then there exists an orthogonal basis. | Proof. We assume first that \( g \) is non-degenerate, and prove our assertion by induction in that case. If the dimension \( n \) is 1, then our assertion is obvious.\n\nAssume \( n > 1 \) . Let \( {v}_{1} \in E \) be such that \( {v}_{1}^{2} \neq 0 \) (such an element exists since \( g \) is assumed non-degenerate). ... | Yes |
Corollary 3.2. Let \( \\left\\{ {{v}_{1},\\ldots ,{v}_{n}}\\right\\} \) be an orthogonal basis of \( E \) . Assume that \( {v}_{i}^{2} \\neq 0 \) for \( i \\leqq r \) and \( {v}_{i}^{2} = 0 \) for \( i > r \) . Then the kernel of \( E \) is equal to \( \\left\\{ {{v}_{r + 1},\\ldots ,{v}_{n}}\\right\\} \) . | Proof. Obvious. | No |
Theorem 4.1. (Sylvester) Let \( k \) be an ordered field and let \( E \) be a finite dimensional vector space over \( k \), with a non-degenerate symmetric form \( g \) . There exists an integer \( r \geqq 0 \) such that, if \( \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) is an orthogonal basis of \( E \) , then precis... | Proof. Let \( {a}_{i} = {v}_{i}^{2} \), for \( i = 1,\ldots, n \) . After renumbering the basis elements, say \( {a}_{1},\ldots ,{a}_{r} > 0 \) and \( {a}_{i} < 0 \) for \( i > r \) . Let \( \left\{ {{w}_{1},\ldots ,{w}_{n}}\right\} \) be any orthogonal basis, and let \( {b}_{i} = {w}_{i}^{2} \) . Say \( {b}_{1},\ldots... | Yes |
Corollary 4.2. Assume that every positive element of \( k \) is a square. Then there exists an orthogonal basis \( \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) of \( E \) such that \( {v}_{i}^{2} = 1 \) for \( i \leqq r \) and \( {v}_{i}^{2} = - 1 \) for \( i > r \), and \( r \) is uniquely determined. | Proof. We divide each vector in an orthogonal basis by the square root of the absolute value of its square. | No |
Theorem 5.1. There exists an orthogonal basis. If the form is non-degenerate, there exists an integer \( r \) having the following property. If \( \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) is an orthogonal basis, then precisely \( r \) among the \( n \) elements | The proofs of Theorem 5.1 and its corollaries are identical with those of the analogous results for symmetric forms, and will be left to the reader. | No |
Corollary 5.3. Assume that the form is non-degenerate. Then \( E \) admits an orthogonal decomposition \( E = {E}^{ + } \bot {E}^{ - } \) such that the form is positive definite on \( {E}^{ + } \) and negative definite on \( {E}^{ - } \) . The dimension of \( {E}^{ + } \) (or \( {E}^{ - } \) ) is the same in all such d... | The proofs of Theorem 5.1 and its corollaries are identical with those of the analogous results for symmetric forms, and will be left to the reader. | No |
Proposition 6.1. \( A \) is hermitian if and only if \( \langle {Ax}, x\rangle \) is real for all \( x \in E \) . | Proof. Let \( A \) be hermitian. Then\n\n\[\n\langle \overline{{Ax}, x}\rangle = \langle \overline{x,{Ax}}\rangle = \langle {Ax}, x\rangle\n\]\n\nwhence \( \langle {Ax}, x\rangle \) is real. Conversely, assume \( \langle {Ax}, x\rangle \) is real for all \( x \) . Then\n\n\[\n\langle {Ax}, x\rangle = \langle \overline{... | Yes |
Proposition 6.2. Let \( A \) be hermitian. Then all eigenvalues belonging to nonzero eigenvectors of \( A \) are real. If \( \xi ,{\xi }^{\prime } \) are eigenvectors \( \neq 0 \) having eigenvalues \( \lambda ,{\lambda }^{\prime } \) respectively, and if \( \lambda \neq {\lambda }^{\prime } \), then \( \xi \bot {\xi }... | Proof. Let \( \lambda \) be an eigenvalue, belonging to the eigenvector \( \xi \neq 0 \) . Then \( \langle {A\xi },\xi \rangle = \langle \xi ,{A\xi }\rangle \), and these two numbers are equal respectively to \( \lambda \langle \xi ,\xi \rangle \) and \( \bar{\lambda }\langle \xi ,\xi \rangle \) . Since \( \xi \neq 0 \... | Yes |
Lemma 6.3. Let \( A : E \rightarrow E \) be a linear map, and \( \dim E \geqq 1 \) . Then there exists at least one non-zero eigenvector of \( A \) . | Proof. We consider \( \mathbf{C}\left\lbrack A\right\rbrack \), i.e. the ring generated by \( A \) over \( \mathbf{C} \) . As a vector space over \( \mathbf{C} \), it is contained in the ring of endomorphisms of \( E \), which is finite dimensional, the dimension being the same as for the ring of all \( n \times n \) m... | Yes |
Theorem 6.4. (Spectral Theorem, Hermitian Case). Let \( E \) be a nonzero finite dimensional vector space over the complex numbers, with a positive definite hermitian form. Let \( A : E \rightarrow E \) be a hermitian linear map. Then \( E \) has an orthogonal basis consisting of eigenvectors of \( A \) . | Proof. Let \( {\xi }_{1} \) be a non-zero eigenvector, with eigenvalue \( {\lambda }_{1} \), and let \( {E}_{1} \) be the subspace generated by \( {\xi }_{1} \) . Then \( A \) maps \( {E}_{1}^{ \bot } \) into itself, because\n\n\[ \left\langle {A{E}_{1}^{ \bot },{\xi }_{1}}\right\rangle = \left\langle {{E}_{1}^{ \bot }... | Yes |
Corollary 6.5. Hypotheses being as in the theorem, there exists an orthonormal basis consisting of eigenvectors of \( A \) . | Proof. Divide each vector in an orthogonal basis by its norm. | No |
Corollary 6.6. Let \( E \) be a non-zero finite dimensional vector space over the complex numbers, with a positive definite hermitian form \( f \) . Let \( g \) be another hermitian form on \( E \) . Then there exists a basis of \( E \) which is orthogonal for both \( f \) and \( g \) . | Proof. We write \( f\left( {x, y}\right) = \langle x, y\rangle \) . Since \( f \) is non-singular, being positive definite, there exists a unique hermitian linear map \( A \) such that \( g\left( {x, y}\right) = \langle {Ax}, y\rangle \) for all \( x, y \in E \) . We apply the theorem to \( A \), and find a basis as in... | Yes |
Theorem 6.7. (Spectral Theorem, Unitary Case). Let \( E \) be a non-zero finite dimensional vector space over the complex numbers, with a positive definite hermitian form. Let \( U : E \rightarrow E \) be a unitary linear map. Then \( E \) has an orthogonal basis consisting of eigenvectors of \( U \) . | Proof. Let \( {\xi }_{1} \neq 0 \) be an eigenvector of \( U \) . It is immediately verified that the subspace of \( E \) orthogonal to \( {\xi }_{1} \) is mapped into itself by \( U \), using the relation \( {U}^{ * } = {U}^{-1} \), because if \( \eta \) is perpendicular to \( {\xi }_{1} \), then\n\n\[ \left\langle {{... | Yes |
Proposition 6.8. Let \( P \) be semipositive. Then \( P \) has a unique semipositive square root \( B : E \rightarrow E \), i.e. a semipositive linear map such that \( {B}^{2} = P \) . | Proof. For simplicity, we assume that \( P \) is positive definite. By the spectral theorem, there exists a basis of \( E \) consisting of eigenvectors. The eigenvalues must be \( > 0 \) (immediate from the condition of positivity). The linear map defined by sending each eigenvector to its multiple by the square root o... | No |
Theorem 6.9. Let \( A : E \rightarrow E \) be an invertible linear map. Then \( A \) can be written in a unique way as a product \( A = {UP} \), where \( U \) is unitary and \( P \) is positive definite. | Proof. Let \( P = {\left( {A}^{ * }A\right) }^{1/2} \), and let \( U = A{P}^{-1} \) . Using the defiitions, it is immediately verified that \( U \) is unitary, so we get the existence of the decomposition. As for uniqueness, suppose \( A = {U}_{1}{P}_{1} \) . Let\n\n\[ \n{U}_{2} = P{P}_{1}^{-1} = {U}^{-1}{U}_{1} \n\]\n... | Yes |
Theorem 7.1. (Spectral Theorem, Symmetric Case). Let \( E \neq 0 \) . Let \( A : E \rightarrow E \) be a symmetric linear map. Then \( E \) has an orthogonal basis consisting of eigenvectors of \( A \) . | Proof. If we select an orthogonal basis for the positive definite form, then the matrix of \( A \) with respect to this basis is a real symmetric matrix, and we are reduced to considering the case when \( E = {\mathbf{R}}^{n} \) . Let \( M \) be the matrix representing \( A \) . We may view \( M \) as operating on \( {... | Yes |
Corollary 7.2. Hypotheses being as in the theorem, there exists an orthonormal basis consisting of eigenvectors of \( A \) . | Proof. Divide each vector in an orthogonal basis by its norm. | No |
Corollary 7.3. Let \( E \) be a non-zero finite dimensional vector space over the reals, with a positive definite symmetric form \( f \) . Let \( g \) be another symmetric form on \( E \) . Then there exists a basis of \( E \) which is orthogonal for both \( f \) and \( g \) . | Proof. We write \( f\left( {x, y}\right) = \langle x, y\rangle \) . Since \( f \) is non-singular, being positive definite, there exists a unique symmetric linear map \( A \) such that\n\n\[ g\left( {x, y}\right) = \langle {Ax}, y\rangle \]\n\nfor all \( x, y \in E \) . We apply the theorem to \( A \), and find a basis... | No |
Theorem 8.1. Let \( f \) be an alternating form on the finite dimensional vector space \( E \) over \( k \) . Then \( E \) is an orthogonal sum of its kernel and a hyperbolic subspace. If \( E \) is non-degenerate, then \( E \) is a hyperbolic space, and its dimension is even. | Proof. A complementary subspace to the kernel is non-degenerate, and hence we may assume that \( E \) is non-degenerate. Let \( w \in E, w \neq 0 \) . There exists \( y \in E \) such that \( w \cdot y \neq 0 \) and \( y \neq 0 \) . Then \( \left( {w, y}\right) \) is non-degenerate, hence is a hyperbolic plane \( P \) .... | Yes |
Corollary 8.2. All alternating non-degenerate forms of a given dimension over a field \( k \) are isometric. | We see from Theorem 8.1 that there exists a basis of \( E \) such that relative to this basis, the matrix of the alternating form is \n\nFor convenience of writing, we reorder the basis elements of our orthogonal sum... | Yes |
Corollary 8.3. Let \( E \) be a finite dimensional vector space over \( k \), with a non-degenerate symmetric form denoted by \( \langle \) , \( \rangle \) . Let \( \Omega \) be a non-degenerate alternating form on \( E \) . Then there exists a direct sum decomposition \( E = {E}_{1} \oplus {E}_{2} \) and a symmetric a... | Proof. Take a basis of \( E \) such that the matrix of \( \Omega \) with respect to this basis is the standard alternating matrix. Let \( f \) be the symmetric non-degenerate form on \( E \) given by the dot product with respect to this basis. Then we obtain a direct sum decomposition of \( E \) into subspaces \( {E}_{... | Yes |
Theorem 9.1. Let \( R \) be a commutative ring. Let \( \left( {g}_{ij}\right) = G \) be an alternating matrix with \( {g}_{ij} \in R \) . Then\n\n\[ \det \left( G\right) = {\left( \operatorname{Pf}\left( G\right) \right) }^{2}. \]\n\nFurthermore, if \( C \) is an \( n \times n \) matrix in \( R \), then\n\n\[ \operator... | Proof. The first statement has been proved above. The second statement will follow if we can prove it over \( \mathbf{Z} \) . Let \( {u}_{ij}\left( {i, j = 1,\ldots, n}\right) \) be algebraically independent over \( \mathbf{Q} \), and such that \( {u}_{ij},{t}_{ij} \) are algebraically independent over \( \mathbf{Q} \)... | Yes |
Lemma 10.1. Let \( E \) be a finite dimensional vector space over \( k \), with a nondegenerate symmetric form \( g \) . Let \( F \) be a subspace, \( {F}_{0} \) the kernel of \( F \), and suppose we have an orthogonal decomposition\n\n\[ F = {F}_{0} \bot U. \]\n\nLet \( \\left\\{ {{w}_{1},\\ldots ,{w}_{s}}\\right\\} \... | Proof. Let\n\n\[ {U}_{1} = \\left( {{w}_{2},\\ldots ,{w}_{s}}\\right) \\oplus U. \]\n\nThen \( {U}_{1} \) is contained in \( {F}_{0} \\oplus U \) properly, and consequently \( {\\left( {F}_{0} \\oplus U\\right) }^{ \\bot } \) is contained in \( {U}_{1}^{ \\bot } \) properly. Hence there exists an element \( {u}_{1} \\i... | Yes |
Corollary 10.3. Let \( E,{E}^{\prime } \) be finite dimensional vector spaces with nondegenerate symmetric forms, and assume that they are isometric. Let \( F,{F}^{\prime } \) be subspaces, and let \( \sigma : F \rightarrow {F}^{\prime } \) be an isometry. Then \( \sigma \) can be extended to an isometry of \( E \) ont... | Proof. Clear. | No |
Corollary 10.5. Let \( E \) be a finite dimensional vector space with a nondegenerate symmetric form. Let \( W \) and \( {W}^{\prime } \) be maximal null subspaces. Let \( H \) , \( {H}^{\prime } \) be hyperbolic enlargements of \( W,{W}^{\prime } \) respectively. Then \( H,{H}^{\prime } \) are isometric and so are \( ... | Proof. We have obviously an isometry of \( H \) on \( {H}^{\prime } \), which can be extended to an isometry of \( E \) onto itself. This isometry maps \( {H}^{ \bot } \) on \( {H}^{\prime \bot } \), as desired. | Yes |
Corollary 10.6. Let \( {g}_{1},{g}_{2} \) , \( h \) be symmetric forms on finite dimensional vector spaces over the field of \( k \) . If \( {g}_{1} \oplus h \) is isometric to \( {g}_{2} \oplus h \), and if \( {g}_{1},{g}_{2} \) are non-degenerate, then \( {g}_{1} \) is isometric to \( {g}_{2} \) . | Proof. Let \( {g}_{1} \) be a form on \( {E}_{1} \) and \( {g}_{2} \) a form on \( {E}_{2} \) . Let \( h \) be a form on \( F \) . Then we have an isometry between \( F \oplus {E}_{1} \) and \( F \oplus {E}_{2} \) . Extend the identity id : \( F \rightarrow F \) to an isometry \( \sigma \) of \( F \oplus {E}_{1} \) to ... | Yes |
Proposition 1.1. Let \( {E}_{1},{E}_{2},{E}_{3} \) be modules. Then there exists a unique isomorphism\n\n\[ \n\\left( {{E}_{1} \\otimes {E}_{2}}\\right) \\otimes {E}_{3} \\rightarrow {E}_{1} \\otimes \\left( {{E}_{2} \\otimes {E}_{3}}\\right) \n\]\n\nsuch that\n\n\[ \n\\left( {x \\otimes y}\\right) \\otimes z \\mapsto ... | Proof. Since elements of type \( \\left( {x \\otimes y}\\right) \\otimes z \) generate the tensor product, the uniqueness of the desired linear map is obvious. To prove its existence, let \( x \\in {E}_{1} \) . The map\n\n\[ \n{\\lambda }_{x} : {E}_{2} \\times {E}_{3} \\rightarrow \\left( {{E}_{1} \\otimes {E}_{2}}\\ri... | Yes |
Proposition 1.2. Let \( E, F \) be modules. Then there is a unique isomorphism\n\n\[ E \otimes F \rightarrow F \otimes E \]\n\nsuch that \( x \otimes y \mapsto y \otimes x \) for \( x \in E \) and \( y \in F \) . | Proof. The map \( E \times F \rightarrow F \otimes E \) such that \( \left( {x, y}\right) \mapsto y \otimes x \) is bilinear, and factors through the tensor product \( E \otimes F \), sending \( x \otimes y \) on \( y \otimes x \) . Since this last map has an inverse (by symmetry) we obtain the desired isomorphism. | Yes |
Proposition 2.1. Let \( E = {\bigoplus }_{i = 1}^{n}{E}_{i} \) be a direct sum. Then we have an isomorphism\n\n\[ F \otimes E \leftrightarrow {\bigoplus }_{i = 1}^{n}\left( {F \otimes {E}_{i}}\right) \] | Proof. The isomorphism is given by abstract nonsense. We keep \( F \) fixed, and consider the functor \( \tau : X \mapsto F \otimes X \) . As we saw above, \( \tau \) is linear. We have projections \( {\pi }_{i} : E \rightarrow E \) of \( E \) on \( {E}_{i} \) . Then\n\n\[ {\pi }_{i} \circ {\pi }_{i} = {\pi }_{i},\;{\p... | No |
Corollary 2.2. Let \( I \) be an indexing set, and \( E = {\bigoplus }_{i \in I}{E}_{i} \) . Then we have an isomorphism\n\n\[ \left( {{\bigoplus }_{i \in I}{E}_{i}}\right) \otimes F \approx {\bigoplus }_{i \in I}\left( {{E}_{i} \otimes F}\right) \] | Proof. Let \( S \) be a finite subset of \( I \) . We have a sequence of maps\n\n\[ \left( {{\bigoplus }_{i \in S}{E}_{i}}\right) \times F \rightarrow {\bigoplus }_{i \in S}\left( {{E}_{i} \otimes F}\right) \rightarrow {\bigoplus }_{i \in I}\left( {{E}_{i} \otimes F}\right) \]\n\nthe first of which is bilinear, and the... | Yes |
Proposition 2.3. Let \( E \) be free over \( R \), with basis \( {\left\{ {v}_{i}\right\} }_{i \in I} \). Then every element of \( F \otimes E \) has a unique expression of the form\n\n\[ \mathop{\sum }\limits_{{i \in I}}{y}_{i} \otimes {v}_{i},\;{y}_{i} \in F \] \n\nwith almost all \( {y}_{i} = 0 \) . | Proof. This follows at once from the discussion of the 1-dimensional case, and the corollary of Proposition 2.1. | No |
Corollary 2.4. Let \( E, F \) be free over \( R \), with bases \( {\left\{ {v}_{i}\right\} }_{i \in I} \) and \( {\left\{ {w}_{j}\right\} }_{j \in J} \) respectively. Then \( E \otimes F \) is free, with basis \( \left\{ {{v}_{i} \otimes {w}_{j}}\right\} \) . We have\n\n\[ \dim \left( {E \otimes F}\right) = \left( {\di... | Proof. Immediate from the proposition. | No |
Proposition 2.5. Let \( E, F \) be free of finite dimension over \( R \) . Then we have an isomorphism\n\n\[ \n{\operatorname{End}}_{R}\left( E\right) \otimes {\operatorname{End}}_{R}\left( F\right) \rightarrow {\operatorname{End}}_{R}\left( {E \otimes F}\right) \n\]\n\nwhich is the unique linear map such that\n\n\[ \n... | Proof. Let \( \left\{ {v}_{i}\right\} \) be a basis of \( E \) and let \( \left\{ {w}_{j}\right\} \) be a basis of \( F \) . Then \( \left\{ {{v}_{i} \otimes {w}_{j}}\right\} \) is a basis of \( E \otimes F \) . For each pair of indices \( \left( {{i}^{\prime },{j}^{\prime }}\right) \) there exists a unique endomorphis... | Yes |
Proposition 2.6. Let\n\n\\[ \n0 \rightarrow {E}^{\prime } \\overset{\\varphi }{ \\rightarrow }E \\overset{\\psi }{ \\rightarrow }{E}^{\prime \\prime } \\rightarrow 0 \n\\]\n\nbe an exact sequence, and \\( F \\) any module. Then the sequence\n\n\\[ \nF \\otimes {E}^{\prime } \\rightarrow F \\otimes E \\rightarrow F \\ot... | Proof. Given \\( {x}^{\prime \\prime } \\in {E}^{\prime \\prime } \\) and \\( y \\in F \\), there exists \\( x \\in E \\) such that \\( {x}^{\prime \\prime } = \\psi \\left( x\\right) \\), and hence \\( y \\otimes {x}^{\prime \\prime } \\) is the image of \\( y \\otimes x \\) under the linear map\n\n\\[ \nF \\otimes E ... | Yes |
Proposition 2.7. Let \( \mathfrak{a} \) be an ideal of \( R \) . Let \( E \) be a module. Then the map \( \left( {R/\mathfrak{a}}\right) \times E \rightarrow E/\mathfrak{a}E \) induced by\n\n\[ \left( {a, x}\right) \mapsto {ax}\;\left( {{\;\operatorname{mod}\;\mathfrak{a}}E}\right) ,\;a \in R, x \in E \]\n\nis bilinear... | Proof. Our map \( \left( {a, x}\right) \mapsto {ax}\left( {{\;\operatorname{mod}\;\mathfrak{a}}E}\right) \) clearly induces a bilinear map of \( R/\mathfrak{a} \times E \) onto \( E/\mathfrak{a}E \), and hence a linear map of \( R/\mathfrak{a} \otimes E \) onto \( E/\mathfrak{a}E \) . We can construct an inverse, for w... | Yes |
Lemma 3.3. Let \( F \) be flat, and suppose that\n\n\[ 0 \rightarrow N \rightarrow M \rightarrow F \rightarrow 0 \]\n\nis an exact sequence. Then for any \( E \), we have an exact sequence\n\n\[ 0 \rightarrow N \otimes E \rightarrow M \otimes E \rightarrow F \otimes E \rightarrow 0. \] | Proof. Represent \( E \) as a quotient of a flat \( L \) by an exact sequence\n\n\[ 0 \rightarrow K \rightarrow L \rightarrow E \rightarrow 0. \]\n\nThen we have the following exact and commutative diagram:\n\n\n\nTh... | No |
Proposition 3.4. Let\n\n\[ \n0 \rightarrow {F}^{\prime } \rightarrow F \rightarrow {F}^{\prime \prime } \rightarrow 0 \]\n\nbe an exact sequence, and assume that \( {F}^{\prime \prime } \) is flat. Then \( F \) is flat if and only if \( {F}^{\prime } \) is flat. More generally, let\n\n\[ \n0 \rightarrow {F}^{0} \righta... | Proof. Let \( 0 \rightarrow {E}^{\prime } \rightarrow E \) be an injection. We have an exact and commutative diagram:\n\n\n\nThe 0 on top is by hypothesis that \( {F}^{\prime \prime } \) is flat, and the two zeros on... | Yes |
Lemma 3.5. Assume that \( F \) is \( E \) -flat. Then \( F \) is also flat for every submodule and every quotient module of \( E \) . | Proof. The submodule part is immediate because if \( {E}_{1}^{\prime } \subset {E}_{2}^{\prime } \subset E \) are submodules, and \( F \otimes {E}_{1}^{\prime } \rightarrow F \otimes E \) is a monomorphism so is \( F \otimes {E}_{1}^{\prime } \rightarrow F \otimes {E}_{2}^{\prime } \) since the composite map with \( F ... | Yes |
Lemma 3.6. Let \( \left\{ {E}_{i}\right\} \) be a family of modules, and suppose that \( F \) is flat for each \( {E}_{i} \) . Then \( F \) is flat for their direct sum. | Proof. Let \( E = \bigoplus {E}_{i} \) be their direct sum. We have to prove that given any submodule \( {E}^{\prime } \) of \( E \), the sequence \[ 0 \rightarrow F \otimes {E}^{\prime } \rightarrow F \otimes E = \bigoplus F \otimes {E}_{i} \] is exact. Note that if an element of \( F \otimes {E}^{\prime } \) becomes ... | Yes |
Proposition 3.7. \( F \) is flat if and only if for every ideal \( \mathfrak{a} \) of \( R \) the natural map\n\n\[ \mathfrak{a} \otimes F \rightarrow \mathfrak{a}F \] \n\nis an isomorphism. In fact, \( F \) is flat if and only for every ideal \( \mathfrak{a} \) of \( R \) tensoring the sequence \n\n\[ 0 \rightarrow \m... | Proof. If \( F \) is flat, then tensoring with \( F \) and using Proposition 2.7 shows that the natural map is an isomorphism, because \( \mathfrak{a}M \) is the kernel of \( M \rightarrow M/\mathfrak{a}M \) . Conversely, assume that this map is an isomorphism for all ideals \( \mathfrak{a} \) . This means that \( F \)... | Yes |
Theorem 3.8. Let \( R \) be a commutative local ring, and let \( M \) be a finite flat module over \( R \) . Then \( M \) is free. In fact, if \( {x}_{1},\ldots ,{x}_{n} \in M \) are elements of \( M \) whose residue classes are a basis of \( M/\mathfrak{m}M \) over \( R/\mathfrak{m} \), then \( {x}_{1},\ldots ,{x}_{n}... | Proof. Let \( {R}^{\left( n\right) } \rightarrow M \) be the map which sends the unit vectors of \( {R}^{\left( n\right) } \) on \( {x}_{1},\ldots ,{x}_{n} \) respectively, and let \( N \) be its kernel. We get an exact sequence\n\n\[ 0 \rightarrow N \rightarrow {R}^{\left( n\right) } \rightarrow M \]\n\nwhence a commu... | Yes |
Lemma 3.9. Assume that \( M \) is finitely presented, and let\n\n\[ 0 \rightarrow N \rightarrow E \rightarrow M \rightarrow 0 \]\n\nbe exact, with \( E \) finite free. Then \( N \) is finitely generated. | Proof. Let\n\n\[ {L}_{1} \rightarrow {L}_{2} \rightarrow M \rightarrow 0 \]\n\nbe a finite presentation of \( M \), that is an exact sequence with \( {L}_{1},{L}_{2} \) finite free. Using the freeness, there exists a commutative diagram\n\n![08abf0d1-5b8b-4939-95a5-72db9ebd5d2f_629_1.jpg](images/08abf0d1-5b8b-4939-95a5... | Yes |
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