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Lemma 4.8. If \( X = {v}_{1},\ldots ,{v}_{n} \) is a list in a vector space \( V \), then \( \langle X\rangle \) depends only on its underlying set. | Proof. If \( \sigma \in {S}_{n} \) is a permutation, then define a list \( {X}^{\sigma } = {v}_{\sigma \left( 1\right) },\ldots ,{v}_{\sigma \left( n\right) } \) . Now a linear combination of \( X \) is a vector \( v = {a}_{1}{v}_{1} + \cdots + {a}_{n}{v}_{n} \) . Since addition in \( V \) is commutative, \( v \) is al... | Yes |
Lemma 4.9. Let \( V \) be a vector space over a field \( k \) .\n\n(i) Every intersection of subspaces of \( V \) is itself a subspace. | (i) Let \( \mathcal{S} \) be a family of subspaces of \( V \), and denote \( \mathop{\bigcap }\limits_{{S \in \mathcal{S}}}S \) by \( W \) . Since \( 0 \in S \) for every \( S \in \mathcal{S} \), we have \( 0 \in W \) . If \( x, y \in W \), then \( x, y \in S \) for every \( S \in \mathcal{S} \) ; as \( S \) is a subsp... | Yes |
Given a subspace \( U \) of a vector space \( V \), we seek a list \( X \) which spans \( U \) . Notice that \( U \) can have many such lists; for example, if \( X = {v}_{1},{v}_{2},\ldots ,{v}_{m} \) spans \( U \) and \( u \) is any vector in \( U \), then \( {v}_{1},{v}_{2},\ldots ,{v}_{m}, u \) also spans \( U \) . ... | Definition. A list \( X = {v}_{1},\ldots ,{v}_{m} \) in a vector space \( V \) is a shortest spanning list (or a minimal spanning list) if no proper sublist \( {v}_{1},\ldots ,\widehat{{v}_{i}}\ldots ,{v}_{m} \) spans \( \left\langle {{v}_{1},\ldots ,{v}_{m}}\right\rangle \subseteq V \) . | Yes |
Proposition 4.11. If \( V \) is a vector space, then the following conditions on a list \( X = {v}_{1},\ldots ,{v}_{m} \) spanning \( V \) are equivalent:\n\n(i) \( X \) is not a shortest spanning list; that is, a proper sublist spans \( \langle X\rangle \) .\n\n(ii) some \( {v}_{i} \) is in the subspace spanned by the... | Proof. (i) \( \Rightarrow \) (ii). If \( X \) is not a shortest spanning list, then one of the vectors in \( X \), say, \( {v}_{i} \), can be thrown out, and \( {v}_{i} \in \left\langle {{v}_{1},\ldots ,\widehat{{v}_{i}},\ldots ,{v}_{m}}\right\rangle \) .\n\n(ii) \( \Rightarrow \) (iii). If \( {v}_{i} = \mathop{\sum }\... | Yes |
Proposition 4.15. Let \( X = {v}_{1},\ldots ,{v}_{n} \) be a list in a vector space \( V \) over a field \( k \) . Then \( X \) is a basis if and only if each vector in \( V \) has a unique expression as a linear combination of vectors in \( X \) . | Proof. If a vector \( v = \sum {a}_{i}{v}_{i} = \sum {b}_{i}{v}_{i} \), then \( \sum \left( {{a}_{i} - {b}_{i}}\right) {v}_{i} = 0 \), and so independence gives \( {a}_{i} = {b}_{i} \) for all \( i \) ; that is, the expression is unique.\n\nConversely, existence of an expression shows that the list of \( {v}_{i} \) spa... | Yes |
Theorem 4.16. Every finite-dimensional vector space \( V \) has a basis. | Proof. A finite spanning list \( X \) exists, since \( V \) is finite-dimensional. If it is linearly independent, it is a basis; if not, \( X \) can be shortened to a spanning sublist \( {X}^{\prime } \), by Proposition 4.11. If \( {X}^{\prime } \) is linearly independent, it is a basis; if not, \( {X}^{\prime } \) can... | Yes |
Lemma 4.17. Let \( {u}_{1},\ldots ,{u}_{n} \) span a vector space \( V \) . If \( {v}_{1},\ldots ,{v}_{m} \in V \) and \( m > n \), then \( {v}_{1},\ldots ,{v}_{m} \) is a linearly dependent list. | Proof. The proof is by induction on \( n \geq 1 \) .\n\nBase Step. If \( n = 1 \), then there are at least two vectors \( {v}_{1},{v}_{2} \), for \( m > n \), and \( {v}_{1} = {a}_{1}{u}_{1} \) and \( {v}_{2} = {a}_{2}{u}_{1} \) . If \( {u}_{1} = 0 \), then \( {v}_{1} = 0 \) and the list of \( v \) ’s is linearly depen... | Yes |
Theorem 4.18 (Invariance of Dimension). If \( X = {x}_{1},\ldots ,{x}_{n} \) and \( Y = \) \( {y}_{1},\ldots ,{y}_{m} \) are bases of a vector space \( V \), then \( m = n \) . | Proof. If \( m \neq n \), then either \( n < m \) or \( m < n \) . In the first case, \( {y}_{1},\ldots ,{y}_{m} \in \) \( \left\langle {{x}_{1},\ldots ,{x}_{n}}\right\rangle \), because \( X \) spans \( V \), and Lemma 4.17 gives \( Y \) linearly dependent, a contradiction. A similar contradiction arises if \( m < n \... | Yes |
Corollary 4.20. A homogeneous system of linear equations over a field \( k \) with more unknowns than equations has a nontrivial solution. | Proof. An \( n \) -tuple \( \left( {{s}_{1},\ldots ,{s}_{n}}\right) \) is a solution of a system\n\n\[ \n{a}_{11}{x}_{1} + \cdots + {a}_{1n}{x}_{n} = 0 \n\]\n\n\[ \n\text{……} \n\]\n\n\[ \n{a}_{m1}{x}_{1} + \cdots + {a}_{mn}{x}_{n} = 0 \n\]\n\nif \( {a}_{i1}{s}_{1} + \cdots + {a}_{in}{s}_{n} = 0 \) for all \( i \) . In ... | Yes |
Lemma 4.21. Let \( V \) be a finite-dimensional vector space.\n\n(i) Let \( {v}_{1},\ldots ,{v}_{m} \) be a linearly independent list in \( V \), and let \( v \in V \) . If \( v \notin \left\langle {{v}_{1},\ldots ,{v}_{m}}\right\rangle \), then \( {v}_{1},\ldots ,{v}_{m}, v \) is linearly independent.\n\n(ii) If a lon... | Proof.\n\n(i) Let \( {av} + \mathop{\sum }\limits_{i}{a}_{i}{v}_{i} = 0 \) . If \( a \neq 0 \), then \( v = - {a}^{-1}\mathop{\sum }\limits_{i}{a}_{i}{v}_{i} \in \left\langle {{v}_{1},\ldots ,{v}_{m}}\right\rangle \), a contradiction. Therefore, \( a = 0 \) and \( \mathop{\sum }\limits_{i}{a}_{i}{v}_{i} = 0 \) . But li... | Yes |
Proposition 4.22. If \( Z = {u}_{1},\ldots ,{u}_{m} \) is a linearly independent list in an \( n \) -dimensional vector space \( V \), then \( Z \) can be extended to a basis; that is, there are vectors \( {v}_{1},\ldots ,{v}_{n - m} \) so that \( {u}_{1},\ldots ,{u}_{m},{v}_{1},\ldots ,{v}_{n - m} \) is a basis of \( ... | Proof. If \( m > n \), then Lemma 4.17 implies that \( Z \) is linearly dependent, a contradiction; therefore, \( m \leq n \) . If the linearly independent list \( Z \) does not span \( V \), there is \( {v}_{1} \in V \) with \( {v}_{1} \notin \langle Z\rangle \), and the longer list \( Z,{v}_{1} = {u}_{1},\ldots ,{u}_... | Yes |
Corollary 4.23. If \( \dim \left( V\right) = n \), then any list of \( n + 1 \) or more vectors is linearly dependent. | Proof. Otherwise, such a list could be extended to a basis having too many elements. | No |
Corollary 4.24. Let \( V \) be a vector space with \( \dim \left( V\right) = n \). (i) A list \( X \) of \( n \) vectors which spans \( V \) must be linearly independent. (ii) Any linearly independent list \( Y \) of \( n \) vectors must span \( V \). | Proof. (i) If the list \( X \) is linearly dependent, then it could be shortened to give a basis of \( V \) which is too small. (ii) If the list \( Y \) does not span \( V \), then it could be lengthened to give a basis of \( V \) which is too large. \( \; \bullet \) | Yes |
Corollary 4.25. Let \( U \) be a subspace of a vector space \( V \) of dimension \( n \) .\n\n(i) Then \( U \) is finite-dimensional and \( \dim \left( U\right) \leq \dim \left( V\right) \).\n\n(ii) If \( \dim \left( U\right) = \dim \left( V\right) \), then \( U = V \) . | Proof.\n\n(i) Take \( {u}_{1} \in U \) . If \( U = \left\langle {u}_{1}\right\rangle \), then \( U \) is finite-dimensional. Otherwise, there is \( {u}_{2} \notin \left\langle {u}_{1}\right\rangle \) . By Lemma 4.21, \( {u}_{1},{u}_{2} \) is linearly independent. If \( U = \left\langle {{u}_{1},{u}_{2}}\right\rangle \)... | Yes |
Proposition 4.28. If \( q = {p}^{n} \) for some prime \( p \), then there exists a projective plane of order \( q \) . | Proof. Let \( X = {\mathbb{P}}_{2}\left( k\right) \), where \( k = {\mathbb{F}}_{q} \) . Now \( \left| {k}^{3}\right| = {q}^{3} \), and so there are \( {q}^{3} - 1 \) nonzero vectors in \( {k}^{3} \) . If \( v \in {k}^{3} \) is nonzero, then \( \left| \left\lbrack v\right\rbrack \right| = q - 1 \), for there are exactl... | Yes |
Proposition 4.29 (= Proposition 3.119). If \( E \) is a finite field, then \( \left| E\right| = {p}^{n} \) for some prime \( p \) and some \( n \geq 1 \) . | Proof. By Lemma 3.120(i), the prime field of \( E \) is isomorphic to \( {\mathbb{F}}_{p} \) for some prime \( p \) . Since \( E \) is finite, it is finite-dimensional; say, \( \dim \left( E\right) = n \) . If \( {v}_{1},\ldots ,{v}_{n} \) is a basis, then there are exactly \( {p}^{n} \) vectors \( {a}_{1}{v}_{1} + \cd... | Yes |
Proposition 4.30. Let \( E/k \) be an extension, let \( z \in E \) be a root of an irreducible polynomial \( p\left( x\right) \in k\left\lbrack x\right\rbrack \), and let \( k\left( z\right) \) be the smallest subfield of \( E \) containing \( k \) and \( z \) . Then \[ \left\lbrack {k\left( z\right) : k}\right\rbrack ... | Proof. Proposition 3.116(iv) says that each element in \( k\left( z\right) \) has a unique expression of the form \[ {b}_{0} + {b}_{1}z + \cdots + {b}_{n - 1}{z}^{n - 1} \] where \( {b}_{i} \in k \) and \( n = \deg \left( p\right) \) . In the language of linear algebra (which was not available in Chapter 3), the list \... | Yes |
Theorem 4.31. Let \( k \subseteq K \subseteq E \) be fields, with \( K \) a finite extension of \( k \) and \( E \) a finite extension of \( K \) . Then \( E \) is a finite extension of \( k \), and \[ \left\lbrack {E : k}\right\rbrack = \left\lbrack {E : K}\right\rbrack \left\lbrack {K : k}\right\rbrack \] | Proof. If \( A = {a}_{1},\ldots ,{a}_{n} \) is a basis of \( K \) over \( k \) and if \( B = {b}_{1},\ldots ,{b}_{m} \) is a basis of \( E \) over \( K \), then it suffices to prove that a list \( X \) of all \( {a}_{i}{b}_{j} \) is a basis of \( E \) over \( k \) . To see that \( X \) spans \( E \), take \( e \in E \)... | Yes |
Proposition 4.32. If \( K/k \) is a finite extension, then every \( z \in K \) is algebraic over \( k \) . | Proof. If \( \left\lbrack {K : k}\right\rbrack = n \), then the list \( 1, z,{z}^{2},\ldots ,{z}^{n} \) has length \( n + 1 \), by Corollary 4.23. Hence, there are \( {a}_{i} \in k \), not all zero, with \( \mathop{\sum }\limits_{{i = 0}}^{n}{a}_{i}{z}^{i} = 0 \) . If we define \( f\left( x\right) = \mathop{\sum }\limi... | Yes |
Proposition 4.33. If \( A \rightarrow {A}^{\prime } \) is an elementary row operation, then \( A \) and \( {A}^{\prime } \) have the same row space: \( \operatorname{Row}\left( A\right) = \operatorname{Row}\left( {A}^{\prime }\right) \) . | Proof. Suppose that \( A \rightarrow {A}^{\prime } \) is an elementary operation of Type I. The row space of \( A \) is \( \operatorname{Row}\left( A\right) = \left\langle {{\alpha }_{1},\ldots ,{\alpha }_{m}}\right\rangle \), where \( {\alpha }_{i} \) is the \( i \) th row of \( A \) ; the row space \( \operatorname{R... | Yes |
Corollary 4.34. If \( A \rightarrow {A}^{\prime } \) is an elementary row operation, then\n\n\[ \operatorname{rank}\left( A\right) = \operatorname{rank}\left( {A}^{\prime }\right) \] | Proof. Even more is true; the row spaces of \( A \) and of \( {A}^{\prime } \) are equal, and so they certainly have the same dimension. - | Yes |
Lemma 4.35. If \( A \) is an \( m \times n \) matrix and if \( A\overset{o}{ \rightarrow }{A}^{\prime } \) is an elementary row operation, then \( o\left( A\right) = o\left( I\right) A \) ; if \( A\overset{o}{ \rightarrow }{A}^{\prime } \) is an elementary column operation, then \( o\left( A\right) = {Ao}\left( I\right... | Proof. We will merely illustrate the result, leaving the proof to the reader.\n\n\[ \text{Type I}\;\left\lbrack \begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ u & 0 & 1 \end{array}\right\rbrack \left\lbrack \begin{array}{lll} a & b & c \\ d & e & f \\ g & h & i \end{array}\right\rbrack = \left\lbrack \begin{matrix} a & b... | No |
Proposition 4.36. Every elementary matrix \( E \) is a nonsingular matrix. In fact, \( {E}^{-1} \) is an elementary matrix of the same type as \( E \) . | Proof. If \( o \) is an elementary row operation of Type I, then \( o \) replaces ROW \( \left( i\right) \) by \( \operatorname{ROW}\left( i\right) + c\operatorname{ROW}\left( j\right) \) . Define \( {o}^{\prime } \) to be the elementary row operation which replaces \( \operatorname{ROW}\left( i\right) \) by \( \operat... | Yes |
Proposition 4.37. If \( A \rightarrow {A}^{\prime } \) is an elementary row operation, then the linear systems \( {Ax} = 0 \) and \( {A}^{\prime }x = 0 \) have the same solution space. | Proof. Let \( S \) and \( {S}^{\prime } \) be the solution spaces of \( {Ax} = 0 \) and \( {A}^{\prime }x = 0 \), respectively. If \( {A}^{\prime } = o\left( A\right) \), then \( {A}^{\prime } = {EA} \), by Lemma 4.35, where \( E \) is the elementary matrix \( o\left( I\right) \) . If \( v \in S \), then \( {Av} = 0 \)... | Yes |
Corollary 4.38. If \( A \) and \( B \) are \( m \times n \) matrices over a field \( k \), and if there is a sequence of elementary row operations\n\n\[ A = {A}_{0} \rightarrow {A}_{1} \rightarrow \cdots \rightarrow {A}_{p} = B, \]\n\nthen there is a nonsingular matrix \( P \) with \( B = {PA} \) . If there is a sequen... | Proof. There are elementary matrices \( {E}_{i} \) with \( {A}_{i} = {E}_{i}{A}_{i - 1} \) for all \( i \geq 1 \) . Therefore, \( B = {E}_{p}\cdots {E}_{2}{E}_{1}A \) . Define \( P = {E}_{p}\cdots {E}_{2}{E}_{1} \), so that \( B = {PA} \) . Now \( P \) is nonsingular, for the product of nonsingular matrices is nonsingu... | Yes |
Corollary 4.40. Let \( A \) be an \( m \times n \) matrix over a field \( k \) . If \( B \) is the row reduced echelon form of \( A \), then a basis of \( \operatorname{Row}\left( A\right) \) consists of the nonzero rows of \( B \) . | Proof. NowRow \( \left( A\right) = \operatorname{Row}\left( B\right) \), by Proposition 4.33, and so it is spanned by the rows of \( B \) . But it is obvious that the nonzero rows of \( B \) are linearly independent, and so they form a basis. - | Yes |
Theorem 4.41. Let \( {Ax} = 0 \) be a system of linear equations, where \( A \) is an \( m \times n \) matrix over a field \( k \), and let \( B \) be the (unique) row reduced echelon form Gaussian equivalent to \( A \) . Let the fixed variables be \( {x}_{{t}_{1}},\ldots ,{x}_{{t}_{r}} \), let the free variables be \(... | Proof. By Proposition 4.37, an \( n \) -tuple \( s = \left( {{c}_{1},\ldots ,{c}_{n}}\right) \) is a solution of \( {Ax} = 0 \) if and only if it is a solution of \( {Bx} = 0 \) . Now the \( i \) th entry of the \( n \times 1 \) matrix \( {Bs} \) is \( {c}_{{t}_{i}} + \mathop{\sum }\limits_{\ell }{b}_{i{p}_{\ell }}{c}_... | Yes |
Theorem 4.42. Let \( A \) be an \( m \times n \) matrix over a field \( k \) . If \( \operatorname{Sol}\left( A\right) \) is the solution space of the homogeneous linear system \( {Ax} = 0 \), then\n\n\[ \dim \left( {\operatorname{Sol}\left( A\right) }\right) = n - r \]\n\nwhere \( r = \operatorname{rank}\left( A\right... | Proof. Let us assume that the variables have been relabeled so that the fixed variables precede all the free variables. For each \( \ell \) with \( 1 \leq \ell \leq n - r \), define \( {s}_{\ell } \) to be the solution \( \left( {{c}_{1},\ldots ,{c}_{n}}\right) \) with \( {c}_{{p}_{\ell }} = 1 \) and \( {c}_{{p}_{v}} =... | Yes |
Lemma 4.45. A complex number \( z = x + {iy} \) is constructible if and only if its real part \( x \) and its imaginary part \( y \) are constructible. | Proof. If \( z \) is constructible, then a standard euclidean construction draws the vertical line \( L \) through \( \left( {x, y}\right) \) which is parallel to the \( y \) -axis. It follows that \( x \) is constructible, for the point \( \left( {x,0}\right) \) is constructible, being the intersection of \( L \) and ... | Yes |
Corollary 4.47. The set \( K \) of all constructible numbers is a subfield of \( \mathbb{C} \) that is closed under square roots. | Proof. If \( z = a + {ib} \) and \( w = c + {id} \) are constructible, then \( a, b, c, d \) are constructible, by Theorem 4.46, and so \( a, b, c, d \in K \cap \mathbb{R} \) . Hence, \( a + c, b + d \in \) \( K \cap \mathbb{R} \), because \( K \cap \mathbb{R} \) is a subfield of \( \mathbb{R} \), and so \( \left( {a +... | Yes |
Corollary 4.48. If \( a, b, c \) are constructible, then the roots of the quadratic \( a{x}^{2} + {bx} + c \) are also constructible. | ## Proof. This follows from the quadratic formula and Corollary 4.47. - | No |
Lemma 4.49. If \( F/k \) is a field extension, then \( \left\lbrack {F : k}\right\rbrack = 2 \) if and only \( F = k\left( u\right) \) , where \( u \in F \) is a root of some quadratic polynomial \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) . | Proof. If \( \left\lbrack {F : k}\right\rbrack = 2 \), then \( F \neq k \) and there is some \( u \in F \) with \( u \notin k \) . By Proposition 4.32, there is some (irreducible) polynomial \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) having \( u \) as a root. We have \( 2 = \left\lbrack {F : k}\right\rbrac... | Yes |
Lemma 4.51. Let \( P = a + {ib}, Q = c + {id} \in \mathcal{P} \). (i) The line \( L\left\lbrack {P, Q}\right\rbrack \) has equation \( x = a \) if it is vertical \( \left( {c = a}\right) \) or \( y = {mx} + q \) if it is not vertical \( \left( {c \neq a}\right) \), where \( m, q \in \mathcal{P} \). (ii) The circle \( C... | Proof. Lemma 4.50 gives \( a, b, c, d \in \mathcal{P} \). (i) If \( L\left\lbrack {P, Q}\right\rbrack \) is not vertical, then its equation is \( y = {mx} + q \), where \( m = \) \( \left( {d - b}\right) /\left( {c - a}\right) \) and \( q = - {ma} + b \) . Hence \( m, q \in \mathcal{P} \). (ii) The circle \( C\left\lbr... | Yes |
Proposition 4.52. Every polyquadratic number \( z \) is constructible. | Proof. If \( z \in \mathcal{P} \), then there is a 2-tower \( \mathbb{Q}\left( i\right) = {F}_{0} \subseteq {F}_{1} \subseteq \cdots \subseteq {F}_{n} \) with \( z \in {F}_{n} \) ; we prove that \( z \in K \) by induction on \( n \geq 0 \) . The base step is true, for \( {F}_{0} = \mathbb{Q}\left( i\right) \subseteq K ... | Yes |
Corollary 4.54. If a complex number \( z \) is constructible, then \( \left\lbrack {\mathbb{Q}\left( z\right) : \mathbb{Q}}\right\rbrack \) is a power of 2. | Proof. This follows from Theorems 4.53 and 4.31. - | No |
Theorem 4.55 (Wantzel). It is impossible to duplicate the cube using only straightedge and compass. | Proof. \( {}^{6} \) The question is whether \( z = \sqrt[3]{2} \) is constructible. Since \( {x}^{3} - 2 \) is irreducible, \( \left\lbrack {\mathbb{Q}\left( z\right) : \mathbb{Q}}\right\rbrack = 3 \), by Corollary 4.54; but 3 is not a power of 2. | Yes |
Theorem 4.56 (Wantzel). It is impossible to trisect a \( {60}^{ \circ } \) angle using only straightedge and compass. | Proof. We may assume that one side of the angle is on the \( x \) -axis, and so the question is whether \( z = \cos {20}^{ \circ } + i\sin {20}^{ \circ } \) is constructible. If \( z \) is constructible, then Lemma 4.45 would show that \( \cos {20}^{ \circ } \) is constructible. Corollary 1.23, the triple angle formula... | Yes |
Theorem 4.57 (Archimedes). Every angle can be trisected using ruler and compass. (Recall that a ruler is a straightedge on which points \( U \) and \( V \) can be marked; moreover, the point \( U \) is allowed to slide along a circle.) | Proof. Since it is easy to construct \( {30}^{ \circ },{60}^{ \circ } \), and \( {90}^{ \circ } \), it suffices to trisect an acute angle \( \alpha \), for if \( {3\beta } = \alpha \), then \( 3\left( {\beta + {30}^{ \circ }}\right) = \alpha + {90}^{ \circ },3\left( {\beta + {60}^{ \circ }}\right) = \alpha + {180}^{ \c... | No |
Theorem 4.58 (Lindemann). It is impossible to square the circle with straightedge and compass. | Proof. The problem is whether one can construct a square whose area is the same as the area of the unit circle. If a side of the square has length \( z \), then one is asking whether \( z = \sqrt{\pi } \) is constructible. Now \( \mathbb{Q}\left( \pi \right) \) is a subspace of \( \mathbb{Q}\left( \sqrt{\pi }\right) \)... | Yes |
Theorem 4.59 (Gauss-Wantzel). Let \( p \) be an odd prime. A regular p-gon is constructible if and only if \( p = {2}^{{2}^{t}} + 1 \) for some \( t \geq 0 \) . | Proof. We only prove necessity; for sufficiency, see Theorem 5.41. The problem is whether \( z = {e}^{{2\pi i}/p} \) is constructible. Now \( z \) is a root of the cyclotomic polynomial \( {\Phi }_{p}\left( x\right) \), which is an irreducible polynomial in \( \mathbb{Q}\left\lbrack x\right\rbrack \) of degree \( p - 1... | Yes |
Theorem 4.61. Let \( {v}_{1},\ldots ,{v}_{n} \) be a basis of a vector space \( V \) over a field \( k \) . If \( W \) is a vector space over \( k \) and \( {w}_{1},\ldots ,{w}_{n} \) is a list in \( W \), then there exists a unique linear transformation \( T : V \rightarrow W \) with \( T\left( {v}_{i}\right) = {w}_{i... | Proof. By Theorem 4.15, each \( v \in V \) has a unique expression of the form \( v = \mathop{\sum }\limits_{i}{a}_{i}{v}_{i} \), and so \( T : V \rightarrow W \), given by \( T\left( v\right) = \sum {a}_{i}{w}_{i} \), is a (well-defined!) function. It is now a routine verification to check that \( T \) is a linear tra... | Yes |
Corollary 4.62. If linear transformations \( S, T : V \rightarrow W \) agree on a basis, then \( S = T \) . | Proof. If \( {v}_{1},\ldots ,{v}_{n} \) is a basis of \( V \) and if \( S\left( {v}_{i}\right) = T\left( {v}_{i}\right) \) for all \( i \), then the uniqueness statement in Theorem 4.61 gives \( S = T \) . | Yes |
Proposition 4.63. If \( T : {k}^{n} \rightarrow {k}^{m} \) is a linear transformation, then there exists a unique \( m \times n \) matrix \( A \) such that\n\n\[ T\left( y\right) = {Ay} \]\n\nfor all \( y \in {k}^{n} \) (here, \( y \) is an \( n \times 1 \) column matrix and \( {Ay} \) is matrix multiplication). | Proof. If \( {e}_{1},\ldots ,{e}_{n} \) is the standard basis of \( {k}^{n} \) and \( {e}_{1}^{\prime },\ldots ,{e}_{m}^{\prime } \) is the standard basis of \( {k}^{m} \), define \( A = \left\lbrack {a}_{ij}\right\rbrack \) to be the matrix whose \( j \) th column is the coordinate list of \( T\left( {e}_{j}\right) \)... | Yes |
Proposition 4.69. Let \( V \) and \( W \) be vector spaces over a field \( k \), and let \( X \) and \( Y \) be bases of \( V \) and \( W \), respectively. The function \[ {\mu }_{X, Y} : {\operatorname{Hom}}_{k}\left( {V, W}\right) \rightarrow {\operatorname{Mat}}_{m \times n}\left( k\right) , \] given by \[ T \mapsto... | Proof. First, let us show that \( {\mu }_{X, Y} \) is surjective. Given a matrix \( A \), its columns define vectors in \( W \) . In more detail, if \( X = {v}_{1},\ldots ,{v}_{n} \) and \( Y = {w}_{1},\ldots ,{w}_{m} \) , then the \( j \) th column of \( A \) is \( {\left( {a}_{1j},\ldots ,{a}_{mj}\right) }^{T} \) ; d... | Yes |
Proposition 4.70. Let \( T : V \rightarrow W \) and \( S : W \rightarrow U \) be linear transformations. Choose bases \( X = {x}_{1},\ldots ,{x}_{n} \) of \( V, Y = {y}_{1},\ldots ,{y}_{m} \) of \( W \), and \( Z = {z}_{1},\ldots ,{z}_{\ell } \) of \( U \) . Then\n\n\[ \n{}_{Z}{\left\lbrack S \circ T\right\rbrack }_{X}... | Proof. Let \( {}_{Y}{\left\lbrack T\right\rbrack }_{X} = \left\lbrack {a}_{ij}\right\rbrack \), so that \( T\left( {x}_{j}\right) = \mathop{\sum }\limits_{p}{a}_{pj}{y}_{p} \), and let \( {}_{Z}{\left\lbrack S\right\rbrack }_{Y} = \left\lbrack {b}_{qp}\right\rbrack \) , so that \( S\left( {y}_{p}\right) = \mathop{\sum ... | Yes |
Corollary 4.71. Matrix multiplication is associative: \( A\left( {BC}\right) = \left( {AB}\right) C \) . | Proof. Let \( A \) be an \( m \times n \) matrix, let \( B \) be an \( n \times p \) matrix, and let \( C \) be a \( p \times q \) matrix. By Theorem 4.61, there are linear transformations\n\n\[ \n{k}^{q}\overset{T}{ \rightarrow }{k}^{p}\overset{S}{ \rightarrow }{k}^{n}\overset{R}{ \rightarrow }{k}^{m} \n\] \n\nwith \(... | Yes |
Corollary 4.72. Let \( T : V \rightarrow W \) be a linear transformation of vector spaces \( V \) and \( W \) over a field \( k \), and let \( X \) and \( Y \) be bases of \( V \) and \( W \), respectively. If \( T \) is nonsingular and \( A = {}_{Y}{\left\lbrack T\right\rbrack }_{X} \), then \( A \) is a nonsingular m... | Proof.\n\n\[ \nI = {}_{Y}{\left\lbrack {1}_{W}\right\rbrack }_{Y} = \left( {{}_{Y}{\left\lbrack T\right\rbrack }_{X}}\right) \left( {{}_{X}{\left\lbrack {T}^{-1}\right\rbrack }_{Y}}\right) \n\]\n\nand\n\n\[ \nI = {}_{X}{\left\lbrack {1}_{V}\right\rbrack }_{X} = \left( {{}_{X}{\left\lbrack {T}^{-1}\right\rbrack }_{Y}}\r... | Yes |
Let \( T : V \rightarrow V \) be a linear transformation on a vector space \( V \) over a field \( k \). If \( X \) and \( Y \) are bases of \( V \), then there is a nonsingular matrix \( P \) with entries in \( k \), namely, \( P = {}_{Y}{\left\lbrack {1}_{V}\right\rbrack }_{X} \), so that\n\n\[ \n{}_{Y}{\left\lbrack ... | The first statement follows from Proposition 4.70 and associativity:\n\n\[ \n{}_{Y}{\left\lbrack T\right\rbrack }_{Y} = {}_{Y}{\left\lbrack {1}_{V}T{1}_{V}\right\rbrack }_{Y} = \left( {{}_{Y}{\left\lbrack {1}_{V}\right\rbrack }_{X}}\right) \left( {{}_{X}{\left\lbrack T\right\rbrack }_{X}}\right) \left( {{}_{X}{\left\lb... | Yes |
Proposition 4.75. Let \( T : V \rightarrow W \) be a linear transformation.\n\n(i) \( \ker T \) is a subspace of \( V \), and \( \operatorname{im}T \) is a subspace of \( W \).\n\n(ii) \( T \) is injective if and only if \( \ker T = \{ 0\} \) . | We can now give a new proof of Corollary 4.20 that a homogeneous system over a field \( k \) with \( r \) equations in \( n \) unknowns has a nontrivial solution if \( r < n \) . If \( A \) is the \( r \times n \) coefficient matrix of the system, then \( T : x \mapsto {Ax} \) is a linear transformation \( T : {k}^{n} ... | No |
Lemma 4.76. Let \( T : V \rightarrow W \) be a linear transformation.\n\n(i) Let \( T \) be nonsingular. For every basis \( X = {v}_{1},{v}_{2},\ldots ,{v}_{n} \) of \( V \), we have \( T\left( X\right) = T\left( {v}_{1}\right), T\left( {v}_{2}\right) ,\ldots, T\left( {v}_{n}\right) \) is a basis of \( W \) .\n\n(ii) C... | Proof.\n\n(i) If \( \sum {c}_{i}T\left( {v}_{i}\right) = 0 \), then \( T\left( {\sum {c}_{i}{v}_{i}}\right) = 0 \), and so \( \sum {c}_{i}{v}_{i} \in \ker T = \{ 0\} \) . Hence each \( {c}_{i} = 0 \), because \( X \) is linearly independent. If \( w \in W \), then the surjectivity of \( T \) provides \( v \in V \) with... | Yes |
Theorem 4.77. If \( V \) is an \( n \) -dimensional vector space over a field \( k \), then \( V \) is isomorphic to \( {k}^{n} \) . | Proof. Choose a basis \( {v}_{1},\ldots ,{v}_{n} \) of \( V \) . If \( {e}_{1},\ldots ,{e}_{n} \) is the standard basis of \( {k}^{n} \) , then Theorem 4.61 says that there is a linear transformation \( T : V \rightarrow {k}^{n} \) with \( T\left( {v}_{i}\right) = {e}_{i} \) for all \( i \) ; by Lemma 4.76, \( T \) is ... | Yes |
Corollary 4.78. Two finite-dimensional vector spaces \( V \) and \( W \) over a field \( k \) are isomorphic if and only if \( \dim \left( V\right) = \dim \left( W\right) \) . | Proof. Assume that there is a nonsingular \( T : V \rightarrow W \) . If \( X = {v}_{1},\ldots ,{v}_{n} \) is a basis of \( V \), then Lemma 4.76 says that \( T\left( {v}_{1}\right) ,\ldots, T\left( {v}_{n}\right) \) is a basis of \( W \) . Therefore, \( \dim \left( W\right) = \left| X\right| = \dim \left( V\right) \) ... | Yes |
Proposition 4.79. Let \( V \) be a finite-dimensional vector space over a field \( k \) with \( \dim \left( V\right) = n \), and let \( T : V \rightarrow V \) be a linear transformation. The following statements are equivalent:\n\n(i) \( T \) is an isomorphism;\n\n(ii) \( T \) is surjective;\n\n(iii) \( T \) is injecti... | Proof.\n\n(i) \( \Rightarrow \) (ii) This implication is obvious.\n\n(ii) \( \Rightarrow \) (iii) Assume that \( T \) is surjective. If \( X = {v}_{1},\ldots ,{v}_{n} \) is a basis of \( V \), we claim that \( T\left( X\right) = T\left( {v}_{1}\right) ,\ldots, T\left( {v}_{n}\right) \) spans \( V \) . If \( w \in V \),... | Yes |
Corollary 4.80. Let \( V \) be a finite-dimensional vector space, and let \( T : V \rightarrow V \) be a linear transformation on \( V \) . Then \( T \) is singular if and only if there exists a nonzero vector \( v \in V \) with \( T\left( v\right) = 0 \) . | Proof. If \( T \) is singular, then \( \ker T \neq \{ 0\} \), by Proposition 4.79. Conversely, if there is a nonzero vector \( v \) with \( T\left( v\right) = 0 \), then \( \ker T \neq \{ 0\} \) and \( T \) is singular. | Yes |
Corollary 4.81. Let \( A \) and \( B \) be \( n \times n \) matrices with entries in a field \( k \) . If \( {AB} = I \), then \( {BA} = I \) . Therefore, \( A \) is nonsingular and \( B = {A}^{-1} \) . | Proof. There are linear transformations \( T, S : {k}^{n} \rightarrow {k}^{n} \) with \( {}_{X}{\left\lbrack T\right\rbrack }_{X} = A \) and \( {}_{X}{\left\lbrack S\right\rbrack }_{X} = B \), where \( X \) is the standard basis. Let us abbreviate \( {}_{X}{\left\lbrack T\right\rbrack }_{X} \) to \( \left\lbrack T\righ... | Yes |
Proposition 4.82. Let \( T : V \rightarrow W \) be a linear transformation, where \( V \) and W are vector spaces over a field \( k \) of dimension \( n \) and \( m \), respectively. Then\n\n\[ \dim \left( {\ker T}\right) + \dim \left( {\operatorname{im}T}\right) = n. \] | Proof. Choose a basis \( {u}_{1},\ldots ,{u}_{p} \) of \( \ker T \), and extend it to a basis of \( V \) by adjoining vectors \( {w}_{1},\ldots ,{w}_{q} \) . Now im \( T \) is spanned by the list \( T\left( {u}_{1}\right) ,\ldots, T\left( {u}_{p}\right) \) , \( T\left( {w}_{1}\right) ,\ldots, T\left( {w}_{q}\right) \) ... | Yes |
Corollary 4.83. If \( A \) is an \( m \times n \) matrix over a field \( k \), then\n\n\[ \operatorname{rank}\left( A\right) = \operatorname{rank}\left( {A}^{T}\right) \]\n\nthe row space \( \operatorname{Row}\left( A\right) \) and the column space \( \operatorname{Col}\left( A\right) \) have the same dimension. | Proof. Define \( {T}_{A} : {k}^{n} \rightarrow {k}^{m} \) by \( {T}_{A}\left( v\right) = {Av} \) . Now \( \ker {T}_{A} \) is the solution space of the linear system \( {Ax} = 0 \), so that Theorem 4.42 gives \( \dim \left( {\ker {T}_{A}}\right) = \) \( n - r \), where \( r = \operatorname{rank}\left( A\right) \) [recal... | Yes |
Proposition 4.84. Let \( V \) be an \( n \) -dimensional vector space over a field \( k \), and let \( X = {v}_{1},\ldots ,{v}_{n} \) be a basis of \( V \) . Then \( \mu : \mathrm{{GL}}\left( V\right) \rightarrow \mathrm{{GL}}\left( {n, k}\right) \), defined by \( T \mapsto {}_{X}{\left\lbrack T\right\rbrack }_{X} \), ... | Proof. By Proposition 4.69, the function \( {\mu }_{X, X} : T \mapsto \left\lbrack T\right\rbrack = {}_{X}{\left\lbrack T\right\rbrack }_{X} \) is an isomorphism of vector spaces\n\n\[ \n{\operatorname{Hom}}_{k}\left( {V, V}\right) \rightarrow {\operatorname{Mat}}_{n}\left( k\right) \n\]\n\nMoreover, Proposition 4.70 s... | Yes |
Proposition 4.86. If \( A \) is an \( n \times n \) matrix with entries in a commutative ring \( R \), then\n\n\[ A\operatorname{adj}\left( A\right) = \det \left( A\right) I = \operatorname{adj}\left( A\right) A. \] | Proof. If \( A = \left\lbrack {a}_{ij}\right\rbrack \), let us write \( {\left( A\right) }_{ij} = {a}_{ij} \) in this proof. Thus, if \( C = \left\lbrack {c}_{ij}\right\rbrack \) , then \( {\left( AC\right) }_{ij} = \mathop{\sum }\limits_{k}{a}_{ik}{c}_{kj} \) . If we now define \( C = \left\lbrack {c}_{ij}\right\rbrac... | Yes |
Corollary 4.87. If \( A \) is an \( n \times n \) matrix with entries in a commutative ring \( R \) , then \( A \) is invertible if and only if \( \det \left( A\right) \) is a unit in \( R \) . | Proof. If \( A \) is invertible, then there is a matrix \( B \) with \( {AB} = I \) . By Eq. (1), \( 1 = \) \( \det \left( I\right) = \det \left( {AB}\right) = \det \left( A\right) \det \left( B\right) \), so that \( \det \left( A\right) \) is a unit in \( R \) . Conversely, assume that \( \det \left( A\right) \) is a ... | Yes |
Proposition 4.88. Let \( P \) and \( M \) be \( n \times n \) matrices with entries in a commutative ring \( R \) . If \( P \) is invertible, then\n\n\[ \det \left( {{PM}{P}^{-1}}\right) = \det \left( M\right) \] | Proof. We have \( \det \left( {P}^{-1}\right) = \det {\left( P\right) }^{-1} \), by Corollary 4.87. Since determinants lie in the commutative ring \( R \) ,\n\n\[ \det \left( {{PM}{P}^{-1}}\right) = \det \left( P\right) \det \left( M\right) \det \left( {P}^{-1}\right) \]\n\n\[ = \det \left( M\right) \det \left( P\right... | Yes |
Corollary 4.89. Let \( T : V \rightarrow V \) be a linear transformation on a vector space \( V \) over a field \( k \), and let \( X \) and \( Y \) be bases of \( V \) . If \( A = {}_{X}{\left\lbrack T\right\rbrack }_{X} \) and \( B = {}_{Y}{\left\lbrack T\right\rbrack }_{Y} \) , then \( \det \left( A\right) = \det \l... | Proof. By Corollary 4.73, \( A \) and \( B \) are similar; that is, there is a nonsingular (hence, invertible) matrix \( P \) with \( B = {PA}{P}^{-1} \) . | Yes |
Proposition 4.90. Let \( T : V \rightarrow V \) be a linear transformation, where \( V \) is a vector space over a field \( k \), and let \( {c}_{1},\ldots ,{c}_{r} \) be distinct eigenvalues of \( T \) lying in \( k \) . If \( {v}_{i} \) is an eigenvector of \( T \) for \( {c}_{i} \), then the list \( X = {v}_{1},\ldo... | Proof. We use induction on \( r \geq 1 \) . The base step \( r = 1 \) is true, for any nonzero vector is a linearly independent list of length 1 , and eigenvectors are, by definition, nonzero. For the inductive step, assume that\n\n\[ \n{a}_{1}{v}_{1} + \cdots + {a}_{r + 1}{v}_{r + 1} = 0.\n\]\n\nApplying \( T \) to th... | Yes |
If \( A \) and \( B \) are similar \( n \times n \) matrices with entries in a field \( k \) , then \[ \det \left( {{xI} - A}\right) = \det \left( {{xI} - B}\right) ,\] and so similar matrices have the same eigenvalues (occurring with the same multiplicities). | Proof. If \( B = {PA}{P}^{-1} \), then \[ P\left( {{xI} - A}\right) {P}^{-1} = {PxI}{P}^{-1} - {PA}{P}^{-1} = {xI} - B. \] Therefore, \( \det \left( {P\left( {{xI} - A}\right) {P}^{-1}}\right) = \det \left( {{xI} - B}\right) \) . But \[ \det \left( {P\left( {{xI} - A}\right) {P}^{-1}}\right) = \det \left( P\right) \det... | Yes |
Proposition 4.92. If \( T : V \rightarrow V \) is a linear transformation, where \( \dim \left( V\right) = n \) , then \( T \) has at most \( n \) eigenvalues. | Proof. If \( \dim \left( V\right) = n \), then \( \deg \left( {h}_{T}\right) = n \), and the result follows from Theorem 3.50. | Yes |
Proposition 4.93. Let \( k \) be a field, and let \( A \) be an \( n \times n \) matrix with entries in \( k \) . Then \( {h}_{A}\left( x\right) \) is a monic polynomial of degree \( n \) . Moreover, the coefficient of \( {x}^{n - 1} \) in \( {h}_{A}\left( x\right) \) is \( - \operatorname{tr}\left( A\right) \) and the... | Proof. Let \( A = \left\lbrack {a}_{ij}\right\rbrack \) and let \( B = {xI} - A \) ; thus, \( B = \left\lbrack {b}_{ij}\right\rbrack = \left\lbrack {x{\delta }_{ij} - {a}_{ij}}\right\rbrack \) (where \( {\delta }_{ij} \) is the Kronecker delta). The complete expansion is\n\n\[ \det \left( B\right) = \mathop{\sum }\limi... | Yes |
Corollary 4.94. If \( A \) and \( B \) are similar \( n \times n \) matrices with entries in a field \( k \) , then \( A \) and \( B \) have the same trace and the same determinant. | Proof. Now \( A \) and \( B \) have the same characteristic polynomial, by Proposition 4.91, and so Proposition 4.93 applies to give \( \operatorname{tr}\left( A\right) = \operatorname{tr}\left( B\right) \) and \( \det \left( A\right) = \) \( \det \left( B\right) \) . | Yes |
Corollary 4.95. Let \( A = \left\lbrack {a}_{ij}\right\rbrack \) be an \( n \times n \) matrix with entries in a field \( k \), and let \( {h}_{A}\left( x\right) = \mathop{\prod }\limits_{{i = 1}}^{n}\left( {x - {\alpha }_{i}}\right) \) be its characteristic polynomial. Then \[ \operatorname{tr}\left( A\right) = \matho... | Proof. By Exercise 3.99 on page 305, if \( f\left( x\right) = \mathop{\sum }\limits_{j}{c}_{j}{x}^{j} \in k\left\lbrack x\right\rbrack \) is a monic polynomial with \( f\left( x\right) = \mathop{\prod }\limits_{{i = 1}}^{n}\left( {x - {\alpha }_{i}}\right) \), then \( {c}_{n - 1} = - \mathop{\sum }\limits_{j}{\alpha }_... | No |
Corollary 4.98. Let \( A \) be an \( n \times n \) matrix over a field \( k \) which contains all the eigenvalues of \( A \) . If the characteristic polynomial of \( A \) has no repeated roots, then \( A \) is diagonalizable. | Proof. Since \( \deg \left( {h}_{A}\right) = n \), the hypothesis gives \( n \) distinct eigenvalues \( {c}_{1},\ldots ,{c}_{n} \) . As these eigenvalues all lie in \( k \), there are corresponding eigenvectors \( {v}_{1},\ldots ,{v}_{n} \) in \( {k}^{n} \) ; that is, \( A{v}_{i} = {c}_{i}{v}_{i} \) . By Proposition 4.... | Yes |
Theorem 4.100 (Principal Axis Theorem). If \( A \) is an \( n \times n \) real symmetric matrix, then all its eigenvalues are real, and \( A \) is diagonalizable. | Proof. Define \( T : {\mathbb{C}}^{n} \rightarrow {\mathbb{C}}^{n} \) by \( T\left( u\right) = {Au} \) . Since \( A \) is a symmetric matrix, we see that \( T = {T}^{ * } \) ; that is, for all \( u, v \in {\mathbb{C}}^{n} \) ,\n\n\[ \n\left( {{Tu}, v}\right) = \left( {u,{Tv}}\right) .\n\]\n\nWe first prove that all the... | No |
Proposition 4.104. If \( \mathcal{A} \) is an alphabet, then the Hamming distance is a metric on \( {\mathcal{A}}^{n} \) . | Proof. Let \( w = \left( {{a}_{1},\ldots ,{a}_{n}}\right) ,{w}^{\prime } = \left( {{a}_{1}^{\prime },\ldots ,{a}_{n}^{\prime }}\right) \in {\mathcal{A}}^{n} \) . Clearly, \( \delta \left( {w,{w}^{\prime }}\right) \geq 0 \) and \( \delta \left( {w, w}\right) = 0 \) . On the other hand, if \( \dot{\delta }\left( {w,{w}^{... | Yes |
Proposition 4.105. Let \( \mathcal{A} \) be an alphabet, and let \( C \subseteq {\mathcal{A}}^{n} \) be an \( \left( {n, M, d}\right) \) - code.\n\n(i) If \( d \geq s + 1 \), then \( C \) can detect up to \( s \) errors.\n\n(ii) If \( d \geq {2t} + 1 \), then \( C \) can correct up to \( t \) errors. | Proof.\n\n(i) If \( w \neq c \) differs from \( c \) in at most \( s \) places, then \( 0 < \delta \left( {c, w}\right) \leq s \) . But if \( w \in C \), then\n\n\[ s \geq \delta \left( {c, w}\right) \geq d > s \]\n\na contradiction.\n\n(ii) If \( w \) is obtained from \( c \) by changing at most \( t \) places, then \... | Yes |
Proposition 4.107. If \( C \) is a linear \( \left( {n, M, d}\right) \) -code over a finite field \( {\mathbb{F}}_{q} \), then\n\n\[ d = \mathop{\min }\limits_{\substack{{c \in C} \\ {c \neq 0} }}\{ \mathrm{{wt}}\left( c\right) : c \in C\} \]\n\nThus, \( d \) is the smallest weight among nonzero codewords. | Proof. Since \( C \) is a subspace, \( w,{w}^{\prime } \in C \) implies \( w - {w}^{\prime } \in C \) . Thus,\n\n\[ d = \mathop{\min }\limits_{\substack{{w,{w}^{\prime } \in C} \\ {w \neq {w}^{\prime }} }}\delta \left( {w,{w}^{\prime }}\right) \]\n\n\[ = \mathop{\min }\limits_{\substack{{w,{w}^{\prime } \in C} \\ {w \n... | Yes |
Proposition 4.108. If \( C \) is a linear \( \left\lbrack {n, m}\right\rbrack \) -code over a field \( k \), then there are a linear code \( {C}^{\prime } \) permutation-equivalent to \( C \) and an \( m \times n \) matrix \( G \) of the form \( G = \left\lbrack {I \mid B}\right\rbrack \), where \( I \) is the \( m \ti... | Proof. If \( {e}_{1},\ldots ,{e}_{m} \) is the standard basis of \( {k}^{m} \) and \( {c}_{1},\ldots ,{c}_{m} \) is some basis of \( C \), define a linear transformation \( E : {k}^{m} \rightarrow {k}^{n} \) by \( E\left( {e}_{i}\right) = {c}_{i} \) . Now \( E \) is an injection, by Lemma 4.76(ii); in fact, \( E \) is ... | Yes |
Proposition 4.111. Let \( k \) be a finite field, let \( I = \left( {{x}^{n} - 1}\right) \) be the principal ideal in \( k\left\lbrack x\right\rbrack \) generated by \( {x}^{n} - 1 \), and let \( z = x + I \) . Then \( C \subseteq k\left\lbrack x\right\rbrack /I \) is a cyclic code if and only if \( C \) is an ideal in... | Proof. Let \( C \) be an ideal in \( k\left\lbrack x\right\rbrack /I \), and let \( c = {a}_{0} + {a}_{1}z + \cdots + {a}_{n - 1}{z}^{n - 1} \in C \) . Since \( C \) is an ideal, \( C \) contains \( {zc} = {a}_{0}z + {a}_{1}{z}^{2} + \cdots + {a}_{n - 2}{z}^{n - 1} + {a}_{n - 1}{z}^{n} \) . But \( {z}^{n} = 1 \) (becau... | Yes |
Corollary 4.112. If \( C \) is a cyclic code of length \( n \) over a field \( k \) with generating polynomial \( g\left( x\right) \), then \( \dim \left( C\right) = n - \deg \left( g\right) \) . | Proof. Since \( g\left( x\right) \mid \left( {{x}^{n} - 1}\right) \), there is an inclusion of ideals \( I = \left( {{x}^{n} - 1}\right) \subseteq \) \( \left( {g\left( x\right) }\right) = J \) . Regarding \( k\left\lbrack x\right\rbrack \) and its quotients merely as vector spaces over \( k \) , we see that \ | No |
Corollary 4.113. If \( C \) is a cyclic code of length \( n \) with generating polynomial \( g\left( x\right) = {g}_{0} + {g}_{1}x + \cdots + {g}_{s}{x}^{s} \), then a generating matrix for \( C \) is the \( \left( {n - s}\right) \times n \) matrix:\n\n\[ G = \left\lbrack \begin{matrix} {g}_{0} & {g}_{1} & {g}_{2} & \c... | Proof. Since \( C \) is an ideal, \( g\left( x\right) ,{xg}\left( x\right) ,{x}^{2}g\left( x\right) ,\ldots ,{x}^{n - s}g\left( x\right) \) are codewords, and these codewords correspond to the rows of \( G \) . Write \( G = \left\lbrack {T \mid B}\right\rbrack \), where \( T \) is the \( \left( {n - s}\right) \times \l... | No |
Lemma 4.114. Let \( {\mathbb{F}}_{q} \) denote the finite field of \( q \) elements. If \( n \) is a positive integer, then there exists a primitive nth root of unity in some extension field of \( {\mathbb{F}}_{q} \) if and only if \( \left( {n, q}\right) = 1 \) . | Proof. Assume that \( \left( {n, q}\right) = 1 \), and let \( E/{\mathbb{F}}_{q} \) be a splitting field of \( f\left( x\right) = \) \( {x}^{n} - 1 \) over \( {\mathbb{F}}_{q} \) (actually, we need only that \( f\left( x\right) \) splits over \( E \), which follows from Kronecker’s Theorem 3.118). Now the derivative \(... | No |
Corollary 4.116. Let \( C \) be a BCH-code of length \( n \) with generating polynomial \( g\left( x\right) \) . If consecutive powers \( {\zeta }^{u},{\zeta }^{u + 1},\ldots ,{\zeta }^{u + \ell } \) occur among the roots of \( g\left( x\right) \) , where \( \ell = {2t} \) or \( \ell = {2t} + 1 \) and \( u + \ell < n \... | Proof. By Theorem 4.115, we have \( d\left( C\right) \geq \ell + 2 \geq {2t} + 1 \), and so Proposition 4.105 applies. | Yes |
Corollary 4.117. For any prime \( p \) and any positive integer \( t \), there exists a \( \mathrm{{BCH}} \) -code \( C \) over \( {\mathbb{F}}_{p} \) which corrects up to \( t \) errors. | Proof. Let \( k = {\mathbb{F}}_{q} \), where \( q \) is a power of \( p \) and \( {2t} + 1 \leq q - 1 \) . By Theorem 3.122, the multiplicative group \( {k}^{ \times } \) is a cyclic group of order \( q - 1 \), and a generator \( \zeta \) is a primitive \( \left( {q - 1}\right) \) th root of unity; hence, \( \zeta ,{\z... | Yes |
Corollary 4.120 (Reed-Solomon). There exists a t-error correcting BCH-code over \( {\mathbb{F}}_{{2}^{{2t} + 1}} \) with rate \( 1 - \left\lbrack {{2t}/\left( {{2}^{{2t} + 1} - 1}\right) }\right\rbrack \) . | Proof. Let \( \zeta \) be a primitive element of \( {\mathbb{F}}_{q} \), where \( q = {2}^{{2t} + 1} \), and define\n\n\[ g\left( x\right) = \left( {x - \zeta }\right) \left( {x - {\zeta }^{2}}\right) \cdots \left( {x - {\zeta }^{2t}}\right) \in {\mathbb{F}}_{q}\left\lbrack x\right\rbrack .\n\]\n\nThe Reed-Solomon-code... | Yes |
Proposition 4.122. If \( G = \left\lbrack {I \mid B}\right\rbrack \) is an \( m \times n \) echelon generating matrix of a linear code \( C \) over a field \( k \), then \( w \in {k}^{n} \) lies in \( C \) if and only if \( w{\left\lbrack -{B}^{T} \mid I\right\rbrack }^{T} = 0 \) . | Proof. If \( w \in C \), then \( w \) is a linear combination of the rows of \( G \) . Now the \( i \) th row of \( G \) is \( {e}_{i}G \), where \( {e}_{1},\ldots ,{e}_{m} \) is the standard basis of \( {k}^{m} \) . Hence, if \( G{\left\lbrack -{B}^{T} \mid I\right\rbrack }^{T} = 0 \), then \( \left( {{e}_{i}G}\right)... | Yes |
Lemma 4.124. Let \( C \) be a Reed-Solomon-code of length \( n \) over a field \( k \) whose generating polynomial has \( 1,\zeta ,{\zeta }^{2},\ldots ,{\zeta }^{{2t} - 1} \) among its roots, where \( \zeta \) is a primitive nth root of unity lying in \( k \), and \( {2t} - 1 < n \) .\n\n(i) The matrices \( U \) and \(... | (i) Now \n\n\[ {\operatorname{ROW}}_{U}\left( i\right) = \left( {1{\zeta }^{i}{\zeta }^{2i}\ldots {\zeta }^{i\left( {n - 1}\right) }}\right) \] \n\nwhere \( 0 \leq i \leq t \), and \n\n\[ {\operatorname{ROW}}_{V}\left( \mu \right) = \left( {1{\zeta }^{\mu }{\zeta }^{2\mu }\ldots {\zeta }^{\left( {n - 1}\right) \mu }}\r... | Yes |
Lemma 5.1. The substitution \( X = x - \frac{1}{n}{a}_{n - 1} \) changes\n\n\[ f\left( X\right) = {X}^{n} + {a}_{n - 1}{X}^{n - 1} + h\left( X\right) ,\]\n\nwhere \( h\left( X\right) = 0 \) or \( \deg \left( h\right) \leq n - 2 \), into a reduced polynomial\n\n\[ \widetilde{f}\left( x\right) = f\left( {x - \frac{1}{n}{... | Proof. The substitution \( X = x - \frac{1}{n}{a}_{n - 1} \) gives\n\n\[ \widetilde{f}\left( x\right) = f\left( {x - \frac{1}{n}{a}_{n - 1}}\right) \]\n\n\[ = {\left( x - \frac{1}{n}{a}_{n - 1}\right) }^{n} + {a}_{n - 1}{\left( x - \frac{1}{n}{a}_{n - 1}\right) }^{n - 1} + h\left( {x - \frac{1}{n}{a}_{n - 1}}\right) \]... | Yes |
Corollary 5.2 (Quadratic Formula). If \( f\left( X\right) = {X}^{2} + {bX} + c \), then its roots are \[ \frac{1}{2}\left( {-b \pm \sqrt{{b}^{2} - {4c}}}\right) \] | Proof. Define \( x \) by \( X = x - \frac{1}{2}b \) . Now \[ \widetilde{f}\left( x\right) = {\left( x - \frac{1}{2}b\right) }^{2} + b\left( {x - \frac{1}{2}b}\right) + c. \] The linear terms cancel, the reduced polynomial is \[ \widetilde{f}\left( x\right) = {x}^{2} - \frac{1}{4}\left( {{b}^{2} - {4c}}\right) \] and th... | Yes |
Corollary 5.3. Given numbers \( c \) and \( d \), there exist numbers \( \alpha \) and \( \beta \) with \( \alpha + \beta = c \) and \( {\alpha \beta } = d \) . | Proof. If \( d = 0 \), choose \( \alpha = 0 \) and \( \beta = c \) . If \( d \neq 0 \), then \( \alpha \neq 0 \) and we may set \( \beta = d/\alpha \) . Substituting, \( c = \alpha + \beta = \alpha + d/\alpha \), so that\n\n\[{\alpha }^{2} - {c\alpha } + d = 0.\]\n\nThe quadratic formula now shows that such an \( \alph... | Yes |
Theorem 5.4 (Cubic Formula). The roots of \( {x}^{3} + {qx} + r \) are\n\n\[ \alpha + \beta ,\;{\omega \alpha } + {\omega }^{2}\beta ,\;\text{ and }\;{\omega }^{2}\alpha + {\omega \beta },\]\n\nwhere \( {\alpha }^{3} = \frac{1}{2}\left( {-r + \sqrt{D}}\right) ,\beta = - \frac{1}{3}\frac{q}{\alpha }, D = {r}^{2} + \frac... | Proof. We have just given the proof when \( \alpha \neq 0 \) . By Eq. (2), we have \( {\alpha \beta } = \) \( - q/3 \), and so \( \alpha = 0 \) forces \( q = 0 \) ; that is, the reduced cubic is \( {x}^{3} + r \) . In this case, \( {\beta }^{3} = - r \), the roots are \( \beta ,{\omega \beta } \), and \( {\omega }^{2}\... | No |
We find the roots of \( {x}^{3} - {15x} - {126} \) . | The polynomial is already reduced, for there is no \( {x}^{2} \) term, and so it is in the form to which the cubic formula applies (were it not reduced, one would first reduce it, as in Lemma 5.1). Here, \( q = - {15} \) , \( r = - {126}, D = {\left( -{126}\right) }^{2} + \frac{4}{27}{\left( -{15}\right) }^{3} = {15},{... | Yes |
Lemma 5.6. The discriminant \( {\Delta }^{2} \) of \( f\left( x\right) = {x}^{3} + {qx} + r \) is\n\n\[ \n{\Delta }^{2} = - {27}{r}^{2} - 4{q}^{3} = - {27D}. \n\] | Proof. If the roots of \( f\left( x\right) \) are \( u, v \), and \( w \), then the cubic formula gives\n\n\[ \nu = \alpha + \beta ;\;v = {\omega \alpha } + {\omega }^{2}\beta ;\;w = {\omega }^{2}\alpha + {\omega \beta }, \]\n\nwhere \( \omega = - \frac{1}{2} - i\frac{\sqrt{3}}{2}, D = {r}^{2} + \frac{4}{27}{q}^{3},\al... | Yes |
Lemma 5.7. Every \( f\left( x\right) \in \mathbb{R}\left\lbrack x\right\rbrack \) of odd degree has a real root. | Proof. The proof is by induction on \( n \geq 0 \), where \( \deg \left( f\right) = {2n} + 1 \) . The base step \( n = 0 \) is obviously true. Let \( n \geq 1 \) and let \( u \) be a complex root of \( f\left( x\right) \) . If \( u \) is real, we are done. Otherwise \( u = a + {ib} \), and Exercise 5.7 on page 447 show... | No |
Theorem 5.8. All the roots \( u, v \), w of \( {x}^{3} + {qx} + r \in \mathbb{R}\left\lbrack x\right\rbrack \) are real numbers if and only if the discriminant \( {\Delta }^{2} \geq 0 \) ; that is, \( {27}{r}^{2} + 4{q}^{3} \leq 0 \) . | Proof. If \( u, v \), and \( w \) are real numbers, then \( \Delta = \left( {u - v}\right) \left( {u - w}\right) \left( {v - w}\right) \) is a real number. Therefore, \( - {27}{r}^{2} - 4{q}^{3} = {\Delta }^{2} \geq 0 \), and \( {27}{r}^{2} + r{q}^{3} \leq 0 \). Conversely, assume that \( w = s + {ti} \) is not real (i... | Yes |
Theorem 5.10 (Quartic Formula). There is a method to compute the four roots of a quartic\n\n\\[ \n{X}^{4} + b{X}^{3} + c{X}^{2} + {dX} + e.\n\\] | Proof. As with the cubic, the quartic can be simplified, by setting \\( X = x - \\frac{1}{4}b \\) , to\n\n\\[ \n{x}^{4} + q{x}^{2} + {rx} + s\n\\]\n\n(8)\n\nmoreover, if a number \\( u \\) is a root of the second polynomial, then \\( u = \\frac{1}{4}b \\) is a root of the first.\n\nFactor the quartic in Eq. (8) into qu... | Yes |
Consider\n\n\\[ \n{x}^{4} - 2{x}^{2} + {8x} - 3 = 0 \n\\]\n\nso that \\( q = - 2, r = 8 \\), and \\( s = - 3 \\) . If we factor this quartic into\n\n\\[ \n\\left( {{x}^{2} + {jx} + \\ell }\\right) \\left( {{x}^{2} - {jx} + m}\\right) \n\\]\n\nthen Eq. (12) gives\n\n\\[ \n{j}^{6} - 4{j}^{4} + {16}{j}^{2} - {64} = 0. \n\... | One could use the cubic formula to find \\( {j}^{2} \\), but this would be very tedious, for one must first get rid of the \\( {j}^{4} \\) term before doing the rest of the calculations. It is simpler, in this case, to observe that \\( j = 2 \\) is a root, for the equation can be rewritten\n\n\\[ \n\\begin{matrix} {j}^... | Yes |
Theorem 5.12 (Viète). Let \( f\left( x\right) = {x}^{3} + {qx} + r \) be a cubic polynomial for which \( {27}{r}^{2} + 4{q}^{3} \leq 0 \) . If \( t = \sqrt{-{4q}/3} \) and \( \cos {3\theta } = - {4r}/{t}^{3} \), then the roots of \( f\left( x\right) \) are\n\n\[ t\cos \theta, t\cos \left( {\theta + {120}^{ \circ }}\rig... | ## Example 5.13.\n\nConsider once again the cubic \( {x}^{3} - {7x} + 6 = \left( {x - 1}\right) \left( {x - 2}\right) \left( {x + 3}\right) \) that was discussed in Example 5.9; of course, its roots are 1, 2, and -3 . The cubic formula gave rather complicated expressions for these roots in terms of cube roots of comple... | Yes |
Proposition 5.16. If \( f\left( x\right) \in k\left\lbrack x\right\rbrack \), where \( k \) is a field, then a splitting field \( E/k \) of \( f\left( x\right) \) exists. | Proof. By Kronecker’s theorem, Theorem 3.118, there exists an extension \( K/k \) with \( f\left( x\right) = a\left( {x - {z}_{1}}\right) \cdots \left( {x - {z}_{n}}\right) \) in \( K\left\lbrack x\right\rbrack \) . If we define \( E = k\left( {{z}_{1},\ldots ,{z}_{n}}\right) \) , where \( {z}_{1},\ldots ,{z}_{n} \) ar... | Yes |
Let \( k \) be a subfield of a field \( K \), let\n\n\[ f\left( x\right) = {x}^{n} + {a}_{n - 1}{x}^{n - 1} + \cdots + {a}_{1}x + {a}_{0} \in k\left\lbrack x\right\rbrack ,\]\n\nand let \( E = k\left( {{z}_{1},\ldots ,{z}_{n}}\right) \) be a splitting field. If \( \sigma : E \rightarrow E \) is an automorphism fixing \... | Proof. If \( z \) is a root of \( f\left( x\right) \), then\n\n\[ 0 = f\left( z\right) = {z}^{n} + {a}_{n - 1}{z}^{n - 1} + \cdots + {a}_{1}z + {a}_{0}. \]\n\nApplying \( \sigma \) to this equation gives\n\n\[ 0 = \sigma {\left( z\right) }^{n} + \sigma \left( {a}_{n - 1}\right) \sigma {\left( z\right) }^{n - 1} + \cdot... | No |
Corollary 5.19. Let \( k \subseteq B \subseteq F \) be a tower of fields, where \( B \) is the splitting field of some polynomial \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) . If \( \sigma : F \rightarrow F \) is an automorphism fixing \( k \), then \( \sigma \left( B\right) = B \) . | Proof. By Proposition 5.18, \( \sigma \) permutes the roots \( {z}_{1},\ldots ,{z}_{n} \) of \( f\left( x\right) \), so that \( \sigma \left( B\right) \subseteq B \) . As vector spaces over \( k \), we have \( B \cong \sigma \left( B\right) \), for \( \sigma \) is an injective linear transformation. Since \( \left\lbra... | No |
Proposition 5.20. Let \( E = k\left( {{z}_{1},\ldots ,{z}_{n}}\right) \) . If \( \sigma : E \rightarrow E \) is an automorphism fixing \( k \) and if \( \sigma \left( {z}_{i}\right) = {z}_{i} \) for all \( i \), then \( \sigma \) is the identity. | Proof. We prove the proposition by induction on \( n \geq 1 \) . If \( n = 1 \), then each \( u \in E \) has the form \( f\left( {z}_{1}\right) /g\left( {z}_{1}\right) \), where \( f\left( x\right), g\left( x\right) \in k\left\lbrack x\right\rbrack \) and \( g\left( {z}_{1}\right) \neq 0 \) . But \( \sigma \) fixes \( ... | Yes |
Theorem 5.21. If \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) has degree \( n \), then its Galois group \( \operatorname{Gal}\left( {E/k}\right) \) is isomorphic to a subgroup of \( {S}_{n} \) . | Proof. Let \( E/k \) be a splitting field of \( f\left( x\right) \) over \( k \), and let \( X = \left\{ {{z}_{1},\ldots ,{z}_{n}}\right\} \) be the set of roots of \( f\left( x\right) \) in \( E \) . If \( \sigma \in \operatorname{Gal}\left( {E/k}\right) \), then Proposition 5.18 shows that its restriction \( \sigma \... | Yes |
(i) There exists an isomorphism \( \Phi : E \rightarrow {E}^{\prime } \) extending \( \varphi \) . | (i) The proof is by induction on \( \left\lbrack {E : k}\right\rbrack \) . If \( \left\lbrack {E : k}\right\rbrack = 1 \), then \( f\left( x\right) \) is a product of linear polynomials in \( k\left\lbrack x\right\rbrack \), and it follows easily that \( {f}^{ * }\left( x\right) \) is also a product of linear polynomia... | Yes |
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