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Theorem 3. Let \( T \) be a linear mapping from a normed linear space \( X \) into another normed linear space \( Y,{T}^{\prime } \) its transpose, mapping \( {Y}^{\prime } \) into \( {X}^{\prime } \) . Then\n\n\[ \left| {T}^{\prime }\right| = \left| T\right| \]\n\nwhere \( {X}^{\prime } \) and \( {Y}^{\prime } \) are ... | Proof. Apply definition (7) to \( m = {\mathrm{T}}^{\prime }l \) :\n\n\[ {\left| {\mathrm{T}}^{\prime }l\right| }^{\prime } = \mathop{\sup }\limits_{{\left| x\right| = 1}}\left( {{\mathrm{T}}^{\prime }l, x}\right) \]\n\nUsing definition (5) of the transpose, we can rewrite the right-hand side as\n\n\[ {\left| {\mathrm{... | Yes |
Theorem 4. Suppose \( X, Y \), and \( Z \) above are normed linear spaces; then\n\n\[ \left| \mathrm{{ST}}\right| \leq \left| \mathrm{S}\right| \left| \mathrm{T}\right| \] | Proof. By definition (4),\n\n\[ \left| {\mathrm{S}y}\right| \leq \left| \mathrm{S}\right| \left| y\right| ,\;\left| {\mathrm{T}x}\right| \leq \left| \mathrm{T}\right| \left| x\right| \]\n\nHence\n\n\[ \left| {\mathrm{{ST}}x}\right| \leq \left| \mathrm{S}\right| \left| {\mathrm{T}x}\right| \leq \left| \mathrm{S}\right| ... | Yes |
Theorem 5. Let \( X \) and \( Y \) be finite-dimensional normed linear spaces of the same dimension, and let \( \mathrm{T} \) be a linear mapping of \( X \) into \( Y \) that is invertible. Let \( \mathrm{S} \) be another linear map of \( X \) into \( Y \) that does not differ too much from \( T \) in the sense that\n\... | Proof. We have to show that \( \mathrm{S} \) is one-to-one and onto. We show first that \( \mathrm{S} \) is one-to-one. We argue indirectly; suppose that for \( {x}_{0} \neq 0 \) ,\n\n\[ \mathrm{S}{x}_{0} = 0\text{.}\]\n\nThen\n\n\[ \mathrm{T}{x}_{0} = \left( {\mathrm{T} - \mathrm{S}}\right) {x}_{0} \]\n\nSince \( \mat... | Yes |
Theorem 6. Let \( X \) be a normed finite-dimensional linear space, \( \mathrm{R} \) a linear map of \( X \) into itself whose norm is less than 1 :\n\n\[ \left| \mathbf{R}\right| < 1\text{.} \]\n\nThen\n\n\[ \mathrm{S} = \mathrm{I} - \mathrm{R} \]\n\n is invertible, and\n\n\[ {\mathrm{S}}^{-1} = \mathop{\sum }\limits_... | Proof. Denote \( \mathop{\sum }\limits_{0}^{n}{\mathrm{R}}^{k} \) as \( {\mathrm{T}}_{n} \), and denote \( {\mathrm{T}}_{n}x \) as \( {y}_{n} \). We claim that \( \left\{ {y}_{n}\right\} \) is a Cauchy sequence; that is, \( \left| {{y}_{n} - {y}_{l}}\right| \) tends to zero as \( n \) and \( l \) tend to \( \infty \). ... | Yes |
Theorem 1 (Perron). Every positive matrix \( \mathrm{P} \) has a dominant eigenvalue, denoted by \( \lambda \left( \mathrm{P}\right) \) which has the following properties:\n\n(i) \( \lambda \left( \mathrm{P}\right) \) is positive and the associated eigenvector \( h \) has positive entries:\n\n\[ \mathrm{P}h = \lambda \... | Proof. We recall from Chapter 13 that inequality between vectors in \( {\mathbb{R}}^{n} \) means that the inequality holds for all corresponding components. We denote by \( p\left( \mathrm{\;P}\right) \) the set of all nonnegative numbers \( \lambda \) for which there is a nonnegative vector \( x \neq 0 \) such that\n\... | Yes |
Lemma 2. For P positive,\n\n(i) \( p\left( \mathrm{P}\right) \) is nonempty, and contains a positive number,\n\n(ii) \( p\left( \mathrm{P}\right) \) is bounded,\n\n(iii) \( p\left( \mathrm{P}\right) \) is closed. | Proof. Take any positive vector \( x \) ; since \( \mathrm{P} \) is positive, \( \mathrm{P}x \) is a positive vector. Clearly, (3) will hold for \( \lambda \) small enough positive; this proves (i) of the lemma.\n\nSince both sides of (3) are linear in \( x \), we can normalize \( x \) so that\n\n\[ \n{\xi x} = \sum {x... | Yes |
Theorem 3. Let \( S \) be a positive stochastic matrix.\n\n(i) The dominant eigenvalue \( \lambda \left( \mathrm{S}\right) = 1 \).\n\n(ii) Let \( x \) be any nonnegative vector; then\n\n\[ \mathop{\lim }\limits_{{N \rightarrow \infty }}{\mathrm{S}}^{N}x = {ch} \]\n\nwhere \( h \) the dominant eigenvector and \( c \) is... | Proof. As remarked earlier, if \( \mathrm{S} \) is a positive matrix, so is its transpose \( {\mathrm{S}}^{T} \). Since, according to Theorem 16, Chapter \( 6,\mathrm{\;S} \) and \( {\mathrm{S}}^{T} \) have the same eigenvalues, it follows that \( \mathrm{S} \) and \( {\mathrm{S}}^{T} \) have the same dominant eigenval... | Yes |
Theorem 4. Every nonnegative \( l \times l \) matrix \( \mathrm{F},\mathrm{F} \neq 0 \), has an eigenvalue \( \lambda \left( \mathrm{F}\right) \) with the following properties:\n\n(i) \( \lambda \left( \mathrm{F}\right) \) is nonnegative, and the associated eigenvector has nonnegative entries:\n\n\[ \mathrm{F}h = \lamb... | Proof. Approximate \( \mathrm{F} \) by a sequence \( {\mathrm{F}}_{n} \) of positive matrices. Since the characteristic equations of \( {\mathrm{F}}_{n} \) tend to the characteristic equations of \( \mathrm{F} \), it follows that the eigenvalues of \( {\mathrm{F}}_{n} \) tend to the eigenvalues of \( \mathrm{F} \) . No... | Yes |
Theorem 1. The solution \( x \) of (1) minimizes the functional\n\n\[ E\left( y\right) = \frac{1}{2}\left( {y,\mathrm{A}y}\right) - \left( {y, b}\right) ; \] | Proof. We add to \( E\left( y\right) \) a constant, that is, a term independent of \( y \) :\n\n\[ F\left( y\right) = E\left( y\right) + \frac{1}{2}\left( {x, b}\right) . \]\n\nSet (13) into (14); using \( \mathrm{A}x = b \) and the self-adjointness of \( \mathrm{A} \) we can express \( F\left( y\right) \) as\n\n\[ F\l... | Yes |
Theorem 2. The sequence of approximations defined by (15), with \( s \) given by (16), converges to the solution \( x \) of (1). | Proof. We need a couple of inequalities. We recall from Chapter 8 that for any vector \( r \) the Rayleigh quotient\n\n\[ \frac{\left( r,\mathrm{A}r\right) }{\left( r, r\right) } \]\n\nof a self-adjoint matrix \( A \) lies between the smallest and largest eigenvalues of \( A \) . In our case these were denoted by \( \a... | Yes |
Theorem 3. Suppose that in the expansion (42) of \( {r}_{0} \) none of the coefficients \( {w}_{j} \) are 0 ; suppose further that the eigenvalues \( {a}_{j} \) of \( \mathrm{A} \) are distinct. Then (44) furnishes a Euclidean structure to the space of all polynomials of degree less than the order \( K \) of the matrix... | Proof. According to Chapter 7, a scalar product needs three properties. The first two-bilinearity and symmetry-are obvious from either (40) or (44). To show positivity, we note that since each \( {a}_{j} > 0 \) ,\n\n\[ \{ Q, Q\} = \sum {w}_{j}^{2}{a}_{j}{Q}^{2}\left( {a}_{j}\right) \]\n\n(45)\n\nis obviously nonnegativ... | Yes |
Theorem 4. Let \( K \) be the order of the matrix \( \mathrm{A} \), and let \( {x}_{K} \) be the \( K \) th term of the sequence (36), the coefficients being defined by (52) and (60). We claim that \( {x}_{K} \) satisfies equation (1), \( \mathrm{A}{x}_{K} = b \) . | Proof. \( {Q}_{K} \) is defined as that polynomial of degree \( K \) which satisfies (37) and minimizes (38). We claim that this polynomial is \( {p}_{\mathrm{A}}/{p}_{\mathrm{A}}\left( 0\right) ,{p}_{\mathrm{A}} \) the characteristic polynomial of \( \mathrm{A} \) ; note that \( {p}_{\mathrm{A}}\left( 0\right) \neq 0 ... | Yes |
Theorem 1. Every real invertible square matrix A can be factored as\n\n\\[ \n\\mathrm{A} = \\mathrm{{QR}}, \n\\] \n\nwhere \\( Q \\) is an orthogonal matrix and \\( R \\) is an upper triangular matrix whose diagonal entries are positive. | Proof. The columns of \\( \\mathrm{Q} \\) are constructed out of the columns of \\( \\mathrm{A} \\) by Gram-Schmidt orthonormalization. So the \\( j \\) th column \\( {q}_{j} \\) of \\( \\mathrm{Q} \\) is a linear combination of the first \\( j \\) columns \\( {a}_{1},\\ldots ,{a}_{j} \\) of \\( \\mathrm{A} \\) :\n\n\\... | Yes |
Theorem 2. When the QR algorithm \( {\left( 7\right) }_{k},{\left( 8\right) }_{k} \) is applied to a real, symmetric, tridiagonal matrix \( \mathrm{L} \), all the matrices \( {\mathrm{L}}_{k} \) produced by the algorithm are real, symmetric, and tridiagonal, and have the same eigenvalues as L. | Proof. We have already shown, see \( {\left( 9\right) }_{k} \), that \( {\mathrm{L}}_{k} \) is symmetric and has the same eigenvalues as \( \mathrm{L} \) . To show that \( {\mathrm{L}}_{k} \) is tridiagonal, we start with \( \mathrm{L} = {\mathrm{L}}_{0} \) tridiagonal and then argue by induction on \( k \) . Suppose \... | Yes |
Theorem 3. Solutions \( \mathrm{L}\left( t\right) \) of equations in commutator form (21), where \( \mathrm{B} \) is antisymmetric, are isospectral. | Proof. Let the matrix \( \mathrm{V}\left( t\right) \) be the solution of the differential equation.\n\n\[ \frac{d}{dt}\mathrm{\;V} = \mathrm{{BV}},\;\mathrm{V}\left( 0\right) = \mathrm{I}. \]\n\n(25)\n\nSince \( B\left( t\right) \) is antisymmetric, the transpose of (25) is\n\n\[ \frac{d}{dt}{\mathrm{V}}^{\mathrm{T}} =... | Yes |
Lemma 4. An off-diagonal entry \( {b}_{k}\left( t\right) \) of \( \mathrm{L}\left( t\right) \) is either nonzero for all \( t \), or zero for all \( t \) . | Proof. Let \( \left\lbrack {{t}_{0},{t}_{1}}\right\rbrack \) be an interval on which \( {b}_{k}\left( t\right) \) is nonzero. Divide the differential equation (24) for \( {b}_{k} \) by \( {b}_{k} \) and integrate it from \( {t}_{0} \) to \( {t}_{1} \) :\n\n\[ \log {b}_{k}\left( {t}_{1}\right) - \log {b}_{k}\left( {t}_{... | Yes |
Lemma 5. Suppose none of the off diagonal terms \( {b}_{k} \) in \( \mathrm{L} \) is zero.\n\n(i) The first component \( {u}_{1k} \) of every eigenvector \( {u}_{k} \) of \( \mathrm{L} \) is nonzero.\n\n(ii) Each eigenvalue of \( L \) is simple. | Proof. (i) The first component of the eigenvalue equation\n\n\[ \mathrm{L}{u}_{k} = {d}_{k}{u}_{k} \]\n\n(30)\n\nis\n\n\[ {a}_{1}{u}_{1k} + {b}_{1}{u}_{2k} = {d}_{k}{u}_{{1k}.} \]\n\n(31)\n\nIf \( {u}_{1k} \) were zero, it would follow from (31), since \( {b}_{1} \neq 0 \), that \( {u}_{2, k} = 0 \) . We can then use t... | Yes |
Lemma 6. The eigenvalues \( {d}_{1},\ldots ,{d}_{n} \) and the first components \( {u}_{1, k} \) , \( k = 1,\ldots, n \), of the normalized eigenvectors of \( \mathrm{L} \) uniquely determine all entries \( {a}_{1},\ldots ,{a}_{n} \) and \( {b}_{1},\ldots ,{b}_{k - 1} \) of \( \mathrm{L} \) . | Proof. From the spectral representation (29), we can express the entry \( {\mathrm{L}}_{11} = {a}_{1} \) of \( \mathrm{L} \) as follows:\n\n\[ \n{a}_{1} = \sum {d}_{k}{u}_{1k}^{2} \n\]\n\n(32)\n\nFrom equation (31) we get\n\n\[ \n{b}_{1}{u}_{2k} = \left( {{d}_{k} - {a}_{1}}\right) {u}_{1k} \n\]\n\n\( {\left( {33}\right... | Yes |
Theorem 7. (Moser). \( \\mathrm{L}\\left( t\\right) \) is a solution of equation (21). Denote the eigenvalues of L by \( {d}_{1},\\ldots ,{d}_{n} \), arranged in decreasing order, and denote by D the diagonal matrix with diagonal entries \( {d}_{1},\\ldots ,{d}_{n} \). Then\n\n\[ \n\\mathop{\\lim }\\limits_{{t \\righta... | Proof. We start with the following lemma.\n\nLemma 8. Denote by \( u | No |
Claim:\n\n\[ u\left( t\right) = \frac{u\left( 0\right) {e}^{\mathrm{D}t}}{\begin{Vmatrix}u\left( 0\right) {e}^{\mathrm{D}t}\end{Vmatrix}}. \] | Proof. We have shown that when \( \mathrm{L}\left( t\right) \) satisfies (21), \( \mathrm{L}\left( t\right) \) and \( \mathrm{L}\left( 0\right) \) are related by (28). Multiplying this relation by \( \mathrm{V}\left( t\right) \) on the left gives\n\n\[ \mathrm{L}\left( t\right) \mathrm{V}\left( t\right) = \mathrm{V}\le... | Yes |
Theorem 1\n\n\[ \n\det \mathrm{V}\left( {{a}_{1},\ldots ,{a}_{n}}\right) = \mathop{\prod }\limits_{{j > i}}\left( {{a}_{j} - {a}_{i}}\right) \n\] | Proof. Using formula (16) of Chapter 5 for the determinant, we conclude that det \( \mathrm{V} \) is a polynomial in the \( {a}_{i} \) of degree less than or equal to \( n\left( {n - 1}\right) /2 \) . Whenever two of the scalars \( {a}_{i} \) and \( {a}_{j}, i \neq j \), are equal, \( \mathrm{V} \) has two equal column... | Yes |
Lemma 1. There is a matrix \( C \) whose entries are polynomials in the entries of the antisymmetric matrix \( A \) such that\n\n\[ \mathrm{B} = {\mathrm{{CAC}}}^{T} \]\n\n(1)\n\nis antisymmetric and tridiagonal, that is, \( {b}_{ij} = 0 \) for \( \left| {i - j}\right| > 1 \) . Furthermore, \( \det C \neq 0 \) . | Proof. We construct \( \mathrm{C} \) as a product\n\n\[ \mathrm{C} = {\mathrm{C}}_{n - 2}\cdots {\mathrm{C}}_{2}{\mathrm{C}}_{1} \]\n\n\( {\mathrm{C}}_{1} \) is required to have the following properties:\n\n(i) \( {\mathrm{B}}_{1} = {\mathrm{C}}_{1}{\mathrm{{AC}}}_{1}^{T} \) has zeros for the last \( \left( {n - 2}\rig... | Yes |
Lemma 2. If a polynomial \( P \) in \( n \) variables is the square of a rational function \( R, R \) is a polynomial. | Proof. For functions of one variable this follows by elementary algebra; so we can conclude that for each fixed variable \( x, R \) is a polynomial in \( x \), with coefficients from the field of rational functions in the remaining variables. It follows that there exists a \( k \) such that the \( k \) th partial deriv... | Yes |
A matrix \( \mathrm{S} \) that preserves \( \left( {x,\mathrm{\;J}y}\right) \) :\n\n\[ \left( {\mathrm{S}x,\mathrm{{JS}}y}\right) = \left( {x,\mathrm{J}y}\right) \]\n\nfor all \( x \) and \( y \), satisfies\n\n\[ {\mathrm{S}}^{T}\mathrm{{JS}} = \mathrm{J} \] | Proof. \( \left( {\mathrm{S}x,\mathrm{{JS}}y}\right) = \left( {x,{\mathrm{\;S}}^{T}\mathrm{{JS}}y}\right) \) . If this is equal to \( \left( {x,\mathrm{J}y}\right) \) for all \( x, y,{\mathrm{\;S}}^{T}\mathrm{{SJ}}y = {\mathrm{J}}_{y} \) for all \( y \) . | Yes |
Theorem 2. (i) Symplectic matrices form a group under matrix multiplication. | Proof. (i) It follows from (4) that every symplectic matrix is invertible. That they form a group follows from (3). | No |
Theorem 3. Let \( \mathrm{S}\left( t\right) \) be a differentiable function of the real variable \( t \), whose values are symplectic matrices. Define \( \mathbf{G}\left( t\right) \) by\n\n\[ \frac{d}{dt}\mathrm{\;S} = \mathrm{{GS}} \]\n\n(5)\n\nThen \( G \) is of the form\n\n\[ \mathrm{G} = \mathrm{{JL}}\left( t\right... | Proof. For each \( t\left( 4\right) \) is satisfied; differentiate it with respect to \( t \) :\n\n\[ \left( {\frac{d}{dt}{\mathrm{\;S}}^{T}}\right) \mathrm{{JS}} + {\mathrm{S}}^{T}\mathrm{\;J}\frac{d}{dt}\mathrm{\;S} = 0. \]\n\nMultiply by \( {\mathrm{S}}^{-1} \) on the right, \( {\left( {\mathrm{S}}^{T}\right) }^{-1}... | Yes |
Theorem 4. For a symplectic matrix \( \mathrm{S},\lambda = 1 \) or -1 cannot be a simple eigenvalue. | Proof. We argue indirectly: suppose, say, that \( \lambda = - 1 \) is a simple eigenvalue, with eigenvector \( h \) :\n\n\[ \mathrm{S}h = - h\text{.} \]\n\n(8)\n\nMultiplying both sides by \( {S}^{T}J \) and using (4) we get\n\n\[ \mathrm{J}h = - {\mathrm{S}}^{T}\mathrm{\;J}h \]\n\n\( {\left( 8\right) }^{\prime } \)\n\... | Yes |
Theorem 5. The determinant of a symplectic matrix \( S \) is 1 . | Proof. Since we already know that \( {\left( \det S\right) }^{2} = 1 \), we only have to exclude the possibility that \( \det S \) is negative. The determinant of a matrix is the product of its eigenvalues. The complex eigenvalues come in conjugate pairs; their product is positive. The real eigenvalues \( \neq 1, - 1 \... | No |
Lemma 2. Let \( u, z \) be a pair of vectors in \( U, v, w \) a pair of vectors in \( V \) . Then\n\n\[ \n\left( {u \otimes v, z \otimes w}\right) = \left( {u, z}\right) \left( {v, w}\right) .\n\] | Proof. Expand \( u \) and \( z \) in terms of the \( {e}_{i}, v \) and \( w \) in terms of \( {f}_{j} \) .\n\n\[ \nu = \sum {a}_{i}{e}_{i},\;z = \sum {b}_{k}{e}_{k} \]\n\n\[ v = \sum {c}_{j}{f}_{j},\;w = \sum {d}_{l}{f}_{l} \]\n\nThen\n\n\[ u \otimes v = \sum {a}_{i}{c}_{j}{e}_{i} \otimes {f}_{j},\;z \otimes w = \sum {... | Yes |
Theorem 1. Every lattice has an integer basis, that is, a collection of vectors in \( L \) such that every vector in the lattice can be expressed uniquely as a linear combination of basis vectors with integer coefficients. | Proof. The dimension of a lattice \( L \) is the dimension of the linear space it spans. Let \( L \) be \( k \) -dimensional, and let \( {p}_{1},\ldots ,{p}_{k} \) be a basis in \( L \) for the span of \( L \) ; that is, every vector \( t \) in \( L \) can be expressed uniquely as\n\n\[ t = \sum {a}_{j}{p}_{j},\;{a}_{j... | No |
Theorem 2. Let \( L \) be an \( n \) -dimensional lattice in \( {\mathbb{R}}^{n} \) . Let \( {q}_{1},\ldots ,{q}_{n} \) and \( {r}_{1},\ldots ,{r}_{n} \) be two integer bases of \( L \) ; denote by \( \mathrm{Q} \) and \( \mathrm{R} \) the matrices whose columns are \( {q}_{1} \) and \( {r}_{i} \), respectively. Then\n... | EXERCISE 2. (i) Prove Theorem 2. | No |
Theorem 1. A Lorentz transformation L either maps each light cone onto itself or onto each other. | Proof. Take any point, say \( {\left( 1,0,0\right) }^{\prime } \) in the flc; since \( L \) preserves the Minkowski metric (2), the image of this point belongs to either the forward or backward light cone, say the fic. We claim that then \( L \) maps every point \( u \) in the fic into the fic. For suppose on the contr... | Yes |
Theorem 2. Suppose L belongs to the proper Lorentz group, and maps the point \( e = \left( {1,0,0}\right) \) onto itself. Then \( \mathrm{L} \) is rotation around the \( t \) axis. | Proof. Le \( = e \) implies that the first column of L is \( {\left( 1,0,0\right) }^{t} \) . According to (4), \( {\mathrm{L}}^{\prime }\mathrm{{ML}} = \mathrm{M} \) ; since \( \mathrm{M}e = e,{\mathrm{\;L}}^{\prime }e = e \) ; therefore the first column of \( {\mathrm{L}}^{\prime } \) is \( {\left( 1,0,0\right) }^{\pr... | Yes |
Theorem 3. (a) For \( \mathrm{W} \) in \( \mathrm{{SL}}\left( {2,\mathbb{R}}\right) ,{\mathrm{L}}_{\mathrm{W}} \) belongs to the proper Lorentz group. | Proof. (a) A symmetric matrix \( \mathrm{U} \) representing a point \( u = {\left( t, x, y\right) }^{\prime } \) in the fic is positive, and the converse is also true. For according to (6), \( \det \mathrm{U} = {t}^{2} - {x}^{2} - {y}^{2} \) , \( \operatorname{tr}U = {2t} \), and the positivity of both is equivalent to... | Yes |
Theorem 4. Every proper Lorentz transformation \( L \) is of the form \( {L}_{Y}, Y \) in \( \mathrm{{SL}}\left( {2,\mathbb{R}}\right) \) . | Proof. Denote by \( u \) the image of \( e = \left( {1,0,0}\right) \) under L:\n\n\[ \mathrm{{Le}} = u\text{.}\]\n\nSince \( e \) lies in the fic, so does \( u \) . According to part (b) of Theorem 3, \( {\mathrm{L}}_{\mathrm{W}}e = u \) for some \( \mathrm{W} \) in \( \mathrm{{SL}}\left( {2,\mathbb{R}}\right) \) . The... | Yes |
Theorem 5. (a) Every plane \( \left( {u, p}\right) = 0 \) that satisfies (13) has a nonempty intersection with \( \mathbb{H} \) . | Proof. (a) Set \( p = \left( {s, a, b}\right) \) ; condition (13) means that \( {s}^{2} < {a}^{2} + {b}^{2} \) . The point \( u = \left( {{a}^{2} + {b}^{2}, - {as}, - {bs}}\right) \) satisfies \( \left( {u, p}\right) = 0 \) . We claim that \( u \) belongs to the fic; for\n\n\[ \left( {u,\mathrm{M}u}\right) = {\left( {a... | Yes |
Theorem 6. (a) For every \( \mathrm{W} \) in \( \mathrm{{SL}}\left( {2,\mathbb{C}}\right) ,{\mathrm{L}}_{\mathrm{W}} \) defined above is a proper Lorentz transformation.\n\n(b) The mapping \( \mathrm{W} \rightarrow {\mathrm{L}}_{\mathrm{W}} \) is a homomorphic map of \( \mathrm{{SL}}\left( {2,\mathbb{C}}\right) \) onto... | We leave it to the reader to prove this theorem using the techniques developed in Section 1. | No |
Theorem 7. Z is a \( 2 \times 2 \) unitary matrix of determinant 1, and \( U \) is as above. Then\n\n\[ \mathrm{V} = {\mathrm{{ZUZ}}}^{ * } \]\n\n(25)\n\nis\n\n(i) self-adjoint,\n\n(ii) \( \det \mathrm{V} = \det \mathrm{U} \) ,\n\n(iii) \( \operatorname{tr}\mathrm{V} = 0 \) . | Proof. (i) is clearly true. To see (ii) take the determinant of (25). To deduce (iii), we use the commutativity of trace-that is, that \( \operatorname{tr}{AB} = \operatorname{tr}{BA} \) . Thus\n\n\[ \operatorname{tr}\mathrm{V} = \operatorname{tr}\mathrm{{ZU}}{\mathrm{Z}}^{ * } = \operatorname{tr}\mathrm{U}{\mathrm{Z}}... | Yes |
Theorem 8. The mapping \( Z \rightarrow {O}_{Z} \) defined above is a homomorphic map of \( \mathrm{{SU}}\left( {2,\mathbb{C}}\right) \) onto \( \mathrm{{SO}}\left( {3,\mathbb{R}}\right) \) . This mapping is 2 to 1 . | We leave it to the reader to prove this theorem. | No |
Theorem 1. The unit ball in \( S \) is compact. | Proof. Multiply equation (1) by \( u \) and integrate over \( \mathrm{G} \) :\n\n\[ 0 = {\int }_{\mathrm{G}}\left( {a{u}^{2} + {u\Delta u}}\right) {dxdy} \]\n\n(5)\n\nUsing the definition (2) of \( \Delta \), we can integrate the second term by parts:\n\n\[ {\int }_{\mathrm{G}}u\left( {{u}_{xx} + {u}_{yy}}\right) {dxdy... | Yes |
Theorem 3. Let \( \mathrm{G} \) be a bounded domain with a smooth boundary, and let \( \mathrm{D} \) be a set of functions in \( G \) whose values and the values of their first derivatives are uniformly bounded in \( \mathrm{G} \) by a common bound \( m \) . Every sequence of functions in \( \mathrm{D} \) contains a su... | EXERCISE I. (i) Show that a set of functions whose first derivatives are uniformly bounded in \( \mathrm{G} \) are equicontinuous in \( \mathrm{G} \) .\n\n(ii) Use (i) and the Arzela-Ascoli theorem to prove Theorem 3. | No |
Theorem 1. An \( n \times n \) matrix \( \mathrm{X} \) is the commutator of two \( n \times n \) matrices \( \mathrm{A} \) and \( B \) ,\n\n\[ \n\mathrm{X} = \mathrm{{AB}} - \mathrm{{BA}} \n\]\n\niff the trace of \( \mathrm{X} \) is zero. | Proof. We have shown in Theorem 7 of Chapter 5 that trace is commutative-that is, that\n\n\[ \n\operatorname{tr}{AB} = \operatorname{tr}{BA}. \n\]\n\nIt follows that for \( \mathrm{X} \) of form (1), \( \operatorname{tr}\mathrm{X} = 0 \) . We show now the converse. | Yes |
Lemma 2. Every matrix \( \mathrm{X} \) all of whose diagonal entries are zero can be represented as a commutator. | Proof. We shall construct explicitly a pair of matrices A and B so that (1) holds. We choose arbitrarily \( n \) distinct numbers \( {a}_{1},\ldots ,{a}_{n} \) and define \( \mathrm{A} \) to be the diagonal matrix with diagonal entries \( {a}_{i} \) :\n\n\[ \n{A}_{ij} = \left\{ \begin{array}{ll} 0 & \text{ for }i \neq ... | Yes |
Lemma 3. Every matrix \( \mathrm{X} \) whose trace is zero is similar to a matrix all whose diagonal entries are zero. | Proof. Suppose not all diagonal entries of \( \mathrm{X} \) are zero, say \( {x}_{11} \neq 0 \) . Then, since \( \operatorname{tr}\mathrm{X} = 0 \), there must be another diagonal entry, say \( {x}_{22} \), that is neither zero nor equal to \( {x}_{11} \) . Therefore the \( 2 \times 2 \) minor in the upper left corner ... | Yes |
Theorem 1. Let \( W \) be a mapping of a finite-dimensional Euclidean space into itself whose self-adjoint part is negative:\n\n\[ \mathrm{W} + {\mathrm{W}}^{ * } < 0. \]\n\n(1)\n\nThen the eigenvalues of \( \mathbf{W} \) have negative real part. | This can be proved the same way as Theorem 20 was in Chapter 10. | No |
Theorem 2. Let \( W \) be a mapping of a finite-dimensional Euclidean space \( X \) into itself. Let \( G \) be a positive self-adjoint map of \( X \) into itself that satisfies the inequality\n\n\[ \mathrm{{GW}} + {\mathrm{W}}^{ * }\mathrm{G} < 0. \]\n\nThen the eigenvalues of \( W \) have negative real part. | Proof. Let \( h \) be an eigenvector of \( \mathrm{W} \), where \( w \) is the corresponding eigenvalue:\n\n\[ \mathrm{W}h = {wh}\text{.} \]\n\nLet the left-hand side of (2) act on \( h \), and take the scalar product with \( h \) ; according to (2),\n\n\[ \left( {\left( {\mathrm{{GW}} + {\mathrm{W}}^{ * }\mathrm{G}}\r... | Yes |
Theorem 3 (Liapunov). Let \( W \) be a mapping of a finite-dimensional Euclidean space \( \mathrm{X} \) into itself, whose eigenvalues have negative real part. Then there exists a positive self-adjoint mapping \( G \) such that inequality (2),\n\n\[ \mathrm{{GW}} + {\mathrm{W}}^{ * }\mathrm{G} < 0, \]\n\n(2)\n\nholds. | Proof. We recall from Chapter 9 the definition of the exponential of a matrix:\n\n\[ {e}^{\mathrm{W}} = \mathop{\sum }\limits_{0}^{\infty }\frac{{\mathrm{W}}^{k}}{k!} \]\n\n(5)\n\nAccording to Exercise 7 there, the eigenvalues of \( {e}^{\mathrm{W}} \) are the exponentials of the eigenvalue of W.\n\nLet \( t \) be any ... | No |
Lemma 4. If the eigenvalues of \( \mathrm{W} \) have negative real part, \( \begin{Vmatrix}{e}^{\mathrm{W}t}\end{Vmatrix} \) tends to zero at an exponential rate as \( t \rightarrow + \infty \) . | Proof. According to Theorem 18 of Chapter 7, whose proof appears in Appendix 10, for any mapping \( A \) of a finite-dimensional Euclidean space \( X \) into itself,\n\n\[ \lim {\begin{Vmatrix}{\mathrm{A}}^{j}\end{Vmatrix}}^{1/j} = r\left( \mathrm{\;A}\right) \]\n\n(6)\n\nwhere \( r\left( \mathrm{\;A}\right) \) is the ... | Yes |
Lemma 2. A maps the quotient space \( {\mathrm{N}}_{i + 1}/{\mathrm{N}}_{i} \) into \( {\mathrm{N}}_{i}/{\mathrm{N}}_{i - 1} \), and this mapping is one-to-one. | Proof. A maps \( {\mathrm{N}}_{i + 1} \) into \( {\mathrm{N}}_{ij} \) ; therefore A maps \( {\mathrm{N}}_{i + 1}/{\mathrm{N}}_{i} \) into \( {\mathrm{N}}_{i}/{\mathrm{N}}_{i - 1} \) . Let \( \{ x\} \) be a nonzero equivalence class in \( {\mathrm{N}}_{i + 1}/{\mathrm{N}}_{ij} \) ; this means that no \( x \) belongs to ... | Yes |
\[ w\left( \mathrm{\;A}\right) \leq \parallel \mathrm{A}\parallel . \] | Proof. By the Schwarz inequality we have \[ \left| \left( {\mathrm{A}x, x}\right) \right| \leq \parallel \mathrm{A}x\parallel \parallel x\parallel \leq \parallel \mathrm{A}\parallel \parallel x{\parallel }^{2}; \] since \( \parallel x\parallel = 1 \), part (i) follows. | Yes |
Theorem 2. For A as above,\n\n\[ w\left( {\mathrm{\;A}}^{n}\right) \leq w{\left( \mathrm{\;A}\right) }^{n} \]\n\n(6)\n\nfor every positive integer \( n \) . | The first proof of this result was given by Charles Berger. The remarkable simple proof presented here is Carl Pearcy's. | No |
Theorem 1. Suppose \( M \subseteq \mathbb{C} \) contains 0 and 1 . Set\n\n\[ K \mathrel{\text{:=}} \mathbb{Q}\left( {M \cup \bar{M}}\right) \]\n\nFor a given \( z \in \mathbb{C} \), the following statements are equivalent:\n\n(i) \( z \in {\Delta M} \), that is, \( z \) is constructible from \( M \) with ruler and comp... | Proof. (ii) \( \Rightarrow \) (i): By assumption, there exists a finite chain of subfields of \( \mathbb{C} \), say\n\n\[ K = {K}_{0} \subseteq {K}_{1} \subseteq \cdots \subseteq {K}_{m} = E, \]\n\nsatisfying \( {K}_{i} = {K}_{i - 1}\left( {w}_{i}\right) \) with \( {w}_{i}^{2} \in {K}_{i - 1} \) for each \( i \), and a... | Yes |
Theorem 2 (Lindemann 1882). The number \( \pi \) is transcendental. | Proof. If it were possible, we would have \( \pi \in \Delta \mathbb{Q} \) ; by Theorem 1 then \( \pi \) would be algebraic, which by Lindemann's Theorem is not the case. | No |
Theorem 1. Let \( M \) be a subset of \( \mathbb{C} \) containing 0 and 1. Let \( K = \mathbb{Q}\left( {M \cup \bar{M}}\right) \) . The field extension \( \bigtriangleup M/K \) is algebraic. | Proof. Take \( z \in \bigtriangleup M \) . From F9 of Chapter 1 we know that \( K\left( z\right) : K < \infty \) . Then F6 says that \( z \) is algebraic over \( K \) . | No |
Theorem 1. Let \( L/K \) be a simple algebraic field extension and \( \alpha \) a primitive element of \( L/K \) . Put \( f = {\operatorname{MiPo}}_{K}\left( \alpha \right) \) . The substitution homomorphism corresponding to \( \alpha \) , \[ K\left\lbrack X\right\rbrack \rightarrow L = K\left( \alpha \right) = K\left\... | In the situation of Theorem 1, the isomorphism \[ K\left( \alpha \right) \simeq K\left\lbrack X\right\rbrack /f \] gives a good description of a simple algebraic field extension: it all boils down to computing in \( K\left\lbrack X\right\rbrack \) modulo \( f \) . Crucially, this description also provides a hint for ho... | Yes |
Theorem 2. Let \( f \) be an irreducible polynomial in \( K\left\lbrack X\right\rbrack \) . If \( f \) divides gh for \( g, h \in K\left\lbrack X\right\rbrack \), then \( f \mid g \) or \( f \mid h \) . | Proof. This assertion, which you're surely familiar with, will follow from general considerations in Chapter 4. Here we give an ad hoc argument: Suppose that \( f, g, h \) contradict the theorem. Division with remainder gives \( g = {qf} + r \) . Since \( f \mid {gh} \) we have \( f \mid {rh} \), so we might as well as... | Yes |
Theorem 4 (Kronecker). Every nonconstant polynomial \( f\left( X\right) \) over a field \( K \) has a root in some appropriate extension of \( K \) . | Proof. Since \( \deg f \geq 1 \), there must be an irreducible polynomial \( g \) dividing \( f \) (consider all nonconstant factors of \( f \) and take one of least degree). If an extension of \( K \) contains a root of \( g \) it will also serve for \( f \) ; therefore we assume without loss of generality that \( f \... | Yes |
Theorem 5. Suppose \( E/K \) is an algebraic extension. Then \( E/K \) is simple if and only if it possesses only finitely many intermediate fields. | Proof. Denote by \( \mathcal{L} \) be the set of all intermediate fields of \( E/K \) .\n\n(i) Assume \( E = K\left( \alpha \right) \), and set \( f = {\operatorname{MiPo}}_{K}\left( \alpha \right) \) . To prove the finiteness of \( \mathcal{L} \) , consider the set\n\n\[ \mathcal{D} = \{ g \in E\left\lbrack X\right\rb... | Yes |
Theorem 1 (Gauss). Let \( R \) be a UFD with fraction field \( K \) . Let \( {\mathcal{P}}_{1} \) be a directory of primes for \( R \) and \( {\mathcal{P}}_{2} \) a directory of primes for \( K\left\lbrack X\right\rbrack \) containing only primitive polynomials of \( R\left\lbrack X\right\rbrack \) . Then \( R\left\lbr... | Proof. Take any nonzero \( g \in R\left\lbrack X\right\rbrack \) . Since \( K\left\lbrack X\right\rbrack \) is a UFD, there is a unique factorization\n\n(11)\n\n\[ g = a\mathop{\prod }\limits_{{f \in {\mathcal{P}}_{2}}}{f}^{{e}_{f}}\;\text{ with }a \in {K}^{ \times } = K{\left\lbrack X\right\rbrack }^{ \times }\text{ a... | Yes |
Theorem 2 (Steinitz). Let \( K \) be a field.\n\n(I) There exists an extension \( C \) of \( K \) with the following properties:\n\n(i) \( C \) is algebraically closed.\n\n(ii) \( C/K \) is algebraic.\n\nSuch a field is called an algebraic closure of \( K \) . | Proof of part (I). Let \( K\left\lbrack {{X}_{n}, n \in \mathbb{N}}\right\rbrack \) be the polynomial ring in countably many inde-terminates \( {X}_{1},{X}_{2},\ldots \) over \( K \) . Consider the set \( I \) of all subsets \( M \subseteq K\left\lbrack {{X}_{n}, n \in \mathbb{N}}\right\rbrack \) such that\n\n\( M \) i... | Yes |
Theorem 3. Let \( \sigma : K \rightarrow {K}^{\prime } \) be an isomorphism of fields and let \( L/K \) be an algebraic field extension. If \( C \) is an algebraically closed extension of \( {K}^{\prime } \), the map \( \sigma \) can be extended to a homomorphism \( \tau : L \rightarrow C \) . | Proof. (a) We deal first with the case \( {K}^{\prime } = K \) and \( \sigma = {\mathrm{{id}}}_{K} \) . An application of Theorem 1 to \( {E}_{1} \mathrel{\text{:=}} L \) and \( {E}_{2} \mathrel{\text{:=}} C \) shows there exists an extension \( E/K \) and \( K \) -homomorphisms \( {\sigma }_{i} : {E}_{i} \rightarrow E... | Yes |
Theorem 4. Let \( E/K \) be an algebraic field extension, and let \( C \) be an algebraic closure of \( E \) (and therefore also of \( K \) ). The following statements are equivalent:\n\n(i) \( E/K \) is normal.\n\n(ii) For every homomorphism \( \sigma : E/K \rightarrow C/K \) we have \( {\sigma E} = E \) (that is, \( ... | Proof. (iii) \( \Rightarrow \left( {\mathrm{{ii}}}^{\prime }\right) \) : Let \( \sigma : C/K \rightarrow C/K \) be an isomorphism. Then \( \sigma \left( N\right) \subseteq N \), so \( \sigma \left( {K\left( N\right) }\right) \subseteq K\left( N\right) \), that is, \( {\sigma E} \subseteq E \) . Now F3 shows that \( {\s... | Yes |
Theorem 1. For an algebraic extension \( E/K \) there is equivalence between:\n\n(i) \( E/K \) is Galois.\n\n(ii) \( E/K \) is normal and separable. | Proof. (ii) \( \Rightarrow \) (i): Let \( \alpha \) be any element of \( E \) not lying in \( K \) . We must show that there exists \( \tau \in G\left( {E/K}\right) \) such that \( {\tau \alpha } \neq \alpha \) . Let \( C \) be an algebraic closure of \( E \) . By assumption, \( \alpha \) is separable over \( K \), so ... | Yes |
Theorem 2. Let \( E/K \) be a Galois extension. For every intermediate field \( F \) of \( E/K \) , the extension \( E/F \) is also Galois: | Proof. Clearly \( E/F \) is algebraic, separable and normal if \( E/K \) has each of these properties. Thus \( E/F \) is a Galois extension, by Theorem 1. | Yes |
Theorem 3 (Primitive element theorem). A field extension \( E/K \) that is finite and separable is also simple, that is, \( E = K\left( \alpha \right) \) for some \( \alpha \in E \) . | Proof. In view of Chapter 3, Theorem 5, it suffices to show that \( E/K \) has only finitely many intermediate fields. Let \( {E}^{\prime } \) be a normal closure of \( E/K \) . If \( {E}^{\prime }/K \) has only finitely many intermediate fields, so does \( E/K \) . By F5, \( {E}^{\prime }/K \) is a Galois extension; i... | Yes |
Theorem 4. Let \( E \) be a field and \( G \) a finite group of automorphisms of \( E \), with fixed field \( K = {E}^{G} \) . The extension \( E/K \) is finite and Galois; moreover, \[ G = G\left( {E/K}\right) \] that is, \( G \) coincides with the Galois group of \( E/K \) . | Proof. In view of F1, \( E/K \) is algebraic, separable and normal. Thus it is a Galois extension. Let \( d \mathrel{\text{:=}} \left| G\right| \) be the order of \( G \) . By F1 we know at first only that \[ K\left( \alpha \right) : K \leq d\;\text{ for each }\alpha \in E. \] There is certainly some \( \alpha \in E \)... | Yes |
Theorem 5 (Fundamental theorem of Galois theory for finite Galois extensions). Let \( E/K \) be a finite Galois extension. Then the map\n\n(9)\n\n\[ F \mapsto G\left( {E/F}\right) \]\n\n is a bijection between the set of intermediate fields \( F \) of \( E/K \) and the set of subgroups \( H \) of \( G \mathrel{\text{:=... | Proof. Theorem 2 says that, for every intermediate field \( F \) of \( E/K \),\n\n(14)\n\n\[ {E}^{G\left( {E/F}\right) } = F. \]\n\nSince \( E/K \) was assumed finite, \( G = G\left( {E/K}\right) \) is finite by \( \mathrm{F}6 \), and hence so is any subgroup of \( G \) . By Theorem 4, then,\n\n(15)\n\n\[ G\left( {E/{E... | Yes |
Theorem 2. The multiplicative group \( {K}^{ \times } \) of a finite field \( K \) is cyclic. (A generator of \( {K}^{ \times } \) is called a primitive root.) | In fact we prove something more general:\n\nTheorem 2’. Let \( K \) be any field. Then any finite subgroup \( G \) of \( {K}^{ \times } \) is cyclic.\n\nThe proof relies on the following characterization of cyclic groups:\n\nLemma. A finite group \( G \) of order \( n \) is cyclic if\n\n\[ \left| \left\{ {x \in G \mid ... | Yes |
Theorem 3 (Gauss). Let \( E = \mathbb{Q}\left( \sqrt[n]{1}\right) \) be the field of \( n \) -th roots of unity over \( \mathbb{Q} \) . The Galois group of \( E/\mathbb{Q} \) is canonically isomorphic to the group of prime residue classes \( {\left( \mathbb{Z}/n\right) }^{ \times } \) . In particular, \( \mathbb{Q}\lef... | Proof. Let \( \zeta \) be a primitive \( n \) -th root of unity in \( E \) . In view of the homomorphism (21), we must show that for each \( k \in \mathbb{N} \) relatively prime to \( n \), there exists \( \sigma \in \) \( G\left( {E/K}\right) \) with \( {\sigma \zeta } = {\zeta }^{k} \) . By F7 in Chapter 8, this boil... | Yes |
Theorem 4. Let \( E/K \) be any extension of finite fields and let \( q \) be the number of elements of \( K \) . Then \( E/K \) is Galois with a cyclic Galois group, and in fact \( G\left( {E/K}\right) \) is generated by the automorphism \( {\sigma }_{q} : \alpha \rightarrow {\alpha }^{q} \) of \( E \) . (We call \( {... | Proof. Let \( G \) be the subgroup of Aut \( E \) generated by \( {\sigma }_{q} \) . Then \( K \) is the fixed field of \( G \) in \( E \), since by Section \( {9.1K} \) contains exactly those \( \alpha \in E \) such that \( {\alpha }^{q} = \alpha \) . By Galois theory (Chapter 8, Theorem 4), \( E/K \) is a Galois exte... | Yes |
F1. Let \( G \) act on \( M \) . Any two distinct orbits (relative to the given group action) are disjoint, and \( M \) is the union of all such orbits. | Proof. For \( x, y \in M \), write \( x \sim y \) if there is \( \sigma \in G \) such that \( {\sigma x} = y \) . You can easily persuade yourself that \( \sim \) defines an equivalence relation on \( M \) . The equivalence class of \( x \in M \) under \( \sim \) is none other than the orbit \( {Gx} \) of \( x \) . Thi... | No |
F2 (Cayley’s Theorem). Every group \( G \) of order \( n \) is isomorphic to a subgroup of the symmetric group \( {S}_{n} \) . | Let \( G \) be a group. Then \( G \) acts on \( M \mathrel{\text{:=}} G \) via the action \( \left( {\sigma ,\tau }\right) \mapsto {\sigma \tau } \). The corresponding homomorphism \( T : G \rightarrow S\left( G\right) \) associates to every \( \sigma \in G \) the left translation \( {T}_{\sigma } \) corresponding to \... | Yes |
F3 (Euler-Lagrange). For any subgroup \( H \) of a finite group \( G \) , \[ \left| G\right| = \left( {G : H}\right) \cdot \left| H\right| . \] | If \( G \) is finite, so are \( H \) and \( H \smallsetminus G \) . Since \( \left| {H\sigma }\right| = \left| H\right| \) for all \( \sigma \in G \), we see that \( {\mathrm{{Fl}}}^{\prime } \) implies the next result: | No |
Example 4. Let \( G \) be a group. As can easily be checked, \( G \) acts on \( M = G \) via the map \( \left( {\sigma ,\tau }\right) \mapsto {\sigma \tau }{\sigma }^{-1} \) . Let \( \sigma \mapsto {T}_{\sigma } \) be the corresponding homomorphism from \( G \) into \( S\left( G\right) \), that is, \( {T}_{\sigma }\lef... | \[ {T}_{\sigma }\left( {{\tau }_{1}{\tau }_{2}}\right) = \sigma \left( {{\tau }_{1}{\tau }_{2}}\right) {\sigma }^{-1} = \sigma {\tau }_{1}{\sigma }^{-1}\sigma {\tau }_{2}{\sigma }^{-1} = {T}_{\sigma }\left( {\tau }_{1}\right) {T}_{\sigma }\left( {\tau }_{2}\right) \] therefore \( {T}_{\sigma } \) is actually an automor... | Yes |
Let \( G \) act on \( M \). Then \( G \) acts also on the power set of \( M \), via\n\n\[ \left( {\sigma, X}\right) \mapsto {\sigma X} = \{ {\sigma x} \mid x \in X\} . \] | We have \( \left| {\sigma X}\right| = \left| X\right| \) for every \( X \subseteq M \) . | No |
In the preceding example, consider in particular the action of \( G \) on \( M = G \) by inner automorphisms. The orbit of an \( X \subseteq G \) is then \( \left\{ {{\sigma X}{\sigma }^{-1} \mid \sigma \in G}\right\} \). We use the notation \( {X}^{\sigma } \mathrel{\text{:=}} {\sigma }^{-1}{X\sigma } \). If now \( X ... | By F4, \( G : {N}_{G}H \) is the number of subgroups of \( G \) conjugate to \( H \) . | Yes |
In the situation of the preceding example let \( U \leq G \) be another subgroup besides \( H \) . Then \( U \) acts on \( G/H \) via \( \left( {\sigma ,{\tau H}}\right) \mapsto {\sigma \tau H} \) . The orbit of \( {\tau H} \) is the set\n\n(7)\n\n\[ \{ {\sigma \tau H} \mid \sigma \in U\} \]\n\nof left cosets \( {\;\op... | If we now assume that \( G \) is finite and let \( m \) be the size of the orbit (7) of \( {\tau H} \), we get \( \left| {U\tau H}\right| = m \cdot \left| H\right| \) . By virtue of F4, then, \( m = U : \left( {{\tau H}{\tau }^{-1} \cap U}\right) \) ; see (8). Thus the cardinality of the double coset \( {U\tau H} \) is... | Yes |
Lemma. Suppose \( H \leq G \) and let \( P \) be a Sylow p-subgroup of \( G \) . Then there exists \( \tau \in G \) such that \( H \cap {P}^{\tau } \) is a Sylow p-subgroup of \( H \) . | Proof. We look at the double coset decomposition of \( G \) relative to \( H \) and \( P \) . By (10) we have (14) \[ \left| G\right| = \mathop{\sum }\limits_{{\tau \in T}}\frac{\left| H\right| \left| P\right| }{\left| H \cap {P}^{\tau }\right| } \] Dividing this equation by the highest possible power of \( p \), namel... | Yes |
Theorem 1 (Sylow's Theorems). First, \( G \) contains a Sylow p-subgroup. Every p-subgroup of \( G \) is contained in some Sylow p-subgroup of \( G \). Second. Any two Sylow p-subgroups of \( G \) are conjugate. Third. Let \( {n}_{p} \) be the number of Sylow p-subgroups of \( G \). Then (a) \( {n}_{p} \) divides \( G ... | Proof of Sylow’s Theorems. (1) Let \( n = \left| G\right| \). By F2, \( G \) is isomorphic to a subgroup of \( {S}_{n} \). By associating to each \( \pi \in {S}_{n} \) the permutation matrix \( {P}_{\pi } \in \mathrm{{GL}}\left( {n,{\mathbb{F}}_{p}}\right) \) that accounts for the effect of \( \pi \) on the canonical b... | Yes |
Theorem 1. Let \( M \subseteq \mathbb{C} \) contain 0 and 1, and set\n\n\[ K \mathrel{\text{:=}} \mathbb{Q}\left( {M \cup \bar{M}}\right) \]\n\nGiven \( z \in \mathbb{C} \), there is equivalence between:\n\n(i) \( z \in \bigtriangleup M \) (that is, \( z \) is constructible with ruler and compass);\n\n(ii) \( z \) is a... | Proof. The implication (ii) \( \Rightarrow \) (i) is especially interesting, and we prove it first. By assumption the extension \( E/K \) is Galois, and its Galois group \( G = G\left( {E/K}\right) \) is a 2-group. Thus, by F8 in Chapter 10, there exists a chain\n\n(1)\n\n\[ G = {H}_{0} \supseteq {H}_{1} \supseteq \ldo... | Yes |
Theorem 2 (Gauss). A regular \( n \) -gon is constructible with ruler and compass if and only if\n\n\[ n = {2}^{e}{p}_{1}{p}_{2}\ldots {p}_{r} \]\n\nwhere \( e \geq 0 \) is arbitrary and \( {p}_{1},\ldots ,{p}_{r} \) are distinct primes of the form\n\n\[ {p}_{i} = 1 + {2}^{{2}^{{k}_{i}}} \] | The only thing in this result that has not yet been proved is:\n\nLemma. For \( m \in \mathbb{N} \), the integer \( 1 + {2}^{m} \) cannot be prime unless \( m \) is a power of 2 .\n\nProof. Suppose \( m = {m}_{1}{m}_{2} \), with \( {m}_{2} > 1 \) odd. Then\n\n\[ 1 + {2}^{m} = 1 - {\left( -{2}^{{m}_{1}}\right) }^{{m}_{2... | No |
For every prime \( p \neq 2 \), the extension \( \mathbb{Q}\left( \sqrt{{p}^{ * }}\right) \) is the unique quadratic subfield of \( \mathbb{Q}\left( {\zeta }_{p}\right) \), and thus \( {p}^{ * } \) is the unique square-free integer (apart from 1) such that \( \sqrt{{p}^{ * }} \in \mathbb{Q}\left( {\zeta }_{p}\right) \)... | Proof. As seen earlier, \( \mathbb{Q}\left( {\zeta }_{p}\right) \) has a unique subfield \( F \) such that \( F : \mathbb{Q} = 2 \), and \( F = \mathbb{Q}\left( \sqrt{d}\right) \) for a unique square-free \( d = d\left( p\right) \in \mathbb{Z} \) . Instead of finding \( d \) \ | No |
Theorem 4 (Quadratic reciprocity law). If \( p \) and \( q \) are distinct odd primes,\n\n\[ \left( \frac{q}{p}\right) = \left( \frac{{p}^{ * }}{q}\right) \]\n\nwhich, taking into account \( \left( {6}^{\prime }\right) ,\left( {16}\right) \) and \( \left( {21}\right) \), also means that\n\n\[ \left( \frac{q}{p}\right) ... | Proof. All that remains to prove is (34). To do this we work in the field of eighth roots of unity over \( \mathbb{Q} \) . Choose\n\n\[ \zeta = {e}^{{\pi i}/4} \]\nso\n\n\[ \sqrt{2} = \zeta + {\zeta }^{-1} \]\n\nand with this one gets a formula analogous to (25). Now, if \( {\varepsilon }_{p} \) is the sign such that\n... | Yes |
Theorem 1 (Translation theorem). Let \( E/K \) be a Galois extension and \( {K}^{\prime }/K \) any field extension. Assume, without loss of generality, that \( E \) and \( {K}^{\prime } \) are subfields of a field \( C \), and let \( E{K}^{\prime } = {K}^{\prime }\left( E\right) \) be their composite in \( C \) .\n\n(a... | Proof. Clearly \( E{K}^{\prime }/{K}^{\prime } \) is algebraic and separable. Because \( E/K \) is normal, \( E \) is the splitting field of some set \( M \subseteq K\left\lbrack X\right\rbrack \) of polynomials over \( K \) . Thus \( E{K}^{\prime } = {K}^{\prime }\left( E\right) \) is the splitting field of \( M \subs... | Yes |
Theorem 2 (Linear independence of field homomorphisms). Let \( E/K \) be a finite separable field extension of degree \( n \), and let \( C \) be an algebraically closed extension of \( K \) . Denote by \( {\sigma }_{1},{\sigma }_{2},\ldots ,{\sigma }_{n} \) the distinct \( K \) -homomorphisms of \( E \) in \( C \) . T... | Proof of Theorem 2. Let \( {\beta }_{1},\ldots ,{\beta }_{n} \) be a basis of \( E/K \) . Clearly,(6) is equivalent to\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{n}{c}_{i}{\sigma }_{i}\left( {\beta }_{j}\right) = 0\;\text{ for }1 \leq j \leq n. \]\n\nThe assertion of Theorem 2 is thus equivalent to the nonvanishing of the ... | Yes |
Theorem 3 (Existence of normal bases). Let \( E/K \) be a finite Galois extension with Galois group \( G \) . There exists an element \( \alpha \) in \( E \) such that the family \[ {\left( \sigma \left( \alpha \right) \right) }_{\sigma \in G} \] is a basis of \( E/K \) . Such a family is called a normal basis of \( E/... | Proof of Theorem 3. We first take an arbitrary element \( \alpha \) of \( E \) and assume there is a relation \[ \mathop{\sum }\limits_{{\sigma \in G}}{a}_{\sigma }\sigma \left( \alpha \right) = 0,\;\text{ with }{a}_{\sigma } \in K. \] For any \( \tau \in G \) we can apply \( {\tau }^{-1} \) to the sum, obtaining \[ \m... | Yes |
Theorem 1. Let \( E/K \) be a finite Galois extension with cyclic Galois group (we call \( E/K \) a cyclic extension). Let \( n \) be the order of the Galois group \( G \) of \( E/K \) . If \( K \) contains a primitive \( n \) -th root of unity, \( E \) is obtained from \( K \) by adjoining an \( n \) -th root of an el... | Proof. Let \( K \) contain a primitive \( n \) -th root of unity \( \zeta \), and let \( \sigma \) be a generator of \( G \) . We seek an element \( \alpha \in {E}^{ \times } \) such that\n\n(5)\n\n\[ \frac{\sigma \alpha }{\alpha } = \zeta \]\n\nSuch an \( \alpha \) exists if and only if \( \zeta \) satisfies \( {N}_{E... | Yes |
Theorem 3 (Artin-Schreier). Let \( K \) be a field of characteristic \( p > 0 \).\n\n(I) If \( E/K \) is a cyclic extension of degree \( p \), then \( E \) arises from \( K \) by adjoining a root of a polynomial of the form\n\n\[ {X}^{p} - X - \gamma ,\;\text{ with }\gamma \in K.\]\n\n(II) Conversely, given a polynomia... | Proof. We prove (II) first. Let \( \alpha \in E \) be a root of \( f\left( X\right) \) . For any integer \( j \geq 0 \) , regarded as an element of the prime field of \( K \), we have\n\n\[ f\left( {\alpha + j}\right) = {\left( \alpha + j\right) }^{p} - \left( {\alpha + j}\right) - \gamma = {\alpha }^{p} + {j}^{p} - \a... | Yes |
Theorem 4 (Kummer theory). Under the assumption that \( K \) has a primitive \( n \) -th root of unity, and keeping the earlier notations, the map\n\n(25)\n\n\[ A \mapsto K\left( \sqrt[n]{A}\right) = : {E}_{A} \]\n\n is a bijection between the set of subgroups \( A \) of \( {K}^{ \times } \) such that \( {K}^{\times n}... | Proof. We must show that the map (25) is one-to-one and onto; everything else then follows from F7.\n\nSurjectivity: Let \( E/K \) be abelian of exponent \( n \) . Setting\n\n\[ A \mathrel{\text{:=}} {E}^{\times n} \cap {K}^{ \times } \]\n\nwe get \( K\left( \sqrt[n]{A}\right) \subseteq E \) . Suppose that \( E \) is n... | Yes |
Theorem 7 (Classification of finitely generated abelian groups). Let \( A \) be a finitely generated abelian group. There exist integers \( {c}_{1},{c}_{2},\ldots ,{c}_{s} \geq 0 \), all distinct and different from 1, satisfying\n\n\[ {c}_{j} \mid {c}_{j + 1}\;\text{ for }1 \leq j \leq s - 1, \]\n\nand such that\n\n\[ ... | Proof. Let \( {A}_{T} \) be the torsion component of \( A \) and set \( \bar{A} = A/{A}_{T} \) (see F13). Being a homomorphic image of \( A \), the group \( \bar{A} \) is finitely generated. It is also torsionfree, so Theorem 6 shows that\n\n\[ \bar{A} \simeq {\mathbb{Z}}^{r} = \mathbb{Z}/0 \times \cdots \times \mathbb... | Yes |
Theorem 8. Let \( F \simeq {R}^{n} \) be a free module in \( n \) generators over a principal ideal domain \( R \), and let \( N \) be a submodule of \( F \) . There exists a basis \( {b}_{1},\ldots ,{b}_{n} \) of \( F \) and elements \( {c}_{1},\ldots ,{c}_{m} \) (where \( m \leq n \) ) in \( R \smallsetminus \{ 0\} \... | Proof. If we take existence as proved, we get an isomorphism\n\n\[ F/N \simeq R/{c}_{1} \times \cdots \times R/{c}_{m} \times {R}^{n - m}; \]\n\nthen uniqueness follows immediately from Theorem 7 (but note that here the early \( {c}_{i} \)’s can be units in \( R \) ).\n\nOne might think that the existence part can also... | Yes |
Theorem 1. If \( E/K \) is an extension solvable by radicals and \( {E}^{\prime } \) is the normal closure of \( E/K \), the automorphism group \( G\left( {{E}^{\prime }/K}\right) \) is solvable. | Proof. By Remark (c) following Definition \( 1,{E}^{\prime }/K \) is solvable by radicals if \( E/K \) is. Thus we may as well assume that \( E/K \) is normal: \( E = {E}^{\prime } \) . By assumption there is a radical extension \( F/K \) such that \( E \) is a subfield of \( F \) . Again by Remark (c) we can assume th... | Yes |
Theorem 2. Let \( E/K \) be a finite field extension and let \( {E}^{\prime } \) be the normal closure of \( E/K \) . If the group \( G\left( {{E}^{\prime }/K}\right) \) is solvable and its order is not divisible by the characteristic of \( K \), the extension \( E/K \) is solvable by radicals. | Proof. Clearly we may as well assume that \( E/K \) is normal \( \left( {{E}^{\prime } = E}\right) \) . Let \( {E}_{s} \) be the separable closure of \( K \) in \( E \) . The extension \( E/{E}_{s} \) is radical (see Remark (e) after Definition 1), so it suffices to show that \( {E}_{s}/K \) is solvable by radicals. Ho... | Yes |
Lemma 1. Let \( K \) be a field and \( n \) a natural number. Let \( {K}^{\prime } \) be the field of \( n \) -th roots of unity over \( K \) . If the characteristic of \( K \) is either 0 or greater than all prime factors of \( n \), the extension \( {K}^{\prime }/K \) is solvable by irreducible radicals. | Proof. We work by induction on \( n \) . The cases \( n = 1,2 \) being trivial, assume \( n > 2 \) . There is a primitive \( \varphi \left( n\right) \) -th root of unity \( \eta \) in the algebraic closure \( C \) of \( {K}^{\prime } \), because no prime factor of \( \varphi \left( n\right) \) is greater than all prime... | Yes |
Theorem 5 (Abel, Ruffini). The general polynomial of degree \( n \) over \( k \) is not solvable by radicals if \( n \geq 5 \) . | Proof. Let \( g \) be as in (20) the general polynomial of degree \( n \) over \( k \) and let \( K = k\left( {{u}_{1},\ldots ,{u}_{n}}\right) \) be its coefficient field. By Theorem 4, the Galois group of \( g \) over \( K \) is isomorphic to the symmetric group \( {S}_{n} \) . If the polynomial \( g \in K\left\lbrack... | Yes |
Theorem 6. The alternating group \( {A}_{n} \) is simple for \( n \geq 5 \) . | Recall that a group \( G \neq 1 \) is called simple if it has no normal subgroup apart from itself and 1. A proof of Theorem 6 is outlined in §15.13 and §15.16 in the Appendix. | No |
Theorem 7 (Galois). (I) Let \( f \in K\left\lbrack X\right\rbrack \) be irreducible of prime degree \( p \) ; also assume that \( f \) is separable (that is, not of the form \( c\left( {{X}^{p} - a}\right) \) if \( p = \operatorname{char}K \) ). As before we regard the Galois group \( G \) of \( f \) over \( K \) as a ... | Proof. Part (I) has been proved above. For part (II), let’s denote an element \( \sigma \in S\left( {\mathbb{F}}_{p}\right) \) of the form (55) by \( {\sigma }_{a, b} \) ; note that \( b \in {\mathbb{F}}_{p} \) and \( a \in {\mathbb{F}}_{p}^{ \times } \) are uniquely determined by \( \sigma \) . A simple calculation sh... | Yes |
Lemma 1. If \( A/R \) and \( B/A \) are finite ring extensions, \( B/R \) is also finite. | Proof. The relations \( A = R{e}_{1} + \cdots + R{e}_{m} \) and \( B = A{f}_{1} + \cdots + A{f}_{n} \) obviously imply \( B = R{e}_{1}{f}_{1} + \cdots + R{e}_{m}{f}_{n} \) . | Yes |
Lemma 2. Let \( A/R \) be a ring extension and \( \sigma : A \rightarrow B \) a ring homomorphism. If \( \alpha \in A \) is integral over \( R \), then \( \sigma \left( \alpha \right) \) is integral over \( \sigma \left( R\right) \) . | Proof. This is clear. | No |
Theorem 2. If \( {x}_{1},{x}_{2},\ldots ,{x}_{m} \) are distinct algebraic numbers, then \( {e}^{{x}_{1}},{e}^{{x}_{2}},\ldots ,{e}^{{x}_{m}} \) are linearly independent over \( \mathbb{Q} \) . | Proof. We pick a finite Galois extension \( K/\mathbb{Q} \) containing all the \( {x}_{i} \) . By supplementing the \( {x}_{1},{x}_{2},\ldots ,{x}_{m} \) with conjugates as needed, we can assume that every \( \sigma \in G = G\left( {K/\mathbb{Q}}\right) \) effects a permutation of the \( {x}_{i} \) ; thus there is a we... | Yes |
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