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Example 8.3.26 In the situation of the above example, find \( {\left\lbrack ax + b\right\rbrack }^{-1} \) assuming \( {a}^{2} + {b}^{2} \neq \) 0 . Note this includes all cases of interest thanks to the above proposition. | You can do it with partial fractions as above.\n\n\[ \frac{1}{\left( {{x}^{2} + 1}\right) \left( {{ax} + b}\right) } = \frac{b - {ax}}{\left( {{a}^{2} + {b}^{2}}\right) \left( {{x}^{2} + 1}\right) } + \frac{{a}^{2}}{\left( {{a}^{2} + {b}^{2}}\right) \left( {{ax} + b}\right) } \]\n\nand so\n\n\[ 1 = \frac{1}{{a}^{2} + {... | Yes |
Theorem 8.3.28 Let \( a \in \mathbb{A} \) . Then there exists a unique monic irreducible polynomial \( p\left( x\right) \) having coefficients in \( \mathbb{F} \) such that \( p\left( a\right) = 0 \) . This is called the minimal polynomial for a. | Proof: By definition, there exists a polynomial \( q\left( x\right) \) having coefficients in \( \mathbb{F} \) such that \( q\left( a\right) = 0 \) . If \( q\left( x\right) \) is irreducible, divide by the leading coefficient and this proves the existence. If \( q\left( x\right) \) is not irreducible, then there exist ... | Yes |
Proposition 8.3.31 Let \( \\left\\{ {{a}_{1},\\cdots ,{a}_{m}}\\right\\} \) be algebraic numbers. Then\n\n\[ \n\\dim \\mathbb{F}\\left\\lbrack {{a}_{1},\\cdots ,{a}_{m}}\\right\\rbrack \\leq \\mathop{\\prod }\\limits_{{j = 1}}^{m}\\deg \\left( {a}_{j}\\right)\n\]\n\nand for an algebraic number \( a \) ,\n\n\[ \n\\dim \... | Proof: First consider the second assertion. Let the minimal polynomial of \( a \) be\n\n\[ \np\\left( x\\right) = {x}^{n} + {a}_{n - 1}{x}^{n - 1} + \\cdots + {a}_{1}x + {a}_{0}.\n\]\n\nSince \( p\\left( a\\right) = 0 \), it follows \( \\left\\{ {1, a,{a}^{2},\\cdots ,{a}^{n}}\\right\\} \) is linearly dependent. Howeve... | Yes |
Theorem 8.3.32 The algebraic numbers \( \mathbb{A} \), those roots of polynomials in \( \mathbb{F}\left\lbrack x\right\rbrack \) which are in \( \mathbb{G} \), are a field. | Proof: Let \( a \) be an algebraic number and let \( p\left( x\right) \) be its minimal polynomial. Then \( p\left( x\right) \) is of the form\n\n\[ \n{x}^{n} + {a}_{n - 1}{x}^{n - 1} + \cdots + {a}_{1}x + {a}_{0} \n\]\n\nwhere \( {a}_{0} \neq 0 \) . Then plugging in \( a \) yields\n\n\[ \na\frac{\left( {{a}^{n - 1} + ... | No |
Theorem 8.3.34 Suppose \( {a}_{1},\cdots ,{a}_{n} \) are algebraic numbers and suppose \( {\alpha }_{1},\cdots ,{\alpha }_{n} \) are distinct algebraic numbers. Then\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}{e}^{{\alpha }_{i}} \neq 0 \]\n\nIn other words, the \( \left\{ {{e}^{{\alpha }_{1}},\cdots ,{e}^{{\alpha ... | There is a proof of this in the appendix. It is long and hard but only depends on elementary considerations other than some algebra involving symmetric polynomials. See Theorem E.3.5. | No |
Lemma 9.2.2 Let \( V \) and \( W \) be vector spaces and suppose \( \left\{ {{v}_{1},\cdots ,{v}_{n}}\right\} \) is a basis for \( V \) . Then if \( L : V \rightarrow W \) is given by \( L{v}_{k} = {w}_{k} \in W \) and\n\n\[ L\left( {\mathop{\sum }\limits_{{k = 1}}^{n}{a}_{k}{v}_{k}}\right) \equiv \mathop{\sum }\limits... | Proof: \( L \) is well defined on \( V \) because, since \( \left\{ {{v}_{1},\cdots ,{v}_{n}}\right\} \) is a basis, there is exactly one way to write a given vector of \( V \) as a linear combination. Next, observe that \( L \) is obviously linear from the definition. If \( L, M \) are equal on the basis, then if \( \... | Yes |
Theorem 9.2.3 Let \( V \) and \( W \) be finite dimensional linear spaces of dimension \( n \) and \( m \) respectively Then \( \dim \left( {\mathcal{L}\left( {V, W}\right) }\right) = {mn} \). | Proof: Let two sets of bases be\n\n\[ \left\{ {{v}_{1},\cdots ,{v}_{n}}\right\} \text{and}\left\{ {{w}_{1},\cdots ,{w}_{m}}\right\} \]\n\nfor \( V \) and \( W \) respectively. Using Lemma 9.2.2, let \( {w}_{i}{v}_{j} \in \mathcal{L}\left( {V, W}\right) \) be the linear transformation defined on the basis, \( \left\{ {{... | Yes |
The matrix of a linear transformation with respect to ordered bases \( \beta ,\gamma \) as described above is characterized by the requirement that multiplication of the components of \( v \) by \( {\left\lbrack L\right\rbrack }_{\gamma \beta } \) gives the components of \( {Lv} \) . | This happens because by definition, if \( v = \mathop{\sum }\limits_{i}{x}_{i}{v}_{i} \), then\n\n\[ \n{Lv} = \mathop{\sum }\limits_{i}{x}_{i}L{v}_{i} \equiv \mathop{\sum }\limits_{i}\mathop{\sum }\limits_{j}{\left\lbrack L\right\rbrack }_{ji}{x}_{i}{w}_{j} = \mathop{\sum }\limits_{j}\mathop{\sum }\limits_{i}{\left\lbr... | Yes |
What is the matrix of this linear transformation with respect to this basis? | Using (9.2),\n\n\[ \left( \begin{array}{llll} 0 & 1 & {2x} & 3{x}^{2} \end{array}\right) = \left( \begin{array}{lll} 1 & x & {x}^{2} \end{array}\right) {\left\lbrack D\right\rbrack }_{\gamma \beta }.\]\n\nIt follows from this that the first column of \( {\left\lbrack D\right\rbrack }_{\gamma \beta } \) is\n\n\[ \left( ... | Yes |
Example 9.3.4 Let \( \\beta \\equiv \\left\\{ {{\\mathbf{v}}_{1},\\cdots ,{\\mathbf{v}}_{n}} \\right\\} \) and \( \\gamma \\equiv \\left\\{ {{\\mathbf{w}}_{1},\\cdots ,{\\mathbf{w}}_{n}} \\right\\} \) be two bases for \( V \) . Let \( L \) be the linear transformation which maps \( {\\mathbf{v}}_{i} \) to \( {\\mathbf{... | Letting \( {\\delta }_{ij} \) be the symbol which equals 1 if \( i = j \) and 0 if \( i \\neq j \), it follows that \( L = \) \( \\mathop{\\sum }\\limits_{{i, j}}{\\delta }_{ij}{\\mathbf{w}}_{i}{\\mathbf{v}}_{j} \) and so \( {\\left\\lbrack L \\right\\rbrack }_{\\gamma \\beta } = I \) the identity matrix. For the secon... | Yes |
In the vector space of \( n \times n \) matrices, define\n\n\[ A \sim B \]\n\nif there exists an invertible matrix \( S \) such that\n\n\[ A = {S}^{-1}{BS} \]\n\nThen \( \sim \) is an equivalence relation and \( A \sim B \) if and only if whenever \( V \) is an \( n \) dimensional vector space, there exists \( L \in \m... | Proof: \( A \sim A \) because \( S = I \) works in the definition. If \( A \sim B \), then \( B \sim A \), because\n\n\[ A = {S}^{-1}{BS} \]\n\nimplies \( B = {SA}{S}^{-1} \) . If \( A \sim B \) and \( B \sim C \), then\n\n\[ A = {S}^{-1}{BS}, B = {T}^{-1}{CT} \]\n\nand so\n\n\[ A = {S}^{-1}{T}^{-1}{CTS} = {\left( TS\r... | Yes |
Proposition 9.3.10 Let \( A \) be an \( m \times n \) matrix and let \( L \) be the linear transformation which is defined by\n\n\[ L\left( {\mathop{\sum }\limits_{{k = 1}}^{n}{x}_{k}{\mathbf{e}}_{k}}\right) \equiv \mathop{\sum }\limits_{{k = 1}}^{n}\left( {A{\mathbf{e}}_{k}}\right) {x}_{k} \equiv \mathop{\sum }\limits... | Proof: Consider the following diagram. \n\nHere the coordinate maps are defined in the usual way. Thus\n\n\[ {q}_{\beta }{\left( \begin{array}{lll} {x}_{1} & \cdots & {x}_{n} \end{array}\right) }^{T} \equiv \mathop{\... | Yes |
Theorem 9.3.12 Let \( A \) be an \( n \times n \) matrix. Then \( A \) is diagonalizable if and only if \( {\mathbb{F}}^{n} \) has a basis of eigenvectors of \( A \). In this case, \( S \) of Definition 9.3.11 consists of the \( n \times n \) matrix whose columns are the eigenvectors of \( A \) and \( D = \operatorname... | Proof: Suppose first that \( {\mathbb{F}}^{n} \) has a basis of eigenvectors, \( \left\{ {{\mathbf{v}}_{1},\cdots ,{\mathbf{v}}_{n}}\right\} \) where \( A{\mathbf{v}}_{i} = {\lambda }_{i}{\mathbf{v}}_{i} \). Then let \( S \) denote the matrix \( \left( \begin{array}{lll} {\mathbf{v}}_{1} & \cdots & {\mathbf{v}}_{n} \en... | Yes |
Corollary 9.3.13 Let \( L \in \mathcal{L}\left( {V, V}\right) \) where \( V \) is an \( n \) dimensional vector space and let \( A \) be the matrix of this linear transformation with respect to a basis on \( V \) . Then it is possible to define\n\n\[ \det \left( L\right) \equiv \det \left( A\right) \] | Proof: Each choice of basis for \( V \) determines a matrix for \( L \) with respect to the basis. If \( A \) and \( B \) are two such matrices, it follows from Theorem 9.3.9 that\n\n\[ A = {S}^{-1}{BS} \]\n\nand so\n\n\[ \det \left( A\right) = \det \left( {S}^{-1}\right) \det \left( B\right) \det \left( S\right) . \]\... | Yes |
Theorem 9.3.15 Let \( A \in \mathcal{L}\left( {X, Y}\right) \) . Then \( \operatorname{rank}\left( A\right) = \operatorname{rank}\left( M\right) \) where \( M \) is the matrix of A taken with respect to a pair of bases for the vector spaces \( X \), and \( Y \) . | Proof: Recall the diagram which describes what is meant by the matrix of \( A \) . Here the two bases are as indicated.\n\n\[ \begin{array}{l} \beta = \left\{ {{v}_{1},\cdots ,{v}_{n}}\right\} \;X\;A\;Y\;\left\{ {{w}_{1},\cdots ,{w}_{m}}\right\} = \gamma \\ {q}_{\beta } \uparrow \; \circ \; \uparrow {q}_{\gamma } \\ {\... | Yes |
Theorem 9.3.16 Let \( L \in \mathcal{L}\left( {V, V}\right) \) where \( V \) is a finite dimensional vector space. Then the following are equivalent.\n\n1. \( L \) is one to one.\n\n2. \( L \) maps a basis to a basis.\n\n3. \( L \) is onto.\n\n4. \( \det \left( L\right) \neq 0 \)\n\n5. If \( {Lv} = 0 \) then \( v = 0 \... | Proof: Suppose first \( L \) is one to one and let \( \beta = {\left\{ {v}_{i}\right\} }_{i = 1}^{n} \) be a basis. Then if \( \mathop{\sum }\limits_{{i = 1}}^{n}{c}_{i}L{v}_{i} = \) 0 it follows \( L\left( {\mathop{\sum }\limits_{{i = 1}}^{n}{c}_{i}{v}_{i}}\right) = 0 \) which means that since \( L\left( 0\right) = 0 ... | Yes |
Determine the matrix for the transformation mapping \( {\mathbb{R}}^{2} \) to \( {\mathbb{R}}^{2} \) which consists of rotating every vector counter clockwise through an angle of \( \theta \) . | Let \( {\mathbf{e}}_{1} \equiv \left( \begin{array}{l} 1 \\ 0 \end{array}\right) \) and \( {\mathbf{e}}_{2} \equiv \left( \begin{array}{l} 0 \\ 1 \end{array}\right) \) . These identify the geometric vectors which point along the positive \( x \) axis and positive \( y \) axis as shown.\n\nFrom Theorem 9.3.18, you only ... | Yes |
Find the matrix of the linear transformation which is obtained by first rotating all vectors through an angle of \( \phi \) and then through an angle \( \theta \) . Thus you want the linear transformation which rotates all angles through an angle of \( \theta + \phi \) . | Let \( {T}_{\theta + \phi } \) denote the linear transformation which rotates every vector through an angle of \( \theta + \phi \) . Then to get \( {T}_{\theta + \phi } \), you could first do \( {T}_{\phi } \) and then do \( {T}_{\theta } \) where \( {T}_{\phi } \) is the linear transformation which rotates through an ... | Yes |
Find the matrix of the linear transformation which rotates vectors in \( {\mathbb{R}}^{3} \) counterclockwise about the positive \( z \) axis. | Let \( T \) be the name of this linear transformation. In this case, \( T{\mathbf{e}}_{3} = {\mathbf{e}}_{3}, T{\mathbf{e}}_{1} = \) \( {\left( \cos \theta ,\sin \theta ,0\right) }^{T} \), and \( T{\mathbf{e}}_{2} = {\left( -\sin \theta ,\cos \theta ,0\right) }^{T} \) . Therefore, the matrix of this transformation is j... | Yes |
Let the projection map be defined above and let \( \mathbf{u} = {\left( 1,2,3\right) }^{T} \) . Find the matrix of this linear transformation with respect to the usual basis. | You can find this matrix in the same way as in earlier examples. \( {\operatorname{proj}}_{\mathbf{u}}\left( {\mathbf{e}}_{i}\right) \) gives the \( {i}^{th} \) column of the desired matrix. Therefore, it is only necessary to find\n\n\[ \n{\operatorname{proj}}_{\mathbf{u}}\left( {\mathbf{e}}_{i}\right) \equiv \left( \f... | Yes |
Find the matrix of the linear transformation which reflects all vectors in \( {\mathbb{R}}^{3} \) through the \( {xz} \) plane. | As illustrated above, you just need to find \( T{\mathbf{e}}_{i} \) where \( T \) is the name of the transformation. But \( T{\mathbf{e}}_{1} = {\mathbf{e}}_{1}, T{\mathbf{e}}_{3} = {\mathbf{e}}_{3} \), and \( T{\mathbf{e}}_{2} = - {\mathbf{e}}_{2} \) so the matrix is\n\n\[ \left( \begin{matrix} 1 & 0 & 0 \\ 0 & - 1 & ... | Yes |
Find the matrix of the linear transformation which first rotates counter clockwise about the positive \( z \) axis and then reflects through the \( {xz} \) plane. | This linear transformation is just the composition of two linear transformations having matrices \[ \left( \begin{matrix} \cos \theta & - \sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{matrix}\right) ,\left( \begin{matrix} 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right) \] respectively. Th... | Yes |
Lemma 9.4.2 When \( \lambda \) is an eigenvalue of \( A \) which is also in \( \mathbb{F} \), the field of scalars, then there exists \( v \neq 0 \) such that \( {Av} = {\lambda v} \) . | Proof: This follows from Theorem 9.3.16. Since \( \lambda \in \mathbb{F} \), \[ {\lambda I} - A \in \mathcal{L}\left( {V, V}\right) \] and since it has zero determinant, it is not one to one. - | Yes |
Lemma 9.4.3 Let \( A \in \mathcal{L}\left( {V, V}\right) \) where \( V \) is a finite dimensional vector space of dimension \( n \) with arbitrary field of scalars. Then there exists a unique polynomial of the form\n\n\[ p\left( \lambda \right) = {\lambda }^{m} + {c}_{m - 1}{\lambda }^{m - 1} + \cdots + {c}_{1}\lambda ... | Proof: Consider the linear transformations, \( I, A,{A}^{2},\cdots ,{A}^{{n}^{2}} \). There are \( {n}^{2} + 1 \) of these transformations and so by Theorem 9.2.3 the set is linearly dependent. Thus there exist constants, \( {c}_{i} \in \mathbb{F} \) such that\n\n\[ {c}_{0}I + \mathop{\sum }\limits_{{k = 1}}^{{n}^{2}}{... | Yes |
Theorem 9.4.4 Let \( V \) be a nonzero finite dimensional vector space of dimension \( n \) with the field of scalars equal to \( \mathbb{F} \). Suppose \( A \in \mathcal{L}\left( {V, V}\right) \) and for \( p\left( \lambda \right) \) the minimal polynomial defined above, let \( \mu \in \mathbb{F} \) be a zero of this ... | Proof: Suppose first \( \mu \) is a zero of \( p\left( \lambda \right) \). Since \( p\left( \mu \right) = 0 \), it follows\n\n\[ \np\left( \lambda \right) = \left( {\lambda - \mu }\right) k\left( \lambda \right)\n\]\n\nwhere \( k\left( \lambda \right) \) is a polynomial having coefficients in \( \mathbb{F} \). Since \(... | Yes |
Lemma 10.1.2 Whenever \( L \in \mathcal{L}\left( {V, W}\right) ,\ker \left( L\right) \) is a subspace. | Proof: If \( a, b \) are scalars and \( v, w \) are in \( \ker \left( L\right) \), then\n\n\[ L\left( {{av} + {bw}}\right) = {aL}\left( v\right) + {bL}\left( w\right) = 0 + 0 = 0\blacksquare \] | No |
Theorem 10.1.3 Let \( A \in \mathcal{L}\left( {V, W}\right) \) and \( B \in \mathcal{L}\left( {W, U}\right) \) where \( V, W, U \) are all vector spaces over a field \( \mathbb{F} \) . Suppose also that \( \ker \left( A\right) \) and \( A\left( {\ker \left( {BA}\right) }\right) \) are finite dimensional subspaces. Then... | Proof: If \( \mathbf{x} \in \ker \left( {BA}\right) \), then \( A\mathbf{x} \in \ker \left( B\right) \) and so \( A\left( {\ker \left( {BA}\right) }\right) \subseteq \ker \left( B\right) \) . The following picture may help.\n\n and if \( {\beta }_{i} = \left\{ {{v}_{1}^{i},\cdots ,{v}_{{m}_{i}}^{i}}\right\} \) is a basis for \( {V}_{i} \), then a basis for \( V \) is \( \left\{ {{\beta }_{1},\cdots ,{\beta }_{r}}\right\} \) . | Proof: Suppose \( \mathop{\sum }\limits_{{i = 1}}^{r}\mathop{\sum }\limits_{{j = 1}}^{{m}_{i}}{c}_{ij}{v}_{j}^{i} = 0 \) . then since it is a direct sum, it follows for each \( i \) ,\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{{m}_{i}}{c}_{ij}{v}_{j}^{i} = 0 \]\n\nand now since \( \left\{ {{v}_{1}^{i},\cdots ,{v}_{{m}_{i}}... | Yes |
Lemma 10.1.6 Let \( {L}_{i} \) be in \( \mathcal{L}\left( {V, V}\right) \) and suppose for \( i \neq j,{L}_{i}{L}_{j} = {L}_{j}{L}_{i} \) and also \( {L}_{i} \) is one to one on \( \ker \left( {L}_{j}\right) \) whenever \( i \neq j \) . Then \[ \ker \left( {\mathop{\prod }\limits_{{i = 1}}^{p}{L}_{i}}\right) = \ker \le... | Proof: Note that since the operators commute, \( {L}_{j} : \ker \left( {L}_{i}\right) \rightarrow \ker \left( {L}_{i}\right) \) . Here is why. If \( {L}_{i}y = 0 \) so that \( y \in \ker \left( {L}_{i}\right) \), then \[ {L}_{i}{L}_{j}y = {L}_{j}{L}_{i}y = {L}_{j}0 = 0 \] and so \( {L}_{j} : \ker \left( {L}_{i}\right) ... | Yes |
Lemma 10.2.2 Let \( L \in \mathcal{L}\left( {V, V}\right) \) where \( V \) is an \( n \) dimensional vector space. Then if \( L \) is one to one, it follows that \( L \) is also onto. In fact, if \( \left\{ {{v}_{1},\cdots ,{v}_{n}}\right\} \) is a basis, then so is \( \left\{ {L{v}_{1},\cdots, L{v}_{n}}\right\} \) . | Proof: Let \( \left\{ {{v}_{1},\cdots ,{v}_{n}}\right\} \) be a basis for \( V \) . Then I claim that \( \left\{ {L{v}_{1},\cdots, L{v}_{n}}\right\} \) is also a basis for \( V \) . First of all, I show \( \left\{ {L{v}_{1},\cdots, L{v}_{n}}\right\} \) is linearly independent. Suppose\n\n\[ \mathop{\sum }\limits_{{k = ... | Yes |
Theorem 10.2.4 In the context of Definition 7.2.3, \[ V = {V}_{1} \oplus \cdots \oplus {V}_{q} \] and each \( {V}_{k} \) is A invariant, meaning \( A\left( {V}_{k}\right) \subseteq {V}_{k}.{\phi }_{l}\left( A\right) \) is one to one on each \( {V}_{k} \) for \( k \neq l \) . If \( {\beta }_{i} = \left\{ {{v}_{1}^{i},\c... | Proof: It is clear \( {V}_{k} \) is a subspace which is \( A \) invariant because \( A \) commutes with \( {\phi }_{k}{\left( A\right) }^{{m}_{k}} \) . It is clear the operators \( {\phi }_{k}{\left( A\right) }^{{r}_{k}} \) commute. Thus if \( v \in {V}_{k} \) , \[ {\phi }_{k}{\left( A\right) }^{{r}_{k}}{\phi }_{l}{\le... | Yes |
Corollary 10.2.5 Let the minimal polynomial of \( A \) be \( p\left( \lambda \right) = \mathop{\prod }\limits_{{k = 1}}^{q}{\phi }_{k}{\left( \lambda \right) }^{{m}_{k}} \) where each \( {\phi }_{k} \) is irreducible. Let \( {V}_{k} = \ker \left( {\phi {\left( A\right) }^{{m}_{k}}}\right) \) . Then\n\n\[ \n{V}_{1} \opl... | Proof: Recall the direct sum, \( {V}_{1} \oplus \cdots \oplus {V}_{q} = V \) where \( {V}_{k} = \ker \left( {{\phi }_{k}{\left( A\right) }^{{m}_{k}}}\right) \) for \( p\left( \lambda \right) = \) \( \mathop{\prod }\limits_{{k = 1}}^{q}{\phi }_{k}{\left( \lambda \right) }^{{m}_{k}} \) the minimal polynomial for \( A \) ... | Yes |
Theorem 10.2.6 Suppose \( V \) is a vector space with field of scalars \( \mathbb{F} \) and \( A \in \mathcal{L}\left( {V, V}\right) \) . Suppose also\n\n\[ V = {V}_{1} \oplus \cdots \oplus {V}_{q} \]\n\nwhere each \( {V}_{k} \) is \( A \) invariant. \( \left( {A{V}_{k} \subseteq {V}_{k}}\right) \) Also let \( {\beta }... | Proof: Recall the matrix \( M \) of a linear transformation \( A \) is defined such that the following diagram commutes.\n\n\[ \left\{ {{v}_{1},\cdots ,{v}_{n}}\right\} \;\begin{matrix} A \\ V \\ q \uparrow \\ {\mathbb{F}}^{n} \end{matrix}\;\begin{matrix} V \\ \rightarrow \\ \circ \\ \uparrow \\ M \end{matrix}\;\begin{... | Yes |
Lemma 10.3.2 Let \( W \) be an \( A \) invariant \( \left( {{AW} \subseteq W}\right) \) subspace of \( \ker \left( {\phi {\left( A\right) }^{m}}\right) \) for \( m \) a positive integer where \( \phi \left( \lambda \right) \) is an irreducible monic polynomial of degree \( d \) . Then if \( \eta \left( \lambda \right) ... | Proof: Consider the first claim. If \( \eta \left( A\right) x = 0 \), then writing\n\n\[ \phi {\left( \lambda \right) }^{m} = \eta \left( \lambda \right) g\left( \lambda \right) + r\left( \lambda \right) \]\n\nwhere either \( r\left( \lambda \right) = 0 \) or the degree of \( r\left( \lambda \right) \) is less than tha... | Yes |
Lemma 10.3.3 Let \( V \) be a vector space and let \( B \in \mathcal{L}\left( {V, V}\right) \) . Then\n\n\[ V = B\left( V\right) \oplus \ker \left( B\right) \] | Proof: Let \( \left\{ {B{v}_{1},\cdots, B{v}_{r}}\right\} \) be a basis for \( B\left( V\right) \) . Now let \( \left\{ {{w}_{1},\cdots ,{w}_{s}}\right\} \) be a basis for \( \ker \left( B\right) \) . Then if \( v \in V \), there exist unique scalars \( {c}_{i} \) such that\n\n\[ {Bv} = \mathop{\sum }\limits_{{i = 1}}^... | Yes |
Theorem 10.3.4 Let \( V = \ker \left( {\phi {\left( A\right) }^{m}}\right) \) for \( m \) a positive integer and \( A \in \mathcal{L}\left( {Z, Z}\right) \) where \( Z \) is some vector space containing \( V \), and \( \phi \left( \lambda \right) \) is an irreducible monic polynomial over the field of scalars. Then the... | Proof: First suppose \( m = 1 \) . Then in Lemma 10.3.2 you can let \( W = \{ 0\} \) and \( U = \) \( V = \ker \left( {\phi \left( A\right) }\right) \) . Then by this lemma, there exist \( {v}_{1},{v}_{2},\cdots ,{v}_{s} \) such that \( \left\{ {{\beta }_{{v}_{1}},\cdots ,{\beta }_{{v}_{s}}}\right\} \) is a basis for \... | Yes |
Lemma 10.4.2 Suppose \( {N}^{k}x \neq 0 \) . Then \( \left\{ {x,{Nx},\cdots ,{N}^{k}x}\right\} \) is linearly independent. Also, the minimal polynomial of \( N \) is \( {\lambda }^{m} \) where \( m \) is the first such that \( {N}^{m} = 0 \) . | Proof: Suppose \( \mathop{\sum }\limits_{{i = 0}}^{k}{c}_{i}{N}^{i}x = 0 \) . There exists \( l \) such that \( k \leq l < m \) and \( {N}^{l + 1}x = 0 \) but \( {N}^{l}x \neq 0 \) . Then multiply both sides by \( {N}^{l} \) to conclude that \( {c}_{0} = 0 \) . Next multiply both sides by \( {N}^{l - 1} \) to conclude ... | No |
Corollary 10.4.5 Let \( J,{J}^{\prime } \) both be matrices of the nilpotent linear transformation \( N \in \) \( \mathcal{L}\left( {W, W}\right) \) which are of the form described in Proposition 10.4.4. Then \( J = {J}^{\prime } \) . In fact, if the rank of \( {J}^{k} \) equals the rank of \( {J}^{\prime k} \) for all... | Proof: Since \( J \) and \( {J}^{\prime } \) are similar, it follows that for each \( k \) an integer, \( {J}^{k} \) and \( {J}^{\prime k} \) are similar. Hence, for each \( k \), these matrices have the same rank. Now suppose \( J \neq {J}^{\prime } \) . Note first that\n\n\[ \n{J}_{r}{\left( 0\right) }^{r} = 0,{J}_{r... | Yes |
Proposition 10.5.1 Let the minimal polynomial of \( A \in \mathcal{L}\left( {V, V}\right) \) be given by\n\n\[ p\left( \lambda \right) = \mathop{\prod }\limits_{{k = 1}}^{r}{\left( \lambda - {\lambda }_{k}\right) }^{{m}_{k}} \]\n\nThen the eigenvalues of \( A \) are \( \left\{ {{\lambda }_{1},\cdots ,{\lambda }_{r}}\ri... | It follows from Corollary 10.2.4 that\n\n\[ V = \ker {\left( A - {\lambda }_{1}I\right) }^{{m}_{1}} \oplus \cdots \oplus \ker {\left( A - {\lambda }_{r}I\right) }^{{m}_{r}} \]\n\n\[ \equiv \;{V}_{1} \oplus \cdots \oplus {V}_{r} \]\n\nwhere \( I \) denotes the identity linear transformation. Without loss of generality, ... | Yes |
Lemma 10.5.3 Suppose \( J \) is of the form \( {J}_{s} \) described above in [10.8] where the constant \( \alpha \), on the main diagonal is less than one in absolute value. Then\n\n\[ \mathop{\lim }\limits_{{k \rightarrow \infty }}{\left( {J}^{k}\right) }_{ij} = 0 \] | Proof: From (10.9), it follows that for large \( k \), and \( j \leq {m}_{s} \), \n\n\[ \left( \begin{array}{l} k \\ j \end{array}\right) \leq \frac{k\left( {k - 1}\right) \cdots \left( {k - {m}_{s} + 1}\right) }{{m}_{s}!}. \]\n\nTherefore, letting \( C \) be the largest value of \( \left| {\left( {N}^{j}\right) }_{pq}... | Yes |
Proposition 10.7.2 Let \( q\left( \lambda \right) \) be a polynomial and let \( C\left( {q\left( \lambda \right) }\right) \) be its companion matrix. Then \( q\left( {C\left( {q\left( \lambda \right) }\right) }\right) = 0 \) . | Proof: Write \( C \) instead of \( C\left( {q\left( \lambda \right) }\right) \) for short. Note that\n\n\[ C{\mathbf{e}}_{1} = {\mathbf{e}}_{2}, C{\mathbf{e}}_{2} = {\mathbf{e}}_{3},\cdots, C{\mathbf{e}}_{n - 1} = {\mathbf{e}}_{n} \]\n\nThus\n\n\[ {\mathbf{e}}_{k} = {C}^{k - 1}{\mathbf{e}}_{1}, k = 1,\cdots, n \]\n\n\(... | Yes |
Theorem 10.7.3 Let \( A \in \mathcal{L}\left( {V, V}\right) \) where \( V \) is a vector space with field of scalars \( \mathbb{F} \) and minimal polynomial \[ \mathop{\prod }\limits_{{i = 1}}^{q}{\phi }_{i}{\left( \lambda \right) }^{{m}_{i}} \] where each \( {\phi }_{i}\left( \lambda \right) \) is irreducible. Letting... | Proof: By Theorem 10.2.6 the matrix of \( A \) with respect to \( \left\{ {{B}_{1},\cdots ,{B}_{q}}\right\} \) is of the form given in (10.13). Now by Theorem 10.3.4 the basis \( {B}_{k} \) may be chosen in the form \( \left\{ {{\beta }_{{v}_{1}},\cdots ,{\beta }_{{v}_{s}}}\right\} \) where each \( {\beta }_{{v}_{k}} \... | Yes |
Theorem 10.8.4 Let \( V \) be a vector space having field of scalars \( \mathbb{F} \) and let \( A \in \mathcal{L}\left( {V, V}\right) \) . Then the rational canonical form of \( A \) is unique up to order of the blocks. | Proof: Let the minimal polynomial of \( A \) be \( \mathop{\prod }\limits_{{k = 1}}^{q}{\phi }_{k}{\left( \lambda \right) }^{{m}_{k}} \) . Then recall from Corollary 10.2.4\n\n\[ V = {V}_{1} \oplus \cdots \oplus {V}_{q} \]\n\nwhere \( {V}_{k} = \ker \left( {{\phi }_{k}{\left( A\right) }^{{m}_{k}}}\right) \) . Also reca... | No |
Find a similarity transformation which will produce the rational canonical form for \( A \) . | The characteristic polynomial is \( {\lambda }^{3} - {24}{\lambda }^{2} + {180\lambda } - {432} \) . This factors as\n\n\[ {\left( \lambda - 6\right) }^{2}\left( {\lambda - {12}}\right) \]\n\nIt turns out this is also the minimal polynomial. You can see this by plugging in \( A \) where you see \( \lambda \) and observ... | No |
Lemma 11.1.2 The property of being a stochastic matrix is preserved by taking products. | Proof: Suppose the sum over a row equals 1 for \( A \) and \( B \) . Then letting the entries be denoted by \( \left( {a}_{ij}\right) \) and \( \left( {b}_{ij}\right) \) respectively,\n\n\[ \mathop{\sum }\limits_{i}\mathop{\sum }\limits_{k}{a}_{ik}{b}_{kj} = \mathop{\sum }\limits_{k}\left( {\mathop{\sum }\limits_{i}{a}... | Yes |
Theorem 11.1.3 Let \( A \) be a real \( p \times p \) matrix having the properties\n\n1. \( {a}_{ij} \geq 0 \)\n\n2. Either \( \mathop{\sum }\limits_{{i = 1}}^{p}{a}_{ij} = 1 \) or \( \mathop{\sum }\limits_{{j = 1}}^{p}{a}_{ij} = 1 \) .\n\n3. The distinct eigenvalues of \( A \) are \( \left\{ {1,{\lambda }_{2},\ldots ,... | Proof. By the existence of the Jordan form for \( A \), it follows that there exists an invertible matrix \( P \) such that\n\n\[ \n{P}^{-1}{AP} = \left( \begin{array}{llll} I + N & & & \\ & {J}_{{r}_{2}}\left( {\lambda }_{2}\right) & & \\ & & \ddots & \\ & & & {J}_{{r}_{m}}\left( {\lambda }_{m}\right) \end{array}\righ... | Yes |
Lemma 11.1.4 Suppose \( A = \left( {a}_{ij}\right) \) is a stochastic matrix. Then \( \lambda = 1 \) is an eigenvalue. If \( {a}_{ij} > 0 \) for all \( i, j \), then if \( \mu \) is an eigenvalue of \( A \), either \( \left| \mu \right| < 1 \) or \( \mu = 1 \) . In addition to this, if \( A\mathbf{v} = \mathbf{v} \) fo... | Proof: Suppose the matrix satisfies\n\n\[ \mathop{\sum }\limits_{j}{a}_{ij} = 1 \]\n\nThen if \( \mathbf{v} = {\left( \begin{array}{lll} 1 & \cdots & 1 \end{array}\right) }^{T} \), it is obvious that \( A\mathbf{v} = \mathbf{v} \) . Therefore, this matrix has \( \lambda = 1 \) as an eigenvalue. Suppose then that \( \mu... | Yes |
Lemma 11.1.5 Let \( A \) be any Markov matrix and let \( \mathbf{v} \) be a vector having all its components non negative with \( \mathop{\sum }\limits_{i}{v}_{i} = c \) . Then if \( \mathbf{w} = A\mathbf{v} \), it follows that \( {w}_{i} \geq 0 \) for all \( i \) and \( \mathop{\sum }\limits_{i}{w}_{i} = c \) . | Proof: From the definition of \( \mathbf{w} \) ,\n\n\[ \n{w}_{i} \equiv \mathop{\sum }\limits_{j}{a}_{ij}{v}_{j} \geq 0 \n\]\n\nAlso\n\n\[ \n\mathop{\sum }\limits_{i}{w}_{i} = \mathop{\sum }\limits_{i}\mathop{\sum }\limits_{j}{a}_{ij}{v}_{j} = \mathop{\sum }\limits_{j}\mathop{\sum }\limits_{i}{a}_{ij}{v}_{j} = \mathop{... | Yes |
Theorem 11.1.6 Suppose \( A \) is a Markov matrix (The sum over a column equals 1) in which \( {a}_{ij} > 0 \) for all \( i, j \) and suppose \( \mathbf{w} \) is a vector. Then for each \( i \) ,\n\n\[ \n\mathop{\lim }\limits_{{k \rightarrow \infty }}{\left( {A}^{k}\mathbf{w}\right) }_{i} = {v}_{i} \n\]\n\nwhere \( A\m... | Proof: By Lemma 11.1.4, since each \( {a}_{ij} > 0 \), the eigenvalues are either 1 or have absolute value less than 1. Therefore, the claimed limit exists by Theorem III.1.3. The assertion that the components are nonnegative and sum to \( c \) follows from Lemma II.L5. That \( A\mathbf{v} = \mathbf{v} \) follows from\... | Yes |
Corollary 11.1.7 Suppose \( A \) is a regular Markov matrix, on for which the entries of \( {A}^{k} \) are all positive for some \( k \), and suppose \( \mathbf{w} \) is a vector. Then for each \( i \) ,\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{\left( {A}^{n}\mathbf{w}\right) }_{i} = {v}_{i} \]\n\nwhere \(... | Proof: Let the entries of \( {A}^{k} \) be all positive. Now suppose that \( {a}_{ij} \geq 0 \) for all \( i, j \) and \( A = \left( {a}_{ij}\right) \) is a transition matrix. Then if \( B = \left( {b}_{ij}\right) \) is a transition matrix with \( {b}_{ij} > 0 \) for all \( {ij} \), it follows that \( {BA} \) is a tran... | Yes |
Proposition 11.3.3 Let \( {p}_{ij}^{n} \) denote the probability that \( {X}_{n} \) is in state \( j \) given that \( {X}_{0} \) was in state \( i \) . Then \( {p}_{ij}^{n} \) is the \( i{j}^{\text{th }} \) entry of the matrix \( {P}^{n} \) where \( P = \left( {p}_{ij}\right) \) . | Proof: This is clearly true if \( n = 1 \) and follows from the definition of the \( {p}_{ij} \) . Suppose true for \( n \) . Then the probability that \( {X}_{n + 1} \) is at \( j \) given that \( {X}_{0} \) was at \( i \) equals \( \mathop{\sum }\limits_{k}{p}_{ik}^{n}{p}_{kj} \) because \( {X}_{n} \) must have some ... | Yes |
Theorem 11.3.4 The eigenvalues of \n\n\\[ \n\\left( \\begin{matrix} 0 & p & 0 & \\cdots & 0 \\\\ q & 0 & p & \\cdots & 0 \\\\ 0 & q & 0 & \\ddots & \\vdots \\\\ \\vdots & 0 & \\ddots & \\ddots & p \\\\ 0 & \\vdots & 0 & q & 0 \\end{matrix}\\right) \n\\]\n\nhave absolute value less than 1. Here \\( p + q = 1 \\) and bot... | Proof: By Gerschgorin’s theorem, if \\( \\lambda \\) is an eigenvalue, then \\( \\left| \\lambda \\right| \\leq 1 \\) . Now suppose \\( \\mathbf{v} \\) is an eigenvector for \\( \\lambda \\) . Then\n\n\\[ \nA\\mathbf{v} = \\left( \\begin{matrix} p{v}_{2} \\\\ q{v}_{1} + p{v}_{3} \\\\ \\vdots \\\\ q{v}_{n - 2} + p{v}_{n... | Yes |
Corollary 11.3.5 Let \( p, q \) be positive numbers and let \( p + q = 1 \) . The eigenvalues of\n\n\[ \left( \begin{matrix} a & p & 0 & \cdots & 0 \\ q & a & p & \cdots & 0 \\ 0 & q & a & \ddots & \vdots \\ \vdots & 0 & \ddots & \ddots & p \\ 0 & \vdots & 0 & q & a \end{matrix}\right) \]\n\nare all strictly closer tha... | Proof: Let \( A \) be the above matrix and suppose \( A\mathbf{x} = \lambda \mathbf{x} \) . Then letting \( {A}^{\prime } \) denote\n\n\[ \left( \begin{matrix} 0 & p & 0 & \cdots & 0 \\ q & 0 & p & \cdots & 0 \\ 0 & q & 0 & \ddots & \vdots \\ \vdots & 0 & \ddots & \ddots & p \\ 0 & \vdots & 0 & q & 0 \end{matrix}\right... | Yes |
Example 12.1.4 Let \( V = {\mathbb{C}}^{n} \) with the inner product given by\n\n\[ \left( {\mathbf{x},\mathbf{y}}\right) \equiv \mathop{\sum }\limits_{{k = 1}}^{n}{x}_{k}{\bar{y}}_{k} \] | This is an example of a complex inner product space already discussed. | No |
Example 12.1.5 Let \( V = {\mathbb{R}}^{n} \) , \[ \left( {\mathbf{x},\mathbf{y}}\right) = \mathbf{x} \cdot \mathbf{y} \equiv \mathop{\sum }\limits_{{j = 1}}^{n}{x}_{j}{y}_{j} \] | This is an example of a real inner product space. | No |
Example 12.1.6 Let \( V \) be any finite dimensional vector space and let \( \left\{ {{v}_{1},\cdots ,{v}_{n}}\right\} \) be a basis. Decree that \[ \left( {{v}_{i},{v}_{j}}\right) \equiv {\delta }_{ij} \equiv \left\{ \begin{array}{ll} 1 & \text{ if }i = j \\ 0 & \text{ if }i \neq j \end{array}\right. \] and define the... | The above is well defined because \( \left\{ {{v}_{1},\cdots ,{v}_{n}}\right\} \) is a basis. Thus the components \( {x}_{i} \) associated with any given \( x \in V \) are uniquely determined. | Yes |
Theorem 12.1.7 (Cauchy Schwarz) In any inner product space\n\n\\[ \n\\left| \\left( {x, y}\\right) \\right| \\leq \\left| x\\right| \\left| y\\right| \n\\]\n\nwhere \\( \\left| x\\right| \\equiv {\\left( x, x\\right) }^{1/2} \\) . | Proof: Let \\( \\omega \\in \\mathbb{C},\\left| \\omega \\right| = 1 \\), and \\( \\bar{\\omega }\\left( {x, y}\\right) = \\left| \\left( {x, y}\\right) \\right| = \\operatorname{Re}\\left( {x,{y\\omega }}\\right) \\) . Let\n\n\\[ \nF\\left( t\\right) = \\left( {x + {ty\\omega }, x + {t\\omega y}}\\right) .\n\\]\n\nThe... | Yes |
Proposition 12.1.8 For an inner product space, \( \left| x\right| \equiv {\left( x, x\right) }^{1/2} \) does specify a norm. | Proof: All the axioms are obvious except the triangle inequality. To verify this,\n\n\[ \n{\left| x + y\right| }^{2} \equiv \left( {x + y, x + y}\right) \equiv {\left| x\right| }^{2} + {\left| y\right| }^{2} + 2\operatorname{Re}\left( {x, y}\right) \n\]\n\n\[ \n\leq {\left| x\right| }^{2} + {\left| y\right| }^{2} + 2\l... | Yes |
Lemma 12.2.1 Let \( X \) be a finite dimensional inner product space of dimension \( n \) whose basis is \( \left\{ {{x}_{1},\cdots ,{x}_{n}}\right\} \) . Then there exists an orthonormal basis for \( X,\left\{ {{u}_{1},\cdots ,{u}_{n}}\right\} \) which has the property that for each \( k \leq n \), span \( \left( {{x}... | Proof: Let \( \left\{ {{x}_{1},\cdots ,{x}_{n}}\right\} \) be a basis for \( X \) . Let \( {u}_{1} \equiv {x}_{1}/\left| {x}_{1}\right| \) . Thus for \( k = 1 \), span \( \left( {u}_{1}\right) = \) \( \operatorname{span}\left( {x}_{1}\right) \) and \( \left\{ {u}_{1}\right\} \) is an orthonormal set. Now suppose for so... | Yes |
Lemma 12.2.3 Suppose \( {\left\{ {u}_{j}\right\} }_{j = 1}^{n} \) is an orthonormal basis for an inner product space \( X \) . Then for all \( x \in X \) , \[ x = \mathop{\sum }\limits_{{j = 1}}^{n}\left( {x,{u}_{j}}\right) {u}_{j} \] | Proof: By assumption that this is an orthonormal basis, \[ \mathop{\sum }\limits_{{j = 1}}^{n}\left( {x,{u}_{j}}\right) \overset{{\delta }_{jl}}{\overbrace{\left( {u}_{j},{u}_{l}\right) }} = \left( {x,{u}_{l}}\right) . \] Letting \( y = \mathop{\sum }\limits_{{k = 1}}^{n}\left( {x,{u}_{k}}\right) {u}_{k} \), it follows... | Yes |
Theorem 12.3.1 Let \( f \in \mathcal{L}\left( {X,\mathbb{F}}\right) \) where \( X \) is an inner product space of dimension \( n \) . Then there exists a unique \( z \in X \) such that for all \( x \in X \) ,\n\n\[ f\left( x\right) = \left( {x, z}\right) . \] | Proof: First I will verify uniqueness. Suppose \( {z}_{j} \) works for \( j = 1,2 \) . Then for all \( x \in X \) ,\n\n\[ 0 = f\left( x\right) - f\left( x\right) = \left( {x,{z}_{1} - {z}_{2}}\right) \]\n\nand so \( {z}_{1} = {z}_{2} \) .\n\nIt remains to verify existence. By Lemma [12.2.1, there exists an orthonormal ... | Yes |
Corollary 12.3.2 Let \( A \in \mathcal{L}\left( {X, Y}\right) \) where \( X \) and \( Y \) are two inner product spaces of finite dimension. Then there exists a unique \( {A}^{ * } \in \mathcal{L}\left( {Y, X}\right) \) such that\n\n\[{\left( Ax, y\right) }_{Y} = {\left( x,{A}^{ * }y\right) }_{X}\]\n\nfor all \( x \in ... | Proof: Let \( {f}_{y} \in \mathcal{L}\left( {X,\mathbb{F}}\right) \) be defined as\n\n\[{f}_{y}\left( x\right) \equiv {\left( Ax, y\right) }_{Y}.\]\n\nThen by the Riesz representation theorem, there exists a unique element of \( X,{A}^{ * }\left( y\right) \) such that\n\n\[{\left( Ax, y\right) }_{Y} = {\left( x,{A}^{ *... | Yes |
Theorem 12.3.4 Let \( M \) be an \( m \times n \) matrix. Then \( {M}^{ * } = {\left( \bar{M}\right) }^{T} \) in words, the transpose of the conjugate of \( M \) is equal to the adjoint. | Proof: Using the definition of the inner product in \( {\mathbb{C}}^{n} \) ,\n\n\[ \left( {M\mathbf{x},\mathbf{y}}\right) = \left( {\mathbf{x},{M}^{ * }\mathbf{y}}\right) \equiv \mathop{\sum }\limits_{i}{x}_{i}\overline{\mathop{\sum }\limits_{j}{\left( {M}^{ * }\right) }_{ij}{y}_{j}} = \mathop{\sum }\limits_{{i, j}}\ov... | Yes |
Theorem 12.3.5 Suppose \( V \) is a subspace of \( {\mathbb{F}}^{n} \) having dimension \( p \leq n \) . Then there exists \( {aQ} \in \mathcal{L}\left( {{\mathbb{F}}^{n},{\mathbb{F}}^{n}}\right) \) such that\n\n\[ \n{QV} \subseteq \operatorname{span}\left( {{\mathbf{e}}_{1},\cdots ,{\mathbf{e}}_{p}}\right)\n\]\n\nand ... | Proof: By Lemma 12.2.1 there exists an orthonormal basis for \( V,{\left\{ {\mathbf{v}}_{i}\right\} }_{i = 1}^{p} \) . By using the Gram Schmidt process this may be extended to an orthonormal basis of the whole space, \( {\mathbb{F}}^{n}, \n\n\[ \n\left\{ {{\mathbf{v}}_{1},\cdots ,{\mathbf{v}}_{p},{\mathbf{v}}_{p + 1},... | Yes |
Lemma 12.4.2 Let \( X, Y, Z \) be inner product spaces. Then for \( \alpha \) a scalar,\n\n\[{\left( \alpha \left( y \otimes x\right) \right) }^{ * } = \bar{\alpha }x \otimes y\] | Proof: Let \( u \in X \) and \( v \in Y \) . Then\n\n\[\\left( {\\alpha \\left( {y \\otimes x}\\right) u, v}\\right) = \\left( {\\alpha \\left( {u, x}\\right) y, v}\\right) = \\alpha \\left( {u, x}\\right) \\left( {y, v}\\right)\]\n\nand\n\n\[\\left( {u,\\bar{\\alpha }x \\otimes y\\left( v\\right) }\\right) = \\left( {... | Yes |
Theorem 12.4.4 Let \( X \) and \( Y \) be finite dimensional inner product spaces. Then \( \mathcal{L}\left( {X, Y}\right) \) is a vector space with the above definition of what it means to multiply by a scalar and add. Let \( \left\{ {{v}_{1},\cdots ,{v}_{n}}\right\} \) be an orthonormal basis for \( X \) and \( \left... | Proof: It is obvious that \( \mathcal{L}\left( {X, Y}\right) \) is a vector space. It remains to verify the given set is a basis. Consider the following: \[ \left( {\left( {A - \mathop{\sum }\limits_{{k, l}}\left( {A{v}_{k},{w}_{l}}\right) {w}_{l} \otimes {v}_{k}}\right) {v}_{p},{w}_{r}}\right) = \left( {A{v}_{p},{w}_{... | Yes |
Theorem 12.4.5 Let \( A = \mathop{\sum }\limits_{{i, j}}{c}_{ij}{w}_{i} \otimes {v}_{j} \in \mathcal{L}\left( {X, Y}\right) \) where as before, the vectors, \( \left\{ {w}_{i}\right\} \) are an orthonormal basis for \( Y \) and the vectors, \( \left\{ {v}_{j}\right\} \) are an orthonormal basis for \( X \) . Then if th... | Proof: Recall\n\n\[ \nA{v}_{i} \equiv \mathop{\sum }\limits_{k}{M}_{ki}{w}_{k} \n\]\n\nAlso\n\n\[ \nA{v}_{i} = \mathop{\sum }\limits_{{k, j}}{c}_{kj}{w}_{k} \otimes {v}_{j}\left( {v}_{i}\right) = \mathop{\sum }\limits_{{k, j}}{c}_{kj}{w}_{k}\left( {{v}_{i},{v}_{j}}\right) \n\]\n\n\[ \n= \mathop{\sum }\limits_{{k, j}}{c... | Yes |
Lemma 12.5.1 Let \( V \) and \( W \) be finite dimensional inner product spaces and let \( A : V \rightarrow W \) be linear. For each \( y \in W \) there exists \( x \in V \) such that\n\n\[ \left| {{Ax} - y}\right| \leq \left| {A{x}_{1} - y}\right| \]\n\nfor all \( {x}_{1} \in V \) . Also, \( x \in V \) is a solution ... | Proof: By Theorem 12.2.4 on Page 291 there exists a point, \( A{x}_{0} \), in the finite dimensional subspace, \( A\left( V\right) \), of \( W \) such that for all \( x \in V,{\left| Ax - y\right| }^{2} \geq {\left| A{x}_{0} - y\right| }^{2} \) . Also, from this theorem, this happens if and only if \( A{x}_{0} - y \) i... | Yes |
Theorem 12.6.2 Let \( A : V \rightarrow W \) where \( A \) is linear and \( V \) and \( W \) are inner product spaces. Then \( A\left( V\right) = \ker {\left( {A}^{ * }\right) }^{ \bot } \) . | Proof: Let \( y = {Ax} \) so \( y \in A\left( V\right) \) . Then if \( {A}^{ * }z = 0 \) ,\n\n\[ \left( {y, z}\right) = \left( {{Ax}, z}\right) = \left( {x,{A}^{ * }z}\right) = 0 \]\n\nshowing that \( y \in \ker {\left( {A}^{ * }\right) }^{ \bot } \) . Thus \( A\left( V\right) \subseteq \ker {\left( {A}^{ * }\right) }^... | Yes |
Corollary 12.6.3 Let \( A, V \), and \( W \) be as described above. If the only solution to \( {A}^{ * }y = 0 \) is \( y = 0 \), then \( A \) is onto \( W \) . | Proof: If the only solution to \( {A}^{ * }y = 0 \) is \( y = 0 \), then \( \ker \left( {A}^{ * }\right) = \{ 0\} \) and so every vector from \( W \) is contained in \( \ker {\left( {A}^{ * }\right) }^{ \bot } \) and by the above theorem, this shows \( A\left( V\right) = W \) . ∎ | Yes |
Lemma 13.1.3 Let \( A \) be an \( n \times n \) matrix and let \( B \) be an \( m \times m \) matrix. Denote by \( C \) the matrix \( C \equiv \left( \begin{matrix} A & 0 \\ 0 & B \end{matrix}\right) \). Then \( C \) is diagonalizable if and only if both \( A \) and \( B \) are diagonalizable. | Proof: Suppose \( {S}_{A}^{-1}A{S}_{A} = {D}_{A} \) and \( {S}_{B}^{-1}B{S}_{B} = {D}_{B} \) where \( {D}_{A} \) and \( {D}_{B} \) are diagonal matrices. You should use block multiplication to verify that \( S \equiv \left( \begin{matrix} {S}_{A} & 0 \\ 0 & {S}_{B} \end{matrix}\right) \) is such that \( {S}^{-1}{CS} = ... | Yes |
Lemma 13.1.6 If \( \mathcal{F} \) is a set of \( n \times n \) matrices which is simultaneously diagonalizable, then \( \mathcal{F} \) is a commuting family of matrices. | Proof: Let \( A, B \in \mathcal{F} \) and let \( S \) be a matrix which has the property that \( {S}^{-1}{AS} \) is a diagonal matrix for all \( A \in \mathcal{F} \) . Then \( {S}^{-1}{AS} = {D}_{A} \) and \( {S}^{-1}{BS} = {D}_{B} \) where \( {D}_{A} \) and \( {D}_{B} \) are diagonal matrices. Since diagonal matrices ... | Yes |
Lemma 13.1.7 Let \( D \) be a diagonal matrix of the form\n\n\[ D \equiv \left( \begin{matrix} {\lambda }_{1}{I}_{{n}_{1}} & 0 & \cdots & 0 \\ 0 & {\lambda }_{2}{I}_{{n}_{2}} & \ddots & \vdots \\ \vdots & \ddots & \ddots & 0 \\ 0 & \cdots & 0 & {\lambda }_{r}{I}_{{n}_{r}} \end{matrix}\right) \]\n\nwhere \( {I}_{{n}_{i}... | Proof: Let \( B = \left( {B}_{ij}\right) \) where \( {B}_{ii} = {B}_{i} \) a block matrix as above in (13.2).\n\n\[ \left( \begin{matrix} {B}_{11} & {B}_{12} & \cdots & {B}_{1r} \\ {B}_{21} & {B}_{22} & \ddots & {B}_{2r} \\ \vdots & \ddots & \ddots & \vdots \\ {B}_{r1} & {B}_{r2} & \cdots & {B}_{rr} \end{matrix}\right)... | Yes |
Lemma 13.1.8 Let \( \mathcal{F} \) denote a commuting family of \( n \times n \) matrices such that each \( A \in \mathcal{F} \) is diagonalizable. Then \( \mathcal{F} \) is simultaneously diagonalizable. | Proof: First note that if every matrix in \( \mathcal{F} \) has only one eigenvalue, there is nothing to prove. This is because for \( A \) such a matrix,\n\n\[ \n{S}^{-1}{AS} = {\lambda I} \n\]\n\nand so\n\n\[ \nA = {\lambda I} \n\]\n\nThus all the matrices in \( \mathcal{F} \) are diagonal matrices and you could pick... | Yes |
Theorem 13.1.9 Let \( \mathcal{F} \) denote a family of matrices which are diagonalizable. Then \( \mathcal{F} \) is simultaneously diagonalizable if and only if \( \mathcal{F} \) is a commuting family. | Proof: If \( \mathcal{F} \) is a commuting family, it follows from Lemma 13.1.8 that it is simultaneously diagonalizable. If it is simultaneously diagonalizable, then it follows from Lemma 13.1.6 that it is a commuting family. | Yes |
Theorem 13.2.4 Let \( L \in \mathcal{L}\left( {H, H}\right) \) where \( H \) is an n dimensional inner product space. If \( L \) is Hermitian, then all of its eigenvalues \( {\lambda }_{k} \) are real and there exists an orthonormal basis of eigenvectors \( \left\{ {\mathbf{w}}_{k}\right\} \) such that\n\n\[ L = \matho... | Proof: By Schur’s theorem, Theorem 13.2.2, there exist \( {l}_{ij} \in \mathbb{F} \) such that\n\n\[ L = \mathop{\sum }\limits_{{j = 1}}^{n}\mathop{\sum }\limits_{{i = 1}}^{j}{l}_{ij}{\mathbf{w}}_{i} \otimes {\mathbf{w}}_{j} \]\n\nThen by Lemma 12.4.2,\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{n}\mathop{\sum }\limits_{{i ... | Yes |
Corollary 13.3.5 Let \( A \in \mathcal{L}\left( {X, X}\right) \) be self adjoint (Hermitian) where \( X \) is a finite dimensional Hilbert space. Then the largest eigenvalue of \( A \) is given by\n\n\[ \max \{ \left( {A\mathbf{x},\mathbf{x}}\right) : \left| \mathbf{x}\right| = 1\} \]\n\n(13.6)\n\nand the minimum eigen... | Proof: The proof of this is just like the proof of Theorem [13,3,3]. Simply replace inf with sup and obtain a decreasing list of eigenvalues. This establishes (13.6). The claim (13.7) follows from Theorem 13.3.3. | No |
Corollary 13.3.6 Let \( A \in \mathcal{L}\left( {X, X}\right) \) where \( A \) is self adjoint. Then \( A = \mathop{\sum }\limits_{i}{\lambda }_{i}{v}_{i} \otimes {v}_{i} \) where \( A{v}_{i} = {\lambda }_{i}{v}_{i} \) and \( {\left\{ {v}_{i}\right\} }_{i = 1}^{n} \) is an orthonormal basis. | Proof : If \( {v}_{k} \) is one of the orthonormal basis vectors, \( A{v}_{k} = {\lambda }_{k}{v}_{k} \) . Also,\n\n\[ \mathop{\sum }\limits_{i}{\lambda }_{i}{v}_{i} \otimes {v}_{i}\left( {v}_{k}\right) = \mathop{\sum }\limits_{i}{\lambda }_{i}{v}_{i}\left( {{v}_{k},{v}_{i}}\right) \]\n\n\[ = \mathop{\sum }\limits_{i}{... | Yes |
Theorem 13.3.7 Let \( A \in \mathcal{L}\left( {X, X}\right) \) be self adjoint where \( X \) is a finite dimensional Hilbert space. Then for \( {\lambda }_{1} \leq {\lambda }_{2} \leq \cdots \leq {\lambda }_{n} \) the eigenvalues of \( A \), there exist orthonormal vectors \( \left\{ {{u}_{1},\cdots ,{u}_{n}}\right\} \... | Proof: From Theorem 13.3.3, there exist eigenvalues and eigenvectors with \( \left\{ {{u}_{1},\cdots ,{u}_{n}}\right\} \) orthonormal and \( {\lambda }_{i} \leq {\lambda }_{i + 1} \) . Therefore, by Corollary 13.3.6\n\n\[ A = \mathop{\sum }\limits_{{j = 1}}^{n}{\lambda }_{j}{u}_{j} \otimes {u}_{j} \]\n\nFix \( \left\{ ... | Yes |
Corollary 13.3.8 Let \( A \in \mathcal{L}\left( {X, X}\right) \) be self adjoint where \( X \) is a finite dimensional Hilbert space. Then for \( {\lambda }_{1} \leq {\lambda }_{2} \leq \cdots \leq {\lambda }_{n} \) the eigenvalues of \( A \), there exist orthonormal vectors \( \left\{ {{u}_{1},\cdots ,{u}_{n}}\right\}... | \[ {\lambda }_{k} \equiv \mathop{\max }\limits_{{{w}_{1},\cdots ,{w}_{k - 1}}}\left\{ {\min \left\{ {\frac{\left( Ax, x\right) }{{\left| x\right| }^{2}} : x \neq 0, x \in {\left\{ {w}_{1},\cdots ,{w}_{k - 1}\right\} }^{ \bot }}\right\} }\right\} \] | Yes |
Corollary 13.3.9 Let \( A \in \mathcal{L}\left( {X, X}\right) \) be self adjoint where \( X \) is a finite dimensional Hilbert space. Then for \( {\lambda }_{1} \leq {\lambda }_{2} \leq \cdots \leq {\lambda }_{n} \) the eigenvalues of \( A \), there exist orthonormal vectors \( \left\{ {{u}_{1},\cdots ,{u}_{n}}\right\}... | \[ {\lambda }_{k} \equiv \mathop{\min }\limits_{{{w}_{1},\cdots ,{w}_{n - k}}}\left\{ {\max \left\{ {\frac{\left( Ax, x\right) }{{\left| x\right| }^{2}} : x \neq 0, x \in {\left\{ {w}_{1},\cdots ,{w}_{n - k}\right\} }^{ \bot }}\right\} }\right\} \]\n\n(13.12)\n\nwhere if \( k = n,{\left\{ {w}_{1},\cdots ,{w}_{n - k}\ri... | Yes |
Corollary 13.4.2 Let \( X \) be a finite dimensional Hilbert space and let \( \left\{ {{v}_{1},\cdots ,{v}_{n}}\right\} \) be an orthonormal basis for \( X \) . Also, let \( q \) be the coordinate map associated with this basis satisfying \( q\left( \mathbf{x}\right) \equiv \mathop{\sum }\limits_{i}{x}_{i}{v}_{i} \) . ... | \[ {\left( Aq\left( \mathbf{x}\right), q\left( \mathbf{y}\right) \right) }_{X} = {\left( M\left( A\right) \mathbf{x},\mathbf{y}\right) }_{{\mathbb{F}}^{n}}. \] | Yes |
Lemma 13.4.4 Let \( X \) be a finite dimensional Hilbert space. A self adjoint \( A \in \mathcal{L}\left( {X, X}\right) \) is positive definite if and only if all its eigenvalues are positive and negative definite if and only if all its eigenvalues are negative. It is positive semidefinite if all the eigenvalues are no... | Proof: Suppose first that \( A \) is positive definite and let \( \lambda \) be an eigenvalue. Then for \( \mathbf{x} \) an eigenvector corresponding to \( \lambda ,\lambda \left( {\mathbf{x},\mathbf{x}}\right) = \left( {\lambda \mathbf{x},\mathbf{x}}\right) = \left( {A\mathbf{x},\mathbf{x}}\right) > 0 \) . Therefore, ... | Yes |
Theorem 13.4.6 Let \( X \) be a finite dimensional Hilbert space and let \( A \in \mathcal{L}\left( {X, X}\right) \) be self adjoint. Then \( A \) is positive definite if and only if \( \det \left( {M{\left( A\right) }_{k}}\right) > 0 \) for every \( k = 1,\cdots, n \) . Here \( M\left( A\right) \) denotes the matrix o... | Proof: This theorem is proved by induction on \( n \) . It is clearly true if \( n = 1 \) . Suppose then that it is true for \( n - 1 \) where \( n \geq 2 \) . Since \( \det \left( {M\left( A\right) }\right) > 0 \), it follows that all the eigenvalues are nonzero. Are they all positive? Suppose not. Then there is some ... | No |
Corollary 13.4.7 Let \( X \) be a finite dimensional Hilbert space and let \( A \in \mathcal{L}\left( {X, X}\right) \) be self adjoint. Then \( A \) is negative definite if and only if \( \det \left( {M{\left( A\right) }_{k}}\right) {\left( -1\right) }^{k} > 0 \) for every \( k = 1,\cdots, n \) . Here \( M\left( A\righ... | Proof: This is immediate from the above theorem by noting that, as in the proof of Lemma [13,4,4, \( A \) is negative definite if and only if \( - A \) is positive definite. Therefore, if \( \det \left( {-M{\left( A\right) }_{k}}\right) > 0 \) for all \( k = 1,\cdots, n \), it follows that \( A \) is negative definite.... | Yes |
Lemma 13.6.1 Suppose \( R \in \mathcal{L}\left( {X, Y}\right) \) where \( X, Y \) are Hilbert spaces and \( R \) preserves distances. Then \( {R}^{ * }R = I \) . | Proof: Since \( R \) preserves distances, \( \left| {R\mathbf{x}}\right| = \left| \mathbf{x}\right| \) for every \( \mathbf{x} \) . Therefore from the axioms of the inner product,\n\n\[ \n{\left| \mathbf{x}\right| }^{2} + {\left| \mathbf{y}\right| }^{2} + \left( {\mathbf{x},\mathbf{y}}\right) + \left( {\mathbf{y},\math... | Yes |
Corollary 13.6.3 Let \( F \in \mathcal{L}\left( {X, Y}\right) \) and suppose \( n \geq m \) where \( X \) is a Hilbert space of dimension \( n \) and \( Y \) is a Hilbert space of dimension \( m \) . Then there exists a Hermitian \( U \in \) \( \mathcal{L}\left( {X, X}\right) \), and an element of \( \mathcal{L}\left( ... | Proof: Recall that \( {L}^{* * } = L \) and \( {\left( ML\right) }^{ * } = {L}^{ * }{M}^{ * } \) . Now apply Theorem 13.6. to \( {F}^{ * } \in \mathcal{L}\left( {Y, X}\right) \) . Thus, \[ {F}^{ * } = {R}^{ * }U \] where \( {R}^{ * } \) and \( U \) satisfy the conditions of that theorem. Then \[ F = {UR} \] and \( R{R}... | Yes |
Theorem 13.6.5 Let \( F \in \mathcal{L}\left( {X, X}\right) \) . Then \( F \) is normal if and only if in Corollary 13.6.4 \( {RU} = {UR} \) and \( {QW} = {WQ} \) . | Proof: I will prove the statement about \( {RU} = {UR} \) and leave the other part as an exercise. First suppose that \( {RU} = {UR} \) and show \( F \) is normal. To begin with,\n\n\[ U{R}^{ * } = {\left( RU\right) }^{ * } = {\left( UR\right) }^{ * } = {R}^{ * }U. \]\n\nTherefore,\n\n\[ {F}^{ * }F = U{R}^{ * }{RU} = {... | No |
Lemma 13.8.1 Let \( A \) be an \( m \times n \) matrix. Then \( {A}^{ * }A \) is self adjoint and all its eigenvalues are nonnegative. | Proof: It is obvious that \( {A}^{ * }A \) is self adjoint. Suppose \( {A}^{ * }A\mathbf{x} = \lambda \mathbf{x} \) . Then \( \lambda {\left| \mathbf{x}\right| }^{2} = \) \( \left( {\lambda \mathbf{x},\mathbf{x}}\right) = \left( {{A}^{ * }A\mathbf{x},\mathbf{x}}\right) = \left( {A\mathbf{x}, A\mathbf{x}}\right) \geq 0 ... | Yes |
Theorem 13.8.3 Let \( A \) be an \( m \times n \) matrix. Then there exist unitary matrices, \( U \) and \( V \) of the appropriate size such that\n\n\[ \n{U}^{ * }{AV} = \left( \begin{array}{ll} \sigma & 0 \\ 0 & 0 \end{array}\right)\n\]\nwhere \( \sigma \) is of the form\n\n\[ \n\sigma = \left( \begin{matrix} {\sigma... | Proof: By the above lemma and Theorem [13,3,3] there exists an orthonormal basis, \( {\left\{ {\mathbf{v}}_{i}\right\} }_{i = 1}^{n} \) such that \( {A}^{ * }A{\mathbf{v}}_{i} = {\sigma }_{i}^{2}{\mathbf{v}}_{i} \) where \( {\sigma }_{i}^{2} > 0 \) for \( i = 1,\cdots, k,\left( {{\sigma }_{i} > 0}\right) \), and equals... | Yes |
Corollary 13.8.4 Let \( A \) be an \( m \times n \) matrix. Then the rank of \( A \) and \( {A}^{ * } \) equals the number of singular values. | Proof: Since \( V \) and \( U \) are unitary, they are each one to one and onto and so it follows\n\nthat\n\[\n\operatorname{rank}\left( A\right) = \operatorname{rank}\left( {{U}^{ * }{AV}}\right) = \operatorname{rank}\left( \begin{array}{ll} \sigma & 0 \\ 0 & 0 \end{array}\right) = \text{ number of singular values. }\... | Yes |
Lemma 13.9.2 Let \( A \) be an \( m \times n \) complex matrix with singular matrix\n\n\[ \sum = \left( \begin{array}{ll} \sigma & 0 \\ 0 & 0 \end{array}\right) \]\n\nwith \( \sigma \) as defined above. Then\n\n\[ \parallel \sum {\parallel }_{F}^{2} = \parallel A{\parallel }_{F}^{2} \] | Proof: From the definition and letting \( U, V \) be unitary and of the right size,\n\n\[ \parallel {UA}{\parallel }_{F}^{2} \equiv \operatorname{trace}\left( {{UA}{A}^{ * }{U}^{ * }}\right) = \operatorname{trace}\left( {A{A}^{ * }}\right) = \parallel A{\parallel }_{F}^{2} \]\n\nAlso,\n\n\[ \parallel {AV}{\parallel }_{... | Yes |
Proposition 13.11.2 \( {A}^{ + }\mathbf{y} \) is the solution to the problem of minimizing \( \left| {A\mathbf{x} - \mathbf{y}}\right| \) for all \( \mathbf{x} \) which has smallest norm. Thus\n\n\[ \left| {A{A}^{ + }\mathbf{y} - \mathbf{y}}\right| \leq \left| {A\mathbf{x} - \mathbf{y}}\right| \text{ for all }\mathbf{x... | Proof: Consider \( \mathbf{x} \) satisfying [13.22], equivalently \( {A}^{ * }A\mathbf{x} = {A}^{ * }\mathbf{y} \), \n\n\[ \left( \begin{matrix} {\sigma }^{2} & 0 \\ 0 & 0 \end{matrix}\right) {V}^{ * }\mathbf{x} = \left( \begin{array}{ll} \sigma & 0 \\ 0 & 0 \end{array}\right) {U}^{ * }\mathbf{y} \]\n\nwhich has smalle... | Yes |
Lemma 13.11.3 The matrix \( {A}^{ + } \) satisfies the following conditions.\n\n\[ A{A}^{ + }A = A,{A}^{ + }A{A}^{ + } = {A}^{ + },{A}^{ + }A\\text{and}A{A}^{ + }\\text{are Hermitian.} \] | Proof: This is routine. Recall\n\n\[ A = U\\left( \\begin{array}{ll} \\sigma & 0 \\\\ 0 & 0 \\end{array}\\right) {V}^{ * }\n\n\\text{and}\n\n{A}^{ + } = V\\left( \\begin{matrix} {\\sigma }^{-1} & 0 \\\\ 0 & 0 \\end{matrix}\\right) {U}^{ * }\n\n\\text{so you just plug in and verify it works.} | No |
Corollary 14.0.7 If \( \left( {X,\parallel \cdot \parallel }\right) \) is a finite dimensional normed linear space with the field of scalars \( \mathbb{F} = \mathbb{C} \) or \( \mathbb{R} \), then \( X \) is complete. | Proof: Let \( \left\{ {\mathbf{x}}^{k}\right\} \) be a Cauchy sequence. Then letting the components of \( {\mathbf{x}}^{k} \) with respect to the given basis be\n\n\[ \n{x}_{1}^{k},\cdots ,{x}_{n}^{k} \n\]\n\nit follows from Theorem 14.0.6, that\n\n\[ \n\left( {{x}_{1}^{k},\cdots ,{x}_{n}^{k}}\right) \n\]\n\nis a Cauch... | Yes |
Corollary 14.0.8 Suppose \( X \) is a finite dimensional linear space with the field of scalars either \( \mathbb{C} \) or \( \mathbb{R} \) and \( \parallel \cdot \parallel \) and \( \parallel \parallel \cdot \parallel \parallel \) are two norms on \( X \) . Then there exist positive constants, \( \delta \) and \( \Del... | Proof: Let \( \left\{ {{\mathbf{v}}_{1},\cdots ,{\mathbf{v}}_{n}}\right\} \) be a basis for \( X \) and let \( \left| \cdot \right| \) be the norm taken with respect to this basis which was described earlier. Then by Theorem 14.0.6, there are positive constants \( {\delta }_{1},{\Delta }_{1},{\delta }_{2},{\Delta }_{2}... | Yes |
Theorem 14.0.10 Let \( X \) and \( Y \) be finite dimensional normed linear spaces of dimension \( n \) and \( m \) respectively and denote by \( \parallel \cdot \parallel \) the norm on either \( X \) or \( Y \) . Then if \( A \) is any linear function mapping \( X \) to \( Y \), then \( A \in \mathcal{L}\left( {X, Y}... | Proof: It is necessary to show the norm defined on linear transformations really is a norm. Again the first and third properties listed above for norms are obvious. It remains to show the second and verify \( \parallel A\parallel < \infty \) . Letting \( \left\{ {{\mathbf{v}}_{1},\cdots ,{\mathbf{v}}_{n}}\right\} \) be... | Yes |
Proposition 14.0.13 The following holds.\n\n\[ \parallel A{\parallel }_{2} = \sup \{ \left| {A\mathbf{x}}\right| : \left| \mathbf{x}\right| = 1\} \equiv \parallel A\parallel . \] | Proof: Note that \( {A}^{ * }A \) is Hermitian and so by Corollary 13.3.5,\n\n\[ \parallel A{\parallel }_{2} = \max \left\{ {{\left( {A}^{ * }A\mathbf{x},\mathbf{x}\right) }^{1/2} : \left| \mathbf{x}\right| = 1}\right\} \]\n\n\[ = \max \left\{ {{\left( A\mathbf{x}, A\mathbf{x}\right) }^{1/2} : \left| \mathbf{x}\right| ... | Yes |
Theorem 14.0.15 Let \( {X}_{i} \) and \( \parallel \cdot {\parallel }_{i} \) be given in the above definition and consider the norms on \( \mathop{\prod }\limits_{{i = 1}}^{n}{X}_{i} \) described there in terms of norms on \( {\mathbb{R}}^{n} \) . Then any two of these norms on \( \mathop{\prod }\limits_{{i = 1}}^{n}{X... | For example, define\n\n\[ \parallel \mathbf{x}{\parallel }_{1} \equiv \mathop{\sum }\limits_{{i = 1}}^{n}\left| {x}_{i}\right| \]\n\n\[ \parallel \mathbf{x}{\parallel }_{\infty } \equiv \max \left\{ {\left| {x}_{i}\right|, i = 1,\cdots, n}\right\} \]\n\nor\n\n\[ \parallel \mathbf{x}{\parallel }_{2} = {\left( \mathop{\s... | No |
Lemma 14.1.3 If \( a, b \geq 0 \) and \( {p}^{\prime } \) is defined by \( \frac{1}{p} + \frac{1}{{p}^{\prime }} = 1 \), then\n\n\[ {ab} \leq \frac{{a}^{p}}{p} + \frac{{b}^{{p}^{\prime }}}{{p}^{\prime }}. \]\n | Proof of the Proposition: If \( \mathbf{x} \) or \( \mathbf{y} \) equals the zero vector there is nothing to prove. Therefore, assume they are both nonzero. Let \( A = {\left( \mathop{\sum }\limits_{{i = 1}}^{n}{\left| {x}_{i}\right| }^{p}\right) }^{1/p} \) and \( B = \) \( {\left( \mathop{\sum }\limits_{{i = 1}}^{n}{\... | Yes |
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