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Lemma 3.3.15 Suppose a matrix is of the form\n\n\\[ \nM = \\left( \\begin{array}{ll} A & * \\\\ \\mathbf{0} & a \\end{array}\\right) \n\\]\n\n\\( \\left( {3.13}\\right) \\)\n\nor\n\\[\nM = \\left( \\begin{array}{ll} A & \\mathbf{0} \\\\ * & a \\end{array}\\right) \n\\]\n\n(3.14)\n\nwhere a is a number and \\( A \\) is ... | Proof: Denote \\( M \\) by \\( \\left( {m}_{ij}\\right) \\) . Thus in the first case, \\( {m}_{nn} = a \\) and \\( {m}_{ni} = 0 \\) if \\( i \\neq n \\) while in the second case, \\( {m}_{nn} = a \\) and \\( {m}_{in} = 0 \\) if \\( i \\neq n \\) . From the definition of the determinant,\n\n\\[\n\\det \\left( M\\right) ... | Yes |
Theorem 3.3.17 Let \( A \) be an \( n \times n \) matrix where \( n \geq 2 \) . Then\n\n\[ \det \left( A\right) = \mathop{\sum }\limits_{{j = 1}}^{n}{a}_{ij}\operatorname{cof}{\left( A\right) }_{ij} = \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{ij}\operatorname{cof}{\left( A\right) }_{ij}. \]\n\n\( \left( {3.15}\right) \)\... | Proof: Let \( \left( {{a}_{i1},\cdots ,{a}_{in}}\right) \) be the \( {i}^{th} \) row of \( A \) . Let \( {B}_{j} \) be the matrix obtained from \( A \) by leaving every row the same except the \( {i}^{\text{th }} \) row which in \( {B}_{j} \) equals \( \left( {0,\cdots ,0,{a}_{ij},0,\cdots ,0}\right) \) . Then by Corol... | Yes |
Theorem 3.3.18 \( {A}^{-1} \) exists if and only if \( \det \left( A\right) \neq 0 \) . If \( \det \left( A\right) \neq 0 \), then \( {A}^{-1} = \left( {a}_{ij}^{-1}\right) \) where\n\n\[ \n{a}_{ij}^{-1} = \det {\left( A\right) }^{-1}\operatorname{cof}{\left( A\right) }_{ji} \n\]\n\nfor \( \operatorname{cof}{\left( A\r... | Proof: By Theorem 3.3.12 and letting \( \left( {a}_{ir}\right) = A \), if \( \det \left( A\right) \neq 0 \) ,\n\n\[ \n\mathop{\sum }\limits_{{i = 1}}^{n}{a}_{ir}\operatorname{cof}{\left( A\right) }_{ir}\det {\left( A\right) }^{-1} = \det \left( A\right) \det {\left( A\right) }^{-1} = 1. \n\]\n\nNow in the matrix \( A \... | Yes |
Corollary 3.3.19 Let \( A \) be an \( n \times n \) matrix and suppose there exists an \( n \times n \) matrix \( B \) such that \( {BA} = I \) . Then \( {A}^{-1} \) exists and \( {A}^{-1} = B \) . Also, if there exists \( C \) an \( n \times n \) matrix such that \( {AC} = I \), then \( {A}^{-1} \) exists and \( {A}^{... | Proof: Since \( {BA} = I \), Theorem 3.3.13 implies\n\n\[ \det B\det A = 1 \]\n\nand so \( \det A \neq 0 \) . Therefore from Theorem 3.3.18, \( {A}^{-1} \) exists. Therefore,\n\n\[ {A}^{-1} = \left( {BA}\right) {A}^{-1} = B\left( {A{A}^{-1}}\right) = {BI} = B. \]\n\nThe case where \( {CA} = I \) is handled similarly. ∎ | Yes |
Theorem 3.3.23 If \( A \), an \( m \times n \) matrix has determinant rank \( r \), then there exist \( r \) rows of the matrix such that every other row is a linear combination of these \( r \) rows. | Proof: Suppose the determinant rank of \( A = \left( {a}_{ij}\right) \) equals \( r \) . Thus some \( r \times r \) submatrix has non zero determinant and there is no larger square submatrix which has non zero determinant. Suppose such a submatrix is determined by the \( r \) columns whose indices are\n\n\[ \n{j}_{1} <... | Yes |
Corollary 3.3.24 The determinant rank equals the row rank. | Proof: From Theorem 3.3.23, every row is in the span of \( r \) rows where \( r \) is the determinant rank. Therefore, the row rank (dimension of the span of the rows) is no larger than the determinant rank. Could the row rank be smaller than the determinant rank? If so, it follows from Theorem 3.3.23 that there exist ... | Yes |
Corollary 3.3.25 If \( A \) has determinant rank \( r \), then there exist \( r \) columns of the matrix such that every other column is a linear combination of these \( r \) columns. Also the column rank equals the determinant rank. | Proof: This follows from the above by considering \( {A}^{T} \) . The rows of \( {A}^{T} \) are the columns of \( A \) and the determinant rank of \( {A}^{T} \) and \( A \) are the same. Therefore, from Corollary 3.3.24, column rank of \( A = \) row rank of \( {A}^{T} = \) determinant rank of \( {A}^{T} = \) determinan... | Yes |
Theorem 3.3.26 Let \( A \) be an \( n \times n \) matrix. Then the following are equivalent.\n\n\[ \text{1.}\det \left( A\right) = 0\text{.} \]\n\n2. \( A,{A}^{T} \) are not one to one.\n\n3. \( A \) is not onto. | Proof: Suppose \( \det \left( A\right) = 0 \) . Then the determinant rank of \( A = r < n \) . Therefore, there exist \( r \) columns such that every other column is a linear combination of these columns by Theorem 3.3.23. In particular, it follows that for some \( m \), the \( {m}^{\text{th }} \) column is a linear co... | Yes |
Corollary 3.3.27 Let \( A \) be an \( n \times n \) matrix. Then the following are equivalent.\n\n1. \( \det \left( A\right) \neq 0 \) .\n\n2. \( A \) and \( {A}^{T} \) are one to one.\n\n3. \( A \) is onto. | Proof: This follows immediately from the above theorem. | No |
Lemma 3.4.2 Suppose for all \( \left| \lambda \right| \) large enough,\n\n\[{A}_{0} + {A}_{1}\lambda + \cdots + {A}_{m}{\lambda }^{m} = 0,\]\n\nwhere the \( {A}_{i} \) are \( n \times n \) matrices. Then each \( {A}_{i} = 0 \) . | Proof: Multiply by \( {\lambda }^{-m} \) to obtain\n\n\[{A}_{0}{\lambda }^{-m} + {A}_{1}{\lambda }^{-m + 1} + \cdots + {A}_{m - 1}{\lambda }^{-1} + {A}_{m} = 0.\]\n\nNow let \( \left| \lambda \right| \rightarrow \infty \) to obtain \( {A}_{m} = 0 \) . With this, multiply by \( \lambda \) to obtain\n\n\[{A}_{0}{\lambda ... | Yes |
Corollary 3.4.3 Let \( {A}_{i} \) and \( {B}_{i} \) be \( n \times n \) matrices and suppose\n\n\[ \n{A}_{0} + {A}_{1}\lambda + \cdots + {A}_{m}{\lambda }^{m} = {B}_{0} + {B}_{1}\lambda + \cdots + {B}_{m}{\lambda }^{m} \n\]\n\nfor all \( \left| \lambda \right| \) large enough. Then \( {A}_{i} = {B}_{i} \) for all \( i ... | Proof: Subtract and use the result of the lemma. - | No |
Theorem 3.4.4 Let \( A \) be an \( n \times n \) matrix and let \( p\left( \lambda \right) \equiv \det \left( {{\lambda I} - A}\right) \) be the characteristic polynomial. Then \( p\left( A\right) = 0 \) . | Proof: Let \( C\left( \lambda \right) \) equal the transpose of the cofactor matrix of \( \left( {{\lambda I} - A}\right) \) for \( \left| \lambda \right| \) large. (If \( \left| \lambda \right| \) is large enough, then \( \lambda \) cannot be in the finite list of eigenvalues of \( A \) and so for such \( \lambda ,{\l... | Yes |
Lemma 3.5.1 Consider the following product.\n\n\\[ \n\\left( \\begin{array}{l} 0 \\\\ I \\\\ 0 \\end{array}\\right) \\left( \\begin{array}{lll} 0 & I & 0 \\end{array}\\right) \n\\]\n\nwhere the first is \( n \times r \) and the second is \( r \times n \) . The small identity matrix \( I \) is an \( r \times r \) matrix... | Proof: From the definition of the way you multiply matrices, the product is\n\n\\[ \n\\left( \\begin{matrix} \\left( \\begin{array}{l} \\mathbf{0} \\\\ I \\\\ \\mathbf{0} \\end{array}\\right) \\mathbf{0} & \\cdots \\left( \\begin{array}{l} \\mathbf{0} \\\\ I \\\\ \\mathbf{0} \\end{array}\\right) \\mathbf{0} & \\left( \... | Yes |
Example 3.5.3 Let an \( n \times n \) matrix have the form \( A = \left( \begin{array}{ll} a & \mathbf{b} \\ \mathbf{c} & P \end{array}\right) \) where \( P \) is \( n - 1 \times n - 1 \) . Multiply it by \( B = \left( \begin{array}{ll} p & \mathbf{q} \\ \mathbf{r} & Q \end{array}\right) \) where \( B \) is also an \( ... | \[ \left( \begin{array}{ll} a & \mathbf{b} \\ \mathbf{c} & P \end{array}\right) \left( \begin{array}{ll} p & \mathbf{q} \\ \mathbf{r} & Q \end{array}\right) = \left( \begin{array}{ll} {ap} + \mathbf{{br}} & a\mathbf{q} + \mathbf{b}Q \\ p\mathbf{c} + P\mathbf{r} & \mathbf{{cq}} + {PQ} \end{array}\right) \] | Yes |
Theorem 3.5.4 Let \( A \) be an \( m \times n \) matrix and let \( B \) be an \( n \times m \) matrix for \( m \leq n \) . Then\n\n\[ \n{p}_{BA}\left( t\right) = {t}^{n - m}{p}_{AB}\left( t\right) \n\] \n\nso the eigenvalues of \( {BA} \) and \( {AB} \) are the same including multiplicities except that \( {BA} \) has \... | Proof: Use block multiplication to write\n\n\[ \n\left( \begin{matrix} {AB} & 0 \\ B & 0 \end{matrix}\right) \left( \begin{array}{ll} I & A \\ 0 & I \end{array}\right) = \left( \begin{matrix} {AB} & {ABA} \\ B & {BA} \end{matrix}\right) \n\] \n\n\[ \n\left( \begin{matrix} I & A \\ 0 & I \end{matrix}\right) \left( \begi... | Yes |
Theorem 4.1.6 To perform any of the three row operations on a matrix \( A \) it suffices to do the row operation on the identity matrix obtaining an elementary matrix \( E \) and then take the product, EA. Furthermore, each elementary matrix is invertible and its inverse is an elementary matrix. | Proof: The first part of this theorem has been proved in Lemmas 4.1.3 - 4.1.5. It only remains to verify the claim about the inverses. Consider first the elementary matrices corresponding to row operation of type three.\n\n\[ E\left( {-c \times i + j}\right) E\left( {c \times i + j}\right) = I \]\n\nThis follows becaus... | Yes |
Lemma 4.2.3 Let \( B \) and \( A \) be two \( m \times n \) matrices and suppose \( B \) results from a row operation applied to \( A \) . Then the \( {k}^{\text{th }} \) column of \( B \) is a linear combination of the \( {i}_{1},\cdots ,{i}_{r} \) columns of \( B \) if and only if the \( {k}^{\text{th }} \) column of... | Proof: Let \( A \) equal the following matrix in which the \( {\mathbf{a}}_{k} \) are the columns\n\n\[ \left( \begin{array}{llll} {\mathbf{a}}_{1} & {\mathbf{a}}_{2} & \cdots & {\mathbf{a}}_{n} \end{array}\right) \]\n\nand let \( B \) equal the following matrix in which the columns are given by the \( {\mathbf{b}}_{k}... | Yes |
Corollary 4.2.4 Let \( A \) and \( B \) be two \( m \times n \) matrices such that \( B \) is obtained by applying a row operation to \( A \) . Then the two matrices have the same rank. | Proof: Lemma 4.2.3 says the linear relationships are the same between the columns of \( A \) and those of \( B \) . Therefore, the column rank of the two matrices is the same. | Yes |
Find the rank of the following matrix and identify columns whose linear combinations yield all the other columns. | Take \( \left( {-1}\right) \) times the first row and add to the second and then take \( \left( {-3}\right) \) times the first row and add to the third. This yields\n\n\[ \left( \begin{matrix} 1 & 2 & 1 & 3 & 2 \\ 0 & 1 & 5 & - 3 & 0 \\ 0 & 1 & 5 & - 3 & 0 \end{matrix}\right) \]\n\nBy the above corollary, this matrix h... | Yes |
Find the rank of the following matrix and identify columns whose linear combinations yield all the other columns. | Take \( \\left( {-1}\\right) \) times the first row and add to the second and then take \( \\left( {-3}\\right) \) times the first row and add to the last row. This yields\n\n\[ \n\\left( \\begin{matrix} 1 & 2 & 1 & 3 & 2 \\ 0 & 0 & 5 & - 3 & 0 \\ 0 & 0 & 5 & - 3 & 0 \\end{matrix}\\right)\n\]\n\nNow multiply the second... | Yes |
Theorem 4.3.2 Let \( A \) be an \( m \times n \) matrix. Then \( A \) has a row reduced echelon form determined by a simple process. | Proof: Viewing the columns of \( A \) from left to right take the first nonzero column. Pick a nonzero entry in this column and switch the row containing this entry with the top row of A. Now divide this new top row by the value of this nonzero entry to get a 1 in this position and then use row operations to make all e... | Yes |
Corollary 4.3.5 The row reduced echelon form is unique. That is if \( B, C \) are two matrices in row reduced echelon form and both are row equivalent to \( A \), then \( B = C \) . | Proof: Suppose \( B \) and \( C \) are both row reduced echelon forms for the matrix \( A \) . Then they clearly have the same zero columns since row operations leave zero columns unchanged. If \( B \) has the sequence \( {\mathbf{e}}_{1},{\mathbf{e}}_{2},\cdots ,{\mathbf{e}}_{r} \) occurring for the first time in the ... | Yes |
Corollary 4.3.6 Let \( A \) be an \( m \times n \) matrix and let \( R \) denote the row reduced echelon form obtained from \( A \) by row operations. Then there exists a sequence of elementary matrices, \( {E}_{1},\cdots ,{E}_{p} \) such that\n\n\[ \left( {{E}_{p}{E}_{p - 1}\cdots {E}_{1}}\right) A = R. \] | Proof: This follows from the fact that row operations are equivalent to multiplication on the left by an elementary matrix. | No |
Corollary 4.3.7 Let \( A \) be an invertible \( n \times n \) matrix. Then \( A \) equals a finite product of elementary matrices. | Proof: Since \( {A}^{-1} \) is given to exist, it follows \( A \) must have rank \( n \) because by Theorem 3.3.18 \( \det \left( A\right) \neq 0 \) which says the determinant rank and hence the column rank of \( A \) is \( n \) and so the row reduced echelon form of \( A \) is \( I \) because the columns of \( A \) fo... | Yes |
Corollary 4.3.8 The rank of a matrix equals the number of nonzero pivot columns. Furthermore, every column is contained in the span of the pivot columns. | Proof: Write the row reduced echelon form for the matrix. From Corollary 4.2.4 this row reduced matrix has the same rank as the original matrix. Deleting all the zero rows and all the columns in the row reduced echelon form which do not correspond to a pivot column, yields an \( r \times r \) identity submatrix in whic... | Yes |
Corollary 4.3.9 Suppose \( A \) is an \( m \times n \) matrix and that \( m < n \). That is, the number of rows is less than the number of columns. Then one of the columns of \( A \) is a linear combination of the preceding columns of \( A \) . | Proof: Since \( m < n \), not all the columns of \( A \) can be pivot columns. That is, in the row reduced echelon form say \( {\mathbf{e}}_{i} \) occurs for the first time at \( {r}_{i} \) where \( {r}_{1} < {r}_{2} < \cdots < {r}_{p} \) where \( p \leq m \). It follows since \( m < n \), there exists some column in t... | Yes |
Theorem 4.3.12 Let \( A \) be an \( m \times n \) matrix. Then \( \operatorname{rank}\left( A\right) + \dim \left( {\ker \left( A\right) }\right) = n \). | Proof: Since \( \ker \left( A\right) \) is a subspace, there exists a basis for \( \ker \left( A\right) ,\left\{ {{\mathbf{x}}_{1},\cdots ,{\mathbf{x}}_{k}}\right\} \). Also let \( \left\{ {A{\mathbf{y}}_{1},\cdots, A{\mathbf{y}}_{l}}\right\} \) be a basis for \( A\left( {\mathbb{F}}^{n}\right) \). Let \( \mathbf{u} \i... | Yes |
Proposition 4.4.1 Let \( A \) be an \( m \times n \) matrix and let \( \mathbf{b} \) be an \( m \times 1 \) column vector. Then there exists a solution to (4.4) if and only if\n\n\[ \operatorname{rank}\left( \begin{array}{lll} A & \mid & \mathbf{b} \end{array}\right) = \operatorname{rank}\left( A\right) . \] | Proof: Place \( \left( {A \mid \mathbf{b}}\right) \) and \( A \) in row reduced echelon form, respectively \( B \) and \( C \) . If the above condition on rank is true, then both \( B \) and \( C \) have the same number of nonzero rows. In particular, you cannot have a row of the form\n\n\[ \left( \begin{array}{llll} 0... | Yes |
Lemma 4.5.2 Let \( A \) be a real \( m \times n \) matrix, let \( \mathbf{x} \in {\mathbb{R}}^{n} \) and \( \mathbf{y} \in {\mathbb{R}}^{m} \) . Then\n\n\[ \left( {A\mathbf{x} \cdot \mathbf{y}}\right) = \left( {\mathbf{x} \cdot {A}^{T}\mathbf{y}}\right) \] | Proof: This follows right away from the definition of the inner product and matrix multiplication.\n\n\[ \left( {A\mathbf{x} \cdot \mathbf{y}}\right) = \mathop{\sum }\limits_{{k, l}}{A}_{kl}{x}_{l}{y}_{k} = \mathop{\sum }\limits_{{k, l}}{\left( {A}^{T}\right) }_{lk}{x}_{l}{y}_{k} = \left( {\mathbf{x} \cdot {A}^{T}\math... | Yes |
Theorem 4.5.3 Let \( A \) be a real \( m \times n \) matrix and let \( \mathbf{b} \in {\mathbb{R}}^{m} \) . There exists a solution, \( \mathbf{x} \) to the equation \( A\mathbf{x} = \mathbf{b} \) if and only if \( \mathbf{b} \in \ker {\left( {A}^{T}\right) }^{ \bot } \) . | Proof: First suppose \( \mathbf{b} \in \ker {\left( {A}^{T}\right) }^{ \bot } \) . Then this says that if \( {A}^{T}\mathbf{x} = \mathbf{0} \), it follows that \( \mathbf{b} \cdot \mathbf{x} = \mathbf{0} \) . In other words, taking the transpose, if\n\n\[ \n{\mathbf{x}}^{T}A = \mathbf{0}\text{, then}{\mathbf{x}}^{T}\ma... | Yes |
Corollary 4.5.4 Let \( A \) be an \( m \times n \) matrix. Then \( A \) maps \( {\mathbb{R}}^{n} \) onto \( {\mathbb{R}}^{m} \) if and only if the only solution to \( {A}^{T}\mathbf{x} = \mathbf{0} \) is \( \mathbf{x} = \mathbf{0} \). | Proof: If the only solution to \( {A}^{T}\mathbf{x} = \mathbf{0} \) is \( \mathbf{x} = \mathbf{0} \), then \( \ker \left( {A}^{T}\right) = \{ \mathbf{0}\} \) and so \( \ker {\left( {A}^{T}\right) }^{ \bot } = \) \( {\mathbb{R}}^{m} \) because every \( \mathbf{b} \in {\mathbb{R}}^{m} \) has the property that \( \mathbf{... | Yes |
Example 4.5.5 Let \( A \) be an \( m \times n \) matrix in which \( m > n \) . Then \( A \) cannot map onto \( {\mathbb{R}}^{m} \) . | The reason for this is that \( {A}^{T} \) is an \( n \times m \) where \( m > n \) and so in the augmented matrix\n\n\[ \left( {{A}^{T} \mid \mathbf{0}}\right) \]\n\n there must be some free variables. Thus there exists a nonzero vector \( \mathbf{x} \) such that \( {A}^{T}\mathbf{x} = \mathbf{0} \) . | Yes |
Can you write \( \left( \begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right) \) in the form LU as just described? | To do so you would need\n\n\[ \left( \begin{array}{ll} 1 & 0 \\ x & 1 \end{array}\right) \left( \begin{array}{ll} a & b \\ 0 & c \end{array}\right) = \left( \begin{matrix} a & b \\ {xa} & {xb} + c \end{matrix}\right) = \left( \begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right) . \]\n\nTherefore, \( b = 1 \) and \( a = ... | Yes |
Find an LU factorization for \( A = \left( \begin{matrix} 1 & 2 & 3 \\ 2 & 1 & - 4 \\ 1 & 5 & 2 \end{matrix}\right) \) | Write the matrix next to the identity matrix as shown.\n\n\[ \left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right) \left( \begin{matrix} 1 & 2 & 3 \\ 2 & 1 & - 4 \\ 1 & 5 & 2 \end{matrix}\right) \]\n\nThe process involves doing row operations to the matrix on the right while simultaneously updat... | Yes |
Find an LU factorization for \( A = \left( \begin{array}{lllll} 1 & 2 & 1 & 2 & 1 \\ 2 & 0 & 2 & 1 & 1 \\ 2 & 3 & 1 & 3 & 2 \\ 1 & 0 & 1 & 1 & 2 \end{array}\right) \) . | This time everything is done at once for a whole column. This saves trouble. First multiply the first row by \( \left( {-1}\right) \) and then add to the last row. Next take \( \left( {-2}\right) \) times the first and add to the second and then \( \left( {-2}\right) \) times the first and add to the third.\n\n\[ \left... | Yes |
Suppose you want to find the solutions to \( \left( \begin{array}{llll} 1 & 2 & 3 & 2 \\ 4 & 3 & 1 & 1 \\ 1 & 2 & 3 & 0 \end{array}\right) \left( \begin{array}{l} x \\ y \\ z \\ w \end{array}\right) = \left( \begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right) \) | Of course one way is to write the augmented matrix and grind away. However, this involves more row operations than the computation of an \( {LU} \) factorization and it turns out that an \( {LU} \) factorization can give the solution quickly. Here is how. The following is an \( {LU} \) factorization for the matrix.\n\n... | Yes |
Example 5.4.1 Find a PLU factorization for the above matrix in (5.1). | Proceed as before trying to find the row echelon form of the matrix. First add -1 times the first row to the second row and then add -4 times the first to the third. This yields\n\n\[ \left( \begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 4 & 0 & 1 \end{matrix}\right) \left( \begin{matrix} 1 & 2 & 3 & 2 \\ 0 & 0 & 0 & - 2 \\ ... | Yes |
Use a PLU factorization of \( M \equiv \left( \begin{array}{llll} 1 & 2 & 3 & 2 \\ 1 & 2 & 3 & 0 \\ 4 & 3 & 1 & 1 \end{array}\right) \) to solve the system \( M\mathbf{x} = \mathbf{b} \) where \( \mathbf{b} = {\left( 1,2,3\right) }^{T} \). | Let \( U\mathbf{x} = \mathbf{y} \) and consider \( {PL}\mathbf{y} = \mathbf{b} \) . In other words, solve,\n\n\[ \left( \begin{array}{lll} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right) \left( \begin{array}{lll} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right) \left( \begin{array}{l} {y}_{1} \\ {y}_{2} \\... | Yes |
One of the most important examples of an orthogonal matrix is the so called Householder matrix. You have \( \mathbf{v} \) a unit vector and you form the matrix\n\n\[ I - 2\mathbf{v}{\mathbf{v}}^{T} \]\n\nThis is an orthogonal matrix which is also symmetric. | To see this, you use the rules of matrix operations.\n\n\[ {\left( I - 2\mathbf{v}{\mathbf{v}}^{T}\right) }^{T} = {I}^{T} - {\left( 2\mathbf{v}{\mathbf{v}}^{T}\right) }^{T} \]\n\n\[ = I - 2\mathbf{v}{\mathbf{v}}^{T} \] so it is symmetric. Now to show it is orthogonal,\n\n\[ \left( {I - 2{\mathbf{{vv}}}^{T}}\right) \lef... | Yes |
Given two vectors \( \mathbf{x},\mathbf{y} \) such that \( \left| \mathbf{x}\right| = \left| \mathbf{y}\right| \neq 0 \) but \( \mathbf{x} \neq \mathbf{y} \) and you want an orthogonal matrix \( Q \) such that \( Q\mathbf{x} = \mathbf{y} \) and \( Q\mathbf{y} = \mathbf{x} \). | The thing which works is the Householder matrix\n\[ Q \equiv I - 2\frac{\mathbf{x} - \mathbf{y}}{{\left| \mathbf{x} - \mathbf{y}\right| }^{2}}{\left( \mathbf{x} - \mathbf{y}\right) }^{T} \]\n\nHere is why this works.\n\n\[ Q\left( {\mathbf{x} - \mathbf{y}}\right) = \left( {\mathbf{x} - \mathbf{y}}\right) - 2\frac{\math... | Yes |
Consider \( z = {x}_{1} - {x}_{2} \) subject to the constraints, \( {x}_{1} + 2{x}_{2} \leq {10},{x}_{1} + 2{x}_{2} \geq 2 \) , and \( 2{x}_{1} + {x}_{2} \leq 6,{x}_{i} \geq 0 \) . Find a simplex tableau for a problem of the form \( \mathbf{x} \geq \mathbf{0}, A\mathbf{x} = \mathbf{b} \) which is equivalent to the abov... | You add in slack variables. These are positive variables, one for each of the first three constraints, which change the first three inequalities into equations. Thus the first three inequalities become \( {x}_{1} + 2{x}_{2} + {x}_{3} = {10},{x}_{1} + 2{x}_{2} - {x}_{4} = 2 \), and \( 2{x}_{1} + {x}_{2} + {x}_{5} = 6,{x... | Yes |
Example 6.2.3 Consider \( z = {x}_{1} - {x}_{2} \) subject to the constraints, \( {x}_{1} + 2{x}_{2} \leq {10},{x}_{1} + 2{x}_{2} \geq 2 \) , and \( 2{x}_{1} + {x}_{2} \leq 6,{x}_{i} \geq 0 \) . Find a simplex tableau. | Adding in slack variables, an augmented matrix which is descriptive of the constraints\n\nis\n\n\[ \left( \begin{matrix} 1 & 2 & 1 & 0 & 0 & {10} \\ 1 & 2 & 0 & - 1 & 0 & 6 \\ 2 & 1 & 0 & 0 & 1 & 6 \end{matrix}\right) \]\n\nThe obvious solution is not feasible because of that -1 in the fourth column. When you let \( {x... | Yes |
Maximize \( z = {x}_{1} - {x}_{2} \) subject to the constraints,\n\n\[ \n{x}_{1} + 2{x}_{2} \leq {10},{x}_{1} + 2{x}_{2} \geq 2 \n\]\n\nand \( 2{x}_{1} + {x}_{2} \leq 6,{x}_{i} \geq 0 \) . | Recall this is the same as maximizing \( z = {x}_{1} - {x}_{2} \) subject to\n\n\[ \n\left( \begin{matrix} 1 & 2 & 1 & 0 & 0 \\ 1 & 2 & 0 & - 1 & 0 \\ 2 & 1 & 0 & 0 & 1 \end{matrix}\right) \left( \begin{array}{l} {x}_{1} \\ {x}_{2} \\ {x}_{3} \\ {x}_{4} \\ {x}_{5} \end{array}\right) = \left( \begin{array}{l} {10} \\ 2 ... | Yes |
How many pounds of each feed per pig should the pig farmer use in order to minimize his cost? | His problem is to minimize \( C \equiv 2{x}_{1} + 3{x}_{2} + 2{x}_{3} + 3{x}_{4} \) subject to the constraints\n\n\[ \n{x}_{1} + 2{x}_{2} + {x}_{3} + 3{x}_{4} \geq 5 \]\n\n\[ \n5{x}_{1} + 3{x}_{2} + 2{x}_{3} + {x}_{4} \geq 8 \]\n\n\[ \n{x}_{1} + 2{x}_{2} + 2{x}_{3} + {x}_{4} \geq 6 \]\n\n\[ \n2{x}_{1} + {x}_{2} + {x}_{... | Yes |
Maximize \( z = {x}_{1} - 3{x}_{2} + {x}_{3} \) subject to the constraints \( {x}_{1} + {x}_{2} + {x}_{3} \leq \) \( {10},{x}_{1} + {x}_{2} + {x}_{3} \geq 2,{x}_{1} + {x}_{2} + 3{x}_{3} \leq 8 \) and \( {x}_{1} + 2{x}_{2} + {x}_{3} \leq 7 \) with all variables nonnegative. | The first part of it is the same. You wind up with the same simplex tableau,\n\n\[ \left( \begin{matrix} 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 8 \\ 1 & 1 & 1 & 0 & - 1 & 0 & 0 & 0 & 2 \\ - 2 & - 2 & 0 & 0 & 3 & 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 5 \\ 0 & 4 & 0 & 0 & - 1 & 0 & 0 & 1 & 2 \end{matrix}\right) \]\n\... | Yes |
Maximize \( {x}_{1} - {x}_{2} + 2{x}_{3} \) subject to the constraints, \( 2{x}_{1} + {x}_{2} - {x}_{3} \geq 3,{x}_{1} + {x}_{2} + {x}_{3} \geq 2,{x}_{1} + {x}_{2} + {x}_{3} \leq 7 \) and \( \mathbf{x} \geq \mathbf{0} \) . | From 6.20 you can immediately assemble an initial simplex tableau. You begin with the first 6 columns and top 3 rows in 6.20 . Then add in the column and row for \( z \) . This yields\n\n\[ \left( \begin{matrix} 1 & \frac{2}{3} & 0 & - \frac{1}{3} & - \frac{1}{3} & 0 & 0 & \frac{5}{3} \\ 0 & \frac{1}{3} & 1 & \frac{1}{... | Yes |
Lemma 6.5.1 Let \( \mathbf{x} \) be a solution of the inequalities of \( A \) .) and let \( \mathbf{y} \) be a solution of the inequalities of \( B \) .). Then\n\n\[ \mathbf{{cx}} \geq \mathbf{{yb}}\text{.} \]\n\nand if equality holds in the above, then \( \mathbf{x} \) is the solution to \( A \) .) and \( \mathbf{y} \... | Proof: This follows immediately. Since \( \mathbf{c} \geq \mathbf{y}A,\mathbf{{cx}} \geq \mathbf{y}A\mathbf{x} \geq \mathbf{{yb}} \) .\n\nIt follows from this lemma that if \( \mathbf{y} \) satisfies the inequalities of \( B \) .) and \( \mathbf{x} \) satisfies the inequalities of \( A \) .) then if equality holds in t... | Yes |
Theorem 6.5.2 Suppose there exists a solution \( \mathbf{x} \) to A.) where \( \mathbf{x} \) is a basic feasible solution of the inequalities of \( A \) .). Then there exists a solution \( \mathbf{y} \) to \( B \) .) and \( \mathbf{{cx}} = \mathbf{{by}} \) . It is also possible to find \( \mathbf{y} \) from \( \mathbf{... | Proof: Since the solution to \( A \) .) is basic and feasible, there exists a simplex tableau like 6.23 such that \( {\mathbf{x}}^{\prime } \) can be split into \( {\mathbf{x}}_{B} \) and \( {\mathbf{x}}_{F} \) such that \( {\mathbf{x}}_{F} = 0 \) and \( {\mathbf{x}}_{B} = {B}^{-1}\mathbf{b} \) . Now since it is a mini... | Yes |
Corollary 6.5.3 Suppose there exists a solution, \( \mathbf{y} \) to \( B \) .) where \( \mathbf{y} \) is a basic feasible solution of the inequalities of \( B \) .). Then there exists a solution, \( \mathbf{x} \) to \( A \) .) and \( \mathbf{{cx}} = \mathbf{{by}} \) . It is also possible to find \( \mathbf{x} \) from ... | In this case, and referring to (6.23), the simple formula is \( \mathbf{x} = {B}_{1}^{-T}{\mathbf{b}}_{{B}_{1}} \) . | Yes |
Example 7.1.4 Let\n\n\[ A = \left( \begin{matrix} 2 & 2 & - 2 \\ 1 & 3 & - 1 \\ - 1 & 1 & 1 \end{matrix}\right) \]\n\nFirst find the eigenvalues. | \[ \det \left( {\lambda \left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right) - \left( \begin{matrix} 2 & 2 & - 2 \\ 1 & 3 & - 1 \\ - 1 & 1 & 1 \end{matrix}\right) }\right) = 0 \]\n\nThis is \( {\lambda }^{3} - 6{\lambda }^{2} + {8\lambda } = 0 \) and the solutions are 0,2, and 4 .\n\n0 Can be a... | Yes |
Example 7.1.5 Let\n\n\[ A = \left( \begin{matrix} 2 & - 2 & - 1 \\ - 2 & - 1 & - 2 \\ {14} & {25} & {14} \end{matrix}\right) \]\n\nFind the eigenvectors and eigenvalues. | In this case the eigenvalues are \( 3,6,6 \) where I have listed 6 twice because it is a zero of algebraic multiplicity two, the characteristic equation being\n\n\[ \left( {\lambda - 3}\right) {\left( \lambda - 6\right) }^{2} = 0.\]\n\nIt remains to find the eigenvectors for these eigenvalues. First consider the eigenv... | Yes |
Theorem 7.1.7 Suppose \( M{\mathbf{v}}_{i} = {\lambda }_{i}{\mathbf{v}}_{i}, i = 1,\cdots, r,{\mathbf{v}}_{i} \neq 0 \), and that if \( i \neq j \), then \( {\lambda }_{i} \neq {\lambda }_{j} \) . Then the set of eigenvectors, \( \left\{ {{\mathbf{v}}_{1},\cdots ,{\mathbf{v}}_{r}}\right\} \) is linearly independent. | Proof. Suppose the claim of the lemma is not true. Then there exists a subset of this set of vectors\n\n\[ \left\{ {{\mathbf{w}}_{1},\cdots ,{\mathbf{w}}_{r}}\right\} \subseteq \left\{ {{\mathbf{v}}_{1},\cdots ,{\mathbf{v}}_{k}}\right\} \]\n\nsuch that\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{r}{c}_{j}{\mathbf{w}}_{j} = ... | Yes |
Find the eigenvalues and eigenvectors of the matrix\n\n\[ A = \left( \begin{matrix} 1 & 0 & 0 \\ 0 & 2 & - 1 \\ 0 & 1 & 2 \end{matrix}\right) \] | You need to find the eigenvalues. Solve\n\n\[ \det \left( {\lambda \left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right) - \left( \begin{matrix} 1 & 0 & 0 \\ 0 & 2 & - 1 \\ 0 & 1 & 2 \end{matrix}\right) }\right) = 0. \]\n\nThis reduces to \( \left( {\lambda - 1}\right) \left( {{\lambda }^{2} - {... | No |
Find the principle directions determined by the matrix\n\n\\[ \n\\left( \\begin{array}{lll} \\frac{29}{11} & \\frac{6}{11} & \\frac{6}{11} \\\\ \\frac{6}{11} & \\frac{41}{44} & \\frac{19}{44} \\\\ \\frac{6}{11} & \\frac{19}{44} & \\frac{41}{44} \\end{array}\\right) \n\\] | The eigenvalues are \\( 3,1 \\), and \\( \\frac{1}{2} \\) .\n\nIt is nice to be given the eigenvalues. The largest eigenvalue is 3 which means that in the direction determined by the eigenvector associated with 3 the stretch is three times as large. The smallest eigenvalue is \\( 1/2 \\) and so in the direction determi... | No |
Example 7.2.2 Find oscillatory solutions to the system of differential equations, \( {\mathbf{x}}^{\prime \prime } = A\mathbf{x} \)\n\nwhere\n\[ A = \left( \begin{matrix} - \frac{5}{3} & - \frac{1}{3} & - \frac{1}{3} \\ - \frac{1}{3} & - \frac{13}{6} & \frac{5}{6} \\ - \frac{1}{3} & \frac{5}{6} & - \frac{13}{6} \end{ma... | The eigenvalues are \( - 1, - 2 \), and -3 .\n\nAccording to the above, you can find solutions by looking for the eigenvectors. Consider the eigenvectors for -3 . The augmented matrix for finding the eigenvectors is\n\n\[ \left( \begin{matrix} - \frac{4}{3} & \frac{1}{3} & \frac{1}{3} & 0 \\ \frac{1}{3} & - \frac{5}{6}... | Yes |
Lemma 7.4.1 Let \( \left\{ {{\mathbf{x}}_{1},\cdots ,{\mathbf{x}}_{n}}\right\} \) be a basis for \( {\mathbb{F}}^{n} \) . Then there exists an orthonormal basis for \( {\mathbb{F}}^{n},\left\{ {{\mathbf{u}}_{1},\cdots ,{\mathbf{u}}_{n}}\right\} \) which has the property that for each \( k \leq n \), span \( \left( {{\m... | Proof: Let \( \left\{ {{\mathbf{x}}_{1},\cdots ,{\mathbf{x}}_{n}}\right\} \) be a basis for \( {\mathbb{F}}^{n} \) . Let \( {\mathbf{u}}_{1} \equiv {\mathbf{x}}_{1}/\left| {\mathbf{x}}_{1}\right| \) . Thus for \( k = 1 \) , \( \operatorname{span}\left( {\mathbf{u}}_{1}\right) = \operatorname{span}\left( {\mathbf{x}}_{1... | Yes |
Proposition 7.4.3 An \( n \times n \) matrix is unitary if and only if the columns are an orthonormal set. | Proof: This follows right away from the way we multiply matrices. If \( U \) is an \( n \times n \) complex matrix, then\n\n\[{\left( {U}^{ * }U\right) }_{ij} = {\mathbf{u}}_{i}^{ * }{\mathbf{u}}_{j} = \overline{\left( {\mathbf{u}}_{i},{\mathbf{u}}_{j}\right) }\n\]\n\nand the matrix is unitary if and only if this equal... | Yes |
Theorem 7.4.4 Let \( A \) be an \( n \times n \) matrix. Then there exists a unitary matrix \( U \) such that\n\n\[ {U}^{ * }{AU} = T \]\n\n(7.11)\n\nwhere \( T \) is an upper triangular matrix having the eigenvalues of \( A \) on the main diagonal listed according to multiplicity as roots of the characteristic equatio... | Proof: The theorem is clearly true if \( A \) is a \( 1 \times 1 \) matrix. Just let \( U = 1 \) the \( 1 \times 1 \) matrix which has 1 down the main diagonal and zeros elsewhere. Suppose it is true for \( \left( {n - 1}\right) \times \left( {n - 1}\right) \) matrices and let \( A \) be an \( n \times n \) matrix. The... | Yes |
Lemma 7.4.5 Let \( A \) be of the form\n\n\[ A = \left( \begin{matrix} {P}_{1} & \cdots & * \\ \vdots & \ddots & \vdots \\ 0 & \cdots & {P}_{s} \end{matrix}\right) \]\n\nwhere \( {P}_{k} \) is an \( {m}_{k} \times {m}_{k} \) matrix. Then\n\n\[ \det \left( A\right) = \mathop{\prod }\limits_{k}\det \left( {P}_{k}\right) ... | Proof: Let \( {U}_{k} \) be an \( {m}_{k} \times {m}_{k} \) unitary matrix such that\n\n\[ {U}_{k}^{ * }{P}_{k}{U}_{k} = {T}_{k} \]\n\nwhere \( {T}_{k} \) is upper triangular. Then it follows that for\n\n\[ U \equiv \left( \begin{matrix} {U}_{1} & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & {U}_{s} \end{matr... | Yes |
Corollary 7.4.7 Let \( A \) be a real \( n \times n \) matrix having only real eigenvalues. Then there exists a real orthogonal matrix \( Q \) and an upper triangular matrix \( T \) such that\n\n\[ {Q}^{T}{AQ} = T \]\n\nand furthermore, if the eigenvalues of \( A \) are listed in decreasing order,\n\n\[ {\lambda }_{1} ... | Proof: Most of this follows right away from Theorem 7.4.6. It remains to verify the claim that the diagonal entries can be arranged in the desired order. However, this follows from a simple modification of the above argument. When you find \( {\mathbf{v}}_{1} \) the eigenvalue of \( {\lambda }_{1} \) , just be sure \( ... | Yes |
Lemma 7.4.10 If \( T \) is upper triangular and normal, then \( T \) is a diagonal matrix. | Proof:This is obviously true if \( T \) is \( 1 \times 1 \) . In fact, it can’t help being diagonal in this case. Suppose then that the lemma is true for \( \left( {n - 1}\right) \times \left( {n - 1}\right) \) matrices and let \( T \) be an upper triangular normal \( n \times n \) matrix. Thus \( T \) is of the form\n... | Yes |
Theorem 7.4.11 Let \( A \) be a normal matrix. Then there exists a unitary matrix \( U \) such that \( {U}^{ * }{AU} \) is a diagonal matrix. | Proof: From Theorem 7.4.4 there exists a unitary matrix \( U \) such that \( {U}^{ * }{AU} \) equals an upper triangular matrix. The theorem is now proved if it is shown that the property of being normal is preserved under unitary similarity transformations. That is, verify that if \( A \) is normal and if \( B = {U}^{... | Yes |
Corollary 7.4.12 If \( A \) is Hermitian, then all the eigenvalues of \( A \) are real and there exists an orthonormal basis of eigenvectors. | Proof: Since \( A \) is normal, there exists unitary, \( U \) such that \( {U}^{ * }{AU} = D \), a diagonal matrix whose diagonal entries are the eigenvalues of \( A \) . Therefore, \( {D}^{ * } = {U}^{ * }{A}^{ * }U = {U}^{ * }{AU} = \) \( D \) showing \( D \) is real.\n\nFinally, let\n\n\[ U = \left( \begin{array}{ll... | Yes |
Corollary 7.4.13 If \( A \) is a real symmetric matrix, then \( A \) is Hermitian and there exists a real unitary matrix \( U \) such that \( {U}^{T}{AU} = D \) where \( D \) is a diagonal matrix whose diagonal entries are the eigenvalues of \( A \) . By arranging the columns of \( U \) the diagonal entries of \( D \) ... | Proof: This follows from Theorem 7.4.6 and Corollary 7.4.12. Let\n\n\[ U = \left( \begin{array}{lll} {\mathbf{u}}_{1} & \cdots & {\mathbf{u}}_{n} \end{array}\right) \]\n\nThen \( {AU} = {UD} \) so\n\n\[ {AU} = \left( \begin{array}{lll} A{\mathbf{u}}_{1} & \cdots & A{\mathbf{u}}_{n} \end{array}\right) = \left( \begin{ar... | Yes |
Theorem 7.5.2 Let \( A \) be an \( n \times n \) matrix. Then \( \operatorname{trace}\left( A\right) \) equals the sum of the eigenvalues of \( A \) and \( \det \left( A\right) \) equals the product of the eigenvalues of \( A \) . | This is proved using Schur's theorem and is in Problem 12 below. | No |
Theorem 7.5.3 Let \( A \) be an \( m \times n \) matrix and let \( B \) be an \( n \times m \) matrix. Then\n\n\[ \operatorname{trace}\left( {AB}\right) = \operatorname{trace}\left( {BA}\right) . \] | \[ \operatorname{trace}\left( {AB}\right) \equiv \mathop{\sum }\limits_{i}\left( {\mathop{\sum }\limits_{k}{A}_{ik}{B}_{ki}}\right) = \mathop{\sum }\limits_{k}\mathop{\sum }\limits_{i}{B}_{ki}{A}_{ik} = \operatorname{trace}\left( {BA}\right) \] | Yes |
Theorem 7.7.1 Suppose \( f : U \subseteq {\mathbb{F}}^{2} \rightarrow \mathbb{R} \) where \( U \) is an open set on which \( {f}_{x},{f}_{y},{f}_{xy} \) and \( {f}_{yx} \) exist. Then if \( {f}_{xy} \) and \( {f}_{yx} \) are continuous at the point \( \left( {x, y}\right) \in U \), it follows\n\n\[ \n{f}_{xy}\left( {x,... | Proof: Since \( U \) is open, there exists \( r > 0 \) such that \( B\left( {\left( {x, y}\right), r}\right) \subseteq U \) . Now let \( \left| t\right| ,\left| s\right| < \) \( r/2, t, s \) real numbers and consider\n\n\[ \n\Delta \left( {s, t}\right) \equiv \frac{1}{st}\{ \overset{h\left( t\right) }{\overbrace{f\left... | Yes |
Corollary 7.7.2 Suppose \( U \) is an open subset of \( {\mathbb{F}}^{n} \) and \( f : U \rightarrow \mathbb{R} \) has the property that for two indices, \( k, l,{f}_{{x}_{k}},{f}_{{x}_{l}},{f}_{{x}_{l}{x}_{k}} \), and \( {f}_{{x}_{k}{x}_{l}} \) exist on \( U \) and \( {f}_{{x}_{k}{x}_{l}} \) and \( {f}_{{x}_{l}{x}_{k}... | Thus the theorem asserts that the mixed partial derivatives are equal at \( \mathbf{x} \) if they are defined near \( \mathbf{x} \) and continuous at \( \mathbf{x} \) . | Yes |
Theorem 7.7.3 Suppose \( f \) has \( n + 1 \) derivatives on an interval, \( \left( {a, b}\right) \) and let \( c \in \left( {a, b}\right) \) . Then if \( x \in \left( {a, b}\right) \), there exists \( \xi \) between \( c \) and \( x \) such that\n\n\[ f\left( x\right) = f\left( c\right) + \mathop{\sum }\limits_{{k = 1... | Proof: If \( n = 0 \) then the theorem is true because it is just the mean value theorem. Suppose the theorem is true for \( n - 1, n \geq 1 \) . It can be assumed \( x \neq c \) because if \( x = c \) there is nothing to show. Then there exists \( K \) such that\n\n\[ f\left( x\right) - \left( {f\left( c\right) + \mat... | Yes |
Theorem 7.7.5 Let \( f : U \rightarrow \mathbb{R} \) and let \( f \in {C}^{2}\left( U\right) \) . Then if\n\n\[ B\left( {\mathbf{x}, r}\right) \subseteq U \]\n\nand \( \parallel \mathbf{v}\parallel < r \), there exists \( t \in \left( {0,1}\right) \) such that.\n\n\[ f\left( {\mathbf{x} + \mathbf{v}}\right) = f\left( \... | Definition 7.7.6 Define the following matrix.\n\n\[ {H}_{ij}\left( {\mathbf{x} + t\mathbf{v}}\right) \equiv \frac{{\partial }^{2}f\left( {\mathbf{x} + t\mathbf{v}}\right) }{\partial {x}_{j}\partial {x}_{i}}. \]\n\nIt is called the Hessian matrix. From Corollary 7.7.2, this is a symmetric matrix. Then in terms of this m... | No |
Theorem 7.7.7 In the above situation, suppose \( {f}_{{x}_{j}}\left( \mathbf{x}\right) = 0 \) for each \( {x}_{j} \) . Then if \( H\left( \mathbf{x}\right) \) has all positive eigenvalues, \( \mathbf{x} \) is a local minimum for \( \dot{f} \) . If \( H\left( \mathbf{x}\right) \) has all negative eigenvalues, then \( \m... | Proof: Since \( {f}_{{x}_{j}}\left( \mathbf{x}\right) = 0 \) for each \( {x}_{j} \), formula (7.19) implies\n\n\[ f\left( {\mathbf{x} + \mathbf{v}}\right) = f\left( \mathbf{x}\right) + \frac{1}{2}{\mathbf{v}}^{T}H\left( \mathbf{x}\right) \mathbf{v} + \frac{1}{2}\left( {{\mathbf{v}}^{T}\left( {H\left( {\mathbf{x} + t\ma... | Yes |
Theorem 7.8.1 Let \( A \) be an \( n \times n \) matrix. Consider the \( n \) Gerschgorin discs defined as\n\n\[ \n{D}_{i} \equiv \left\{ {\lambda \in \mathbb{C} : \left| {\lambda - {a}_{ii}}\right| \leq \mathop{\sum }\limits_{{j \neq i}}\left| {a}_{ij}\right| }\right\} .\n\]\n\nThen every eigenvalue is contained in so... | Proof: Suppose \( A\mathbf{x} = \lambda \mathbf{x} \) where \( \mathbf{x} \neq \mathbf{0} \) . Then for \( A = \left( {a}_{ij}\right) \)\n\n\[ \n\mathop{\sum }\limits_{{j \neq i}}{a}_{ij}{x}_{j} = \left( {\lambda - {a}_{ii}}\right) {x}_{i}\n\]\n\nTherefore, picking \( k \) such that \( \left| {x}_{k}\right| \geq \left|... | Yes |
Example 7.8.2 Here is a matrix. Estimate its eigenvalues. | According to Gerschgorin's theorem the eigenvalues are contained in the disks \[ {D}_{1} = \{ \lambda \in \mathbb{C} : \left| {\lambda - 2}\right| \leq 2\} ,{D}_{2} = \{ \lambda \in \mathbb{C} : \left| {\lambda - 5}\right| \leq 3\} ,\] \[ {D}_{3} = \{ \lambda \in \mathbb{C} : \left| {\lambda - 9}\right| \leq 1\} \] It ... | Yes |
Theorem 7.9.1 Let \( U \) be a region and let \( \gamma : \left\lbrack {a, b}\right\rbrack \rightarrow U \) be closed, continuous, bounded variation, and the winding number, \( n\left( {\gamma, z}\right) = 0 \) for all \( z \notin U \) . Suppose also that \( f \) is analytic on \( U \) having zeros \( {a}_{1},\cdots ,{... | Proof: It is given that \( f\left( z\right) = \mathop{\prod }\limits_{{j = 1}}^{m}\left( {z - {a}_{j}}\right) g\left( z\right) \) where \( g\left( z\right) \neq 0 \) on \( U \) . Hence using the product rule,\n\n\[ \n\frac{{f}^{\prime }\left( z\right) }{f\left( z\right) } = \mathop{\sum }\limits_{{j = 1}}^{m}\frac{1}{z... | Yes |
Theorem 7.9.4 Suppose \( A\left( t\right) \) is an \( n \times n \) matrix and that \( t \rightarrow A\left( t\right) \) is continuous for \( t \in \left\lbrack {0,1}\right\rbrack \) . Let \( \lambda \left( 0\right) \in \sigma \left( {A\left( 0\right) }\right) \) and define \( \sum \equiv { \cup }_{t \in \left\lbrack {... | Proof: Let \( S \equiv \left\{ {t \in \left\lbrack {0,1}\right\rbrack : {K}_{0} \cap \sigma \left( {A\left( s\right) }\right) \neq \varnothing }\right. \) for all \( \left. {s \in \left\lbrack {0, t}\right\rbrack }\right\} \) . Then \( 0 \in S \) . Let \( {t}_{0} = \) \( \sup \left( S\right) \) . Say \( \sigma \left( {... | Yes |
Corollary 7.9.5 Suppose one of the Gerschgorin discs, \( {D}_{i} \) is disjoint from the union of the others. Then \( {D}_{i} \) contains an eigenvalue of \( A \) . Also, if there are \( n \) disjoint Gerschgorin discs, then each one contains an eigenvalue of \( A \) . | Proof: Denote by \( A\left( t\right) \) the matrix \( \left( {a}_{ij}^{t}\right) \) where if \( i \neq j,{a}_{ij}^{t} = t{a}_{ij} \) and \( {a}_{ii}^{t} = {a}_{ii} \) . Thus to get \( A\left( t\right) \) multiply all non diagonal terms by \( t \) . Let \( t \in \left\lbrack {0,1}\right\rbrack \) . Then \( A\left( 0\rig... | Yes |
Corollary 7.9.7 Suppose one of the Gerschgorin discs, \( {D}_{i} \) is disjoint from the union of the others. Then \( {D}_{i} \) contains exactly one eigenvalue of \( A \) and this eigenvalue is a simple root to the characteristic polynomial of \( A \) . | Proof: In the proof of Corollary 7.9.5, note that \( {a}_{ii} \) is a simple root of \( A\left( 0\right) \) since otherwise the \( {i}^{th} \) Gerschgorin disc would not be disjoint from the others. Also, \( K \), the connected component determined by \( {a}_{ii} \) must be contained in \( {D}_{i} \) because it is conn... | Yes |
Example 7.9.8 Consider the matrix \[ \left( \begin{array}{lll} 5 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 0 \end{array}\right) \] The Gerschgorin discs are \( D\left( {5,1}\right), D\left( {1,2}\right) \), and \( D\left( {0,1}\right) \). Observe \( D\left( {5,1}\right) \) is disjoint from the other discs. Therefore, there shoul... | The actual eigenvalues are not easy to find. They are the roots of the characteristic equation, \( {t}^{3} - 6{t}^{2} + \) \( {3t} + 5 = 0 \). The numerical values of these are \( - {.66966},{1.4231} \), and 5.24655, verifying the predictions of Gerschgorin's theorem. | Yes |
Theorem 7.10.1 Suppose \( \Phi \left( t\right) \) is an \( n \times n \) matrix which satisfies \( {\Phi }^{\prime }\left( t\right) = {A\Phi }\left( t\right) \) . Then the general solution to (7.24) is \( \Phi \left( t\right) \mathbf{c} \) if and only if \( \Phi {\left( t\right) }^{-1} \) exists for some \( t \) . Furt... | Hint: Suppose first the general solution is of the form \( \Phi \left( t\right) \mathbf{c} \) where \( \mathbf{c} \) is an arbitrary constant vector in \( {\mathbb{F}}^{n} \) . You need to verify \( \Phi {\left( t\right) }^{-1} \) exists for some \( t \) . In fact, show \( \Phi {\left( t\right) }^{-1} \) exists for eve... | No |
Corollary 8.2.5 If \( \left\{ {{\mathbf{u}}_{1},\cdots ,{\mathbf{u}}_{m}}\right\} \) and \( \left\{ {{\mathbf{v}}_{1},\cdots ,{\mathbf{v}}_{n}}\right\} \) are two bases for \( V \), then \( m = n \) . | Proof: By Theorem 8.2.4, \( m \leq n \) and \( n \leq m \) . ∎ | Yes |
Example 8.2.7 Consider the polynomials defined on \( \mathbb{R} \) of degree no more than 3, denoted here as \( {P}_{3} \) . Then show that a basis for \( {P}_{3} \) is \( \left\{ {1, x,{x}^{2},{x}^{3}}\right\} \) . Here \( {x}^{k} \) symbolizes the function \( x \mapsto {x}^{k} \) . | It is obvious that the span of the given vectors yields \( {P}_{3} \) . Why is this set of vectors linearly independent? Suppose\n\n\[ \n{c}_{0} + {c}_{1}x + {c}_{2}{x}^{2} + {c}_{3}{x}^{3} = 0 \n\]\n\nwhere 0 is the zero function which maps everything to 0 . Then you could differentiate three times and obtain the foll... | Yes |
Theorem 8.2.9 Let \( \\left\\{ {{f}_{1},\\cdots ,{f}_{n}}\\right\\} \) be smooth functions defined on \( \\left\\lbrack {a, b}\\right\\rbrack \) . Then they are linearly independent if there exists some point \( t \\in \\left\\lbrack {a, b}\\right\\rbrack \) where \( W\\left( {{f}_{1},\\cdots ,{f}_{n}}\\right\\) \\left... | Proof: Form the linear combination of these vectors (functions) and suppose it equals 0 . Thus\n\n\[ \n{a}_{1}{f}_{1} + {a}_{2}{f}_{2} + \\cdots + {a}_{n}{f}_{n} = 0 \n\]\n\nThe question you must answer is whether this requires each \( {a}_{j} \) to equal zero. If they all must equal 0 , then this means these vectors (... | Yes |
Lemma 8.2.10 Suppose \( \mathbf{v} \notin \operatorname{span}\left( {{\mathbf{u}}_{1},\cdots ,{\mathbf{u}}_{k}}\right) \) and \( \left\{ {{\mathbf{u}}_{1},\cdots ,{\mathbf{u}}_{k}}\right\} \) is linearly independent. Then \( \left\{ {{\mathbf{u}}_{1},\cdots ,{\mathbf{u}}_{k},\mathbf{v}}\right\} \) is also linearly inde... | Proof: Suppose \( \mathop{\sum }\limits_{{i = 1}}^{k}{c}_{i}{\mathbf{u}}_{i} + d\mathbf{v} = 0 \) . It is required to verify that each \( {c}_{i} = 0 \) and that \( d = 0 \) . But if \( d \neq 0 \), then you can solve for \( \mathbf{v} \) as a linear combination of the vectors, \( \left\{ {{\mathbf{u}}_{1},\cdots ,{\ma... | Yes |
If \( V = \operatorname{span}\left( {{\mathbf{u}}_{1},\cdots ,{\mathbf{u}}_{n}}\right) \) then some subset of \( \left\{ {{\mathbf{u}}_{1},\cdots ,{\mathbf{u}}_{n}}\right\} \) is a basis for V. | Let\n\n\[ S = \left\{ {E \subseteq \left\{ {{\mathbf{u}}_{1},\cdots ,{\mathbf{u}}_{n}}\right\} \text{ such that }\operatorname{span}\left( E\right) = V}\right\} .\n\]\n\nFor \( E \in S \), let \( \left| E\right| \) denote the number of elements of \( E \) . Let\n\n\[ m \equiv \min \{ \left| E\right| \text{ such that }E... | Yes |
Theorem 8.2.12 Let \( V \) be a nonzero subspace of a finite dimensional vector space, \( W \) of dimension, \( n \) . Then \( V \) has a basis with no more than \( n \) vectors. | Proof: Let \( {\mathbf{v}}_{1} \in V \) where \( {\mathbf{v}}_{1} \neq 0 \) . If \( \operatorname{span}\left\{ {\mathbf{v}}_{1}\right\} = V \), stop. \( \left\{ {\mathbf{v}}_{1}\right\} \) is a basis for \( V \) . Otherwise, there exists \( {\mathbf{v}}_{2} \in V \) which is not in span \( \left\{ {\mathbf{v}}_{1}\righ... | Yes |
Lemma 8.3.3 Let \( f\left( \lambda \right) \) and \( g\left( \lambda \right) \neq 0 \) be polynomials. Then there exists a polynomial, \( q\left( \lambda \right) \) such that\n\n\[ f\left( \lambda \right) = q\left( \lambda \right) g\left( \lambda \right) + r\left( \lambda \right) \]\n\nwhere the degree of \( r\left( \l... | Proof: Consider the polynomials of the form \( f\left( \lambda \right) - g\left( \lambda \right) l\left( \lambda \right) \) and out of all these polynomials, pick one which has the smallest degree. This can be done because of the well ordering of the natural numbers. Let this take place when \( l\left( \lambda \right) ... | Yes |
Proposition 8.3.5 The greatest common divisor is unique. | Proof: Suppose both \( q\left( \lambda \right) \) and \( {q}^{\prime }\left( \lambda \right) \) work. Then \( q\left( \lambda \right) \) divides \( {q}^{\prime }\left( \lambda \right) \) and the other way around and so\n\n\[ \n{q}^{\prime }\left( \lambda \right) = q\left( \lambda \right) l\left( \lambda \right), q\left... | Yes |
Theorem 8.3.6 Let \( \psi \left( \lambda \right) \) be the greatest common divisor of \( \left\{ {{\phi }_{i}\left( \lambda \right) }\right\} \), not all of which are zero polynomials. Then there exist polynomials \( {r}_{i}\left( \lambda \right) \) such that\n\n\[ \psi \left( \lambda \right) = \mathop{\sum }\limits_{{... | Proof: Let \( S \) denote the set of monic polynomials which are of the form\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{p}{r}_{i}\left( \lambda \right) {\phi }_{i}\left( \lambda \right) \]\n\nwhere \( {r}_{i}\left( \lambda \right) \) is a polynomial. Then \( S \neq \varnothing \) because some \( {\phi }_{i}\left( \lambda \... | Yes |
Lemma 8.3.7 Suppose \( \phi \left( \lambda \right) \) and \( \psi \left( \lambda \right) \) are monic polynomials which are irreducible and not equal. Then they are relatively prime. | Proof: Suppose \( \eta \left( \lambda \right) \) is a nonconstant polynomial. If \( \eta \left( \lambda \right) \) divides \( \phi \left( \lambda \right) \), then since \( \phi \left( \lambda \right) \) is irreducible, \( \eta \left( \lambda \right) \) equals \( {a\phi }\left( \lambda \right) \) for some \( a \in \math... | Yes |
Lemma 8.3.8 Let \( \psi \left( \lambda \right) \) be an irreducible monic polynomial not equal to 1 which divides \[ \mathop{\prod }\limits_{{i = 1}}^{p}{\phi }_{i}{\left( \lambda \right) }^{{k}_{i}},{k}_{i}\text{ a positive integer,} \] where each \( {\phi }_{i}\left( \lambda \right) \) is an irreducible monic polynom... | Proof : Suppose \( \psi \left( \lambda \right) \neq {\phi }_{i}\left( \lambda \right) \) for all \( i \) . Then by Lemma 8.3.7, there exist polynomials \( {m}_{i}\left( \lambda \right) ,{n}_{i}\left( \lambda \right) \) such that \[ 1 = \psi \left( \lambda \right) {m}_{i}\left( \lambda \right) + {\phi }_{i}\left( \lambd... | Yes |
Lemma 8.3.9 Suppose \( p\left( \lambda \right) \) is a monic polynomial and \( q\left( \lambda \right) \) is a polynomial such that\n\n\[ p\left( \lambda \right) q\left( \lambda \right) = 0. \]\n\nThen \( q\left( \lambda \right) = 0 \) . Also if\n\n\[ p\left( \lambda \right) {q}_{1}\left( \lambda \right) = p\left( \lam... | Proof: Let\n\n\[ p\left( \lambda \right) = \mathop{\sum }\limits_{{j = 1}}^{k}{p}_{j}{\lambda }^{j}, q\left( \lambda \right) = \mathop{\sum }\limits_{{i = 1}}^{n}{q}_{i}{\lambda }^{i},{p}_{k} = 1. \]\n\nThen the product equals\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{k}\mathop{\sum }\limits_{{i = 1}}^{n}{p}_{j}{q}_{i}{\l... | Yes |
Theorem 8.3.10 Let \( f\left( \lambda \right) \) be a nonconstant polynomial with coefficients in \( \mathbb{F} \) . Then there is some \( a \in \mathbb{F} \) such that \( f\left( \lambda \right) = a\mathop{\prod }\limits_{{i = 1}}^{n}{\phi }_{i}\left( \lambda \right) \) where \( {\phi }_{i}\left( \lambda \right) \) is... | Proof: That such a factorization exists is obvious. If \( f\left( \lambda \right) \) is irreducible, you are done. Factor out the leading coefficient. If not, then \( f\left( \lambda \right) = a{\phi }_{1}\left( \lambda \right) {\phi }_{2}\left( \lambda \right) \) where these are monic polynomials. Continue doing this ... | Yes |
Corollary 8.3.11 Let \( q\left( \lambda \right) = \mathop{\prod }\limits_{{i = 1}}^{p}{\phi }_{i}{\left( \lambda \right) }^{{k}_{i}} \) where the \( {k}_{i} \) are positive integers and the \( {\phi }_{i}\left( \lambda \right) \) are irreducible monic polynomials. Suppose also that \( p\left( \lambda \right) \) is a mo... | Proof: Using Theorem 8.3.10, let \( p\left( \lambda \right) = b\mathop{\prod }\limits_{{i = 1}}^{s}{\psi }_{i}{\left( \lambda \right) }^{{r}_{i}} \) where the \( {\psi }_{i}\left( \lambda \right) \) are each irreducible and monic and \( b \in \mathbb{F} \) . Since \( p\left( \lambda \right) \) is monic, \( b = 1 \) . T... | Yes |
Proposition 8.3.16 In the above definition, \( \sim \) is an equivalence relation. | Proof: First of all, note that \( a\left( x\right) \sim a\left( x\right) \) because their difference equals \( {0p}\left( x\right) \) . If \( a\left( x\right) \sim b\left( x\right) \), then \( a\left( x\right) - b\left( x\right) = k\left( x\right) p\left( x\right) \) for some \( k\left( x\right) \) . But then \( b\left... | Yes |
Proposition 8.3.18 In the situation of Definition 8.3.17, \( p\left( x\right) \) and \( q\left( x\right) \) are relatively prime for any \( q\left( x\right) \in \mathbb{F}\left\lbrack x\right\rbrack \) which is not a multiple of \( p\left( x\right) \) . Also the definitions of addition and multiplication are well defin... | Proof: First consider the claim about \( p\left( x\right), q\left( x\right) \) being relatively prime. If \( \psi \left( x\right) \) is the greatest common divisor, it follows \( \psi \left( x\right) \) is either equal to \( p\left( x\right) \) or 1 . If it is \( p\left( x\right) \), then \( q\left( x\right) \) is a mu... | Yes |
Proposition 8.3.21 Let \( F \subseteq K \subseteq L \) be fields. Then \( \left\lbrack {L : F}\right\rbrack = \left\lbrack {L : K}\right\rbrack \left\lbrack {K : F}\right\rbrack \) . | Proof: Let \( {\left\{ {l}_{i}\right\} }_{i = 1}^{n} \) be a basis for \( L \) over \( K \) and let \( {\left\{ {k}_{j}\right\} }_{j = 1}^{m} \) be a basis of \( K \) over \( F \) . Then if \( l \in L \), there exist unique scalars \( {x}_{i} \) in \( K \) such that\n\n\[ l = \mathop{\sum }\limits_{{i = 1}}^{n}{x}_{i}{... | Yes |
The set of all equivalence classes \( \mathbb{G} \equiv \mathbb{F}/\left( {p\left( x\right) }\right) \) described above with the multiplicative identity given by [1] and the additive identity given by [0] along with the operations of Definition 8.3.17, is a field and \( p\left( \left\lbrack x\right\rbrack \right) = \le... | Proof: Everything is obvious except for the existence of the multiplicative inverse and the assertion that \( p\left( \left\lbrack x\right\rbrack \right) = 0 \) . Suppose then that \( \left\lbrack {a\left( x\right) }\right\rbrack \neq \left\lbrack 0\right\rbrack \) . That is, \( a\left( x\right) \) is not a multiple of... | Yes |
Theorem 8.3.23 Let \( p\left( x\right) = {x}^{n} + {a}_{n - 1}{x}^{n - 1} + \cdots + {a}_{1}x + {a}_{0} \) be a polynomial with coefficients in a field of scalars \( \mathbb{F} \) . There exists a larger field \( \mathbb{G} \) such that there exist \( \left\{ {{z}_{1},\cdots ,{z}_{n}}\right\} \) listed according to mul... | Proof: From Theorem 8.3.22, there exists a field \( {\mathbb{F}}_{1} \) such that \( p\left( x\right) \) has a root, \( {z}_{1}( = \left\lbrack x\right\rbrack \) if \( p \) is irreducible.) Then by the Euclidean algorithm\n\n\[ p\left( x\right) = \left( {x - {z}_{1}}\right) {q}_{1}\left( x\right) + r \]\n\nwhere \( r \... | Yes |
The polynomial \( {x}^{2} + 1 \) is irreducible in \( \mathbb{R}\left( x\right) \), polynomials having real coefficients. | To see this is the case, suppose \( \psi \left( x\right) \) divides \( {x}^{2} + 1 \) . Then\n\n\[ \n{x}^{2} + 1 = \psi \left( x\right) q\left( x\right) \n\]\n\nIf the degree of \( \psi \left( x\right) \) is less than 2, then it must be either a constant or of the form \( {ax} + b \) . In the latter case, \( - b/a \) m... | Yes |
Proposition 8.3.25 Suppose \( p\left( x\right) \in \mathbb{F}\left\lbrack x\right\rbrack \) is irreducible and has degree \( n \) . Then every element of \( \mathbb{G} = \mathbb{F}\left\lbrack x\right\rbrack /\left( {p\left( x\right) }\right) \) is of the form \( \left\lbrack 0\right\rbrack \) or \( \left\lbrack {r\lef... | Proof: This follows right away from the Euclidean algorithm for polynomials. If \( k\left( x\right) \) has degree larger than \( n - 1 \), then\n\n\[ k\left( x\right) = q\left( x\right) p\left( x\right) + r\left( x\right) \]\n\nwhere \( r\left( x\right) \) is either equal to 0 or has degree less than \( n \) . Hence\n\... | Yes |
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