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Corollary 5.1.5 Let \( \\left\\{ {{\\mathbf{x}}_{1},\\cdots ,{\\mathbf{x}}_{r}}\\right\\} \) and \( \\left\\{ {{\\mathbf{y}}_{1},\\cdots ,{\\mathbf{y}}_{s}}\\right\\} \) be two bases \( {}^{1} \) of \( {\\mathbb{F}}^{n} \) . Then \( r = s = n \) . | Proof: From the exchange theorem, \( r \\leq s \) and \( s \\leq r \) . Now note the vectors,\n\n\[ \n{\\mathbf{e}}_{i} = \\overset{1\\text{ is in the }{i}^{th}\\text{ slot }}{\\overbrace{\\left( 0,\\cdots ,0,1,0\\cdots ,0\\right) }} \n\]\n\nfor \( i = 1,2,\\cdots, n \) are a basis for \( {\\mathbb{F}}^{n} \) . This pr... | Yes |
Lemma 5.1.6 Let \( \left\{ {{\mathbf{v}}_{1},\cdots ,{\mathbf{v}}_{r}}\right\} \) be a set of vectors. Then \( V \equiv \operatorname{span}\left( {{\mathbf{v}}_{1},\cdots ,{\mathbf{v}}_{r}}\right) \) is a subspace. | Proof: Suppose \( \alpha ,\beta \) are two scalars and let \( \mathop{\sum }\limits_{{k = 1}}^{r}{c}_{k}{\mathbf{v}}_{k} \) and \( \mathop{\sum }\limits_{{k = 1}}^{r}{d}_{k}{\mathbf{v}}_{k} \) are two elements of \( V \) . What about\n\n\[ \alpha \mathop{\sum }\limits_{{k = 1}}^{r}{c}_{k}{\mathbf{v}}_{k} + \beta \matho... | Yes |
Corollary 5.1.8 Let \( \\left\\{ {{\\mathbf{x}}_{1},\\cdots ,{\\mathbf{x}}_{r}}\\right\\} \) and \( \\left\\{ {{\\mathbf{y}}_{1},\\cdots ,{\\mathbf{y}}_{s}}\\right\\} \) be two bases for \( V \) . Then \( r = \) \( s \) . | Proof: From the exchange theorem, \( r \\leq s \) and \( s \\leq r \) . Therefore, this proves the corollary. | Yes |
Lemma 5.1.10 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 \( \mathop{\sum }\limits_{{i = 1}}^{k}{c}_{i}{\mathbf{u}}_{i} + d\mathbf{v} = \mathbf{0} \) . 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},\cd... | Yes |
Theorem 5.1.11 Let \( V \) be a nonzero subspace of \( {\mathbb{F}}^{n} \). Then \( V \) has a basis. | 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 \( \operatorname{span}\left\{ {\mathbf{... | Yes |
Corollary 5.1.12 Let \( V \) be a subspace of \( {\mathbb{F}}^{n} \) and let \( \left\{ {{\mathbf{v}}_{1},\cdots ,{\mathbf{v}}_{r}}\right\} \) be a linearly independent set of vectors in \( V \) . Then either it is a basis for \( V \) or there exist vectors, \( {\mathbf{v}}_{r + 1},\cdots ,{\mathbf{v}}_{s} \) such that... | Proof: This follows immediately from the proof of Theorem 59.16.4. You do exactly the same argument except you start with \( \left\{ {{\mathbf{v}}_{1},\cdots ,{\mathbf{v}}_{r}}\right\} \) rather than \( \left\{ {\mathbf{v}}_{1}\right\} \) . | No |
Theorem 5.1.13 Let \( V \) be a subspace of \( {\mathbb{F}}^{n} \) and suppose \( \operatorname{span}\left( {{\mathbf{u}}_{1}\cdots ,{\mathbf{u}}_{p}}\right) = V \) where the \( {\mathbf{u}}_{i} \) are nonzero vectors. Then there exist vectors, \( \left\{ {{\mathbf{v}}_{1}\cdots ,{\mathbf{v}}_{r}}\right\} \) such that ... | Proof: Let \( r \) be the smallest positive integer with the property that for some set, \( \left\{ {{\mathbf{v}}_{1}\cdots ,{\mathbf{v}}_{r}}\right\} \subseteq \left\{ {{\mathbf{u}}_{1}\cdots ,{\mathbf{u}}_{p}}\right\} \), \[ \operatorname{span}\left( {{\mathbf{v}}_{1}\cdots ,{\mathbf{v}}_{r}}\right) = V. \] Then \( r... | Yes |
Lemma 5.3.1 There exists a function, \( {\operatorname{sgn}}_{n} \) which maps each ordered list of numbers from \( \{ 1,\cdots, n\} \) to one of the three numbers, \( 0,1 \), or -1 which also has the following properties.\n\n\[{\operatorname{sgn}}_{n}\left( {1,\cdots, n}\right) = 1\]\n\n\[{\operatorname{sgn}}_{n}\left... | Proof: Define \( \operatorname{sign}\left( x\right) = 1 \) if \( x > 0, - 1 \) if \( x < 0 \) and 0 if \( x = 0 \) . If \( n = 1 \), there is only one list and it is just the number 1 . Thus one can define \( {\operatorname{sgn}}_{1}\left( 1\right) \equiv 1 \) . For the general case where \( n > 1 \), simply define\n\n... | Yes |
Lemma 5.3.2 Every ordered list of distinct numbers from \( \{ 1,2,\cdots, n\} \) can be obtained from every other such ordered list by a finite number of switches. Also, \( {\operatorname{sgn}}_{n} \) is unique. | Proof: This is obvious if \( n = 1 \) or 2 . Suppose then that it is true for sets of \( n - 1 \) elements. Take two ordered lists of numbers, \( {P}_{1},{P}_{2} \) . Make one switch in both to place \( n \) at the end. Call the result \( {P}_{1}^{n} \) and \( {P}_{2}^{n} \) . Then using induction, there are finitely m... | Yes |
Proposition 5.4.3 Let\n\n\\[ \n\\left( {{r}_{1},\\cdots ,{r}_{n}}\\right) \n\\]\n\nbe an ordered list of numbers from \\( \\{ 1,\\cdots, n\\} \\) . Then\n\n\\[ \n\\operatorname{sgn}\\left( {{r}_{1},\\cdots ,{r}_{n}}\\right) \\det \\left( A\\right) \n\\]\n\n\\[\n= \\mathop{\\sum }\\limits_{\\left( {k}_{1},\\cdots ,{k}_{... | Proof: Let \\( \\left( {1,\\cdots, n}\\right) = \\left( {1,\\cdots, r,\\cdots s,\\cdots, n}\\right) \\) so \\( r < s \\) .\n\n\\[\n\\det \\left( {A\\left( {1,\\cdots, r,\\cdots, s,\\cdots, n}\\right) }\\right) =\n\\]\n\n\\( \\left( {5.4.10}\\right) \\)\n\n\\[\n\\mathop{\\sum }\\limits_{\\left( {k}_{1},\\cdots ,{k}_{n}\... | Yes |
Corollary 5.4.5 The following formula for \( \det \left( A\right) \) is valid.\n\n\[ \det \left( A\right) = \frac{1}{n!} \]\n\n\[ \mathop{\sum }\limits_{\left( {r}_{1},\cdots ,{r}_{n}\right) }\mathop{\sum }\limits_{\left( {k}_{1},\cdots ,{k}_{n}\right) }\operatorname{sgn}\left( {{r}_{1},\cdots ,{r}_{n}}\right) \operato... | Proof: From Proposition 5.4.3, if the \( {r}_{i} \) are distinct,\n\n\[ \det \left( A\right) = \mathop{\sum }\limits_{\left( {k}_{1},\cdots ,{k}_{n}\right) }\operatorname{sgn}\left( {{r}_{1},\cdots ,{r}_{n}}\right) \operatorname{sgn}\left( {{k}_{1},\cdots ,{k}_{n}}\right) {a}_{{r}_{1}{k}_{1}}\cdots {a}_{{r}_{n}{k}_{n}}... | Yes |
If two rows or two columns in an \( n \times n \) matrix \( A \), are switched, the determinant of the resulting matrix equals \( \left( {-1}\right) \) times the determinant of the original matrix. If \( A \) is an \( n \times n \) matrix in which two rows are equal or two columns are equal then \( \det \left( A\right)... | Proof: By Proposition 5.4.3 when two rows are switched, the determinant of the resulting matrix is \( \left( {-1}\right) \) times the determinant of the original matrix. By Corollary 5.4.5 the same holds for columns because the columns of the matrix equal the rows of the transposed matrix. Thus if \( {A}_{1} \) is the ... | Yes |
Corollary 5.4.8 Suppose \( A \) is an \( n \times n \) matrix and some column (row) is a linear combination of \( r \) other columns (rows). Then \( \det \left( A\right) = 0 \) . | Proof: Let \( A = \left( \begin{array}{lll} {\mathbf{a}}_{1} & \cdots & {\mathbf{a}}_{n} \end{array}\right) \) be the columns of \( A \) and suppose the condition that one column is a linear combination of \( r \) of the others is satisfied. Then by using Corollary 5.4.6 the determinant of \( A \) is zero if and only i... | Yes |
Theorem 5.4.10 Let \( A \) and \( B \) be \( n \times n \) matrices. Then\n\n\[ \det \left( {AB}\right) = \det \left( A\right) \det \left( B\right) \] | Proof: Let \( {c}_{ij} \) be the \( i{j}^{\text{th }} \) entry of \( {AB} \) . Then by Proposition 5.4.3,\n\n\[ \det \left( {AB}\right) = \]\n\n\[ \mathop{\sum }\limits_{\left( {k}_{1},\cdots ,{k}_{n}\right) }\operatorname{sgn}\left( {{k}_{1},\cdots ,{k}_{n}}\right) {c}_{1{k}_{1}}\cdots {c}_{n{k}_{n}} \]\n\n\[ = \matho... | Yes |
Lemma 5.4.11 Suppose a matrix is of the form\n\n\[ \nM = \left( \begin{array}{ll} A & * \\ \mathbf{0} & a \end{array}\right) \]\n\n\( \left( {5.4.13}\right) \)\n\nor\n\[ \nM = \left( \begin{array}{ll} A & 0 \\ * & a \end{array}\right) \]\n\n(5.4.14)\n\nwhere \( a \) is a number and \( A \) is an \( \left( {n - 1}\right... | 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) \equiv \mathop{\sum }\l... | Yes |
Theorem 5.4.13 Let \( A \) be an \( n \times n \) matrix where \( n \geq 2 \) . Then \[ \det \left( A\right) = \mathop{\sum }\limits_{{j = 1}}^{n}{a}_{ij}\operatorname{cof}{\left( A\right) }_{ij} = \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{ij}\operatorname{cof}{\left( A\right) }_{ij}. \] | Proof: Let \( \left( {{a}_{i1},\cdots ,{a}_{in}}\right) \) be the \( {i}^{\text{th }} \) row of \( A \) . Let \( {B}_{j} \) be the matrix obtained from \( A \) by leaving every row the same except the \( {i}^{th} \) row which in \( {B}_{j} \) equals \[ \left( {0,\cdots ,0,{a}_{ij},0,\cdots ,0}\right) \text{.} \] Then b... | Yes |
Theorem 5.4.14 \( {A}^{-1} \) 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}{\lef... | Proof: By Theorem 5.4.13 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 consider\n\n\[ \n\ma... | Yes |
Corollary 5.4.15 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 5.4.10 implies\n\n\[ \n\det B\det A = 1 \n\]\n\nand so \( \det A \neq 0 \) . Therefore from Theorem 5.4.14, \( {A}^{-1} \) exists. Therefore,\n\n\[ \n{A}^{-1} = \left( {BA}\right) {A}^{-1} = B\left( {A{A}^{-1}}\right) = {BI} = B. \n\]\n\nThe case where \( {CA} = I \) is handled simi... | Yes |
Lemma 5.5.2 Suppose for all \( \\left| \\lambda \\right| \) large enough,\n\n\[ \n{A}_{0} + {A}_{1}\\lambda + \\cdots + {A}_{m}{\\lambda }^{m} = 0, \n\]\n\nwhere the \( {A}_{i} \) are \( n \\times n \) matrices. Then each \( {A}_{i} = 0 \) . | Proof: Multiply by \( {\\lambda }^{-m} \) to obtain\n\n\[ \n{A}_{0}{\\lambda }^{-m} + {A}_{1}{\\lambda }^{-m + 1} + \\cdots + {A}_{m - 1}{\\lambda }^{-1} + {A}_{m} = 0. \n\]\n\nNow let \( \\left| \\lambda \\right| \\rightarrow \\infty \) to obtain \( {A}_{m} = 0 \) . With this, multiply by \( \\lambda \) to obtain\n\n\... | Yes |
Corollary 5.5.3 Let \( {A}_{i} \) and \( {B}_{i} \) be \( n \times n \) matrices and suppose\n\n\[ {A}_{0} + {A}_{1}\lambda + \cdots + {A}_{m}{\lambda }^{m} = {B}_{0} + {B}_{1}\lambda + \cdots + {B}_{m}{\lambda }^{m} \]\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. The last claim is obvious by matching terms. | No |
Theorem 5.5.4 Let \( A \) be an \( n \times n \) matrix and let \( q\left( \lambda \right) \equiv \det \left( {{\lambda I} - A}\right) \) be the characteristic polynomial. Then \( q\left( A\right) = 0 \) . | 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 |
Theorem 5.5.5 Both the left and the right sides in the following yield the same polynomial in the variables \( {a}_{i},{b}_{i} \) for \( i \leq n \) .\n\n\[ \mathop{\prod }\limits_{{i, j}}\left( {{a}_{i} + {b}_{j}}\right) \left| \begin{matrix} \frac{1}{{a}_{1} + {b}_{1}} & \cdots & \frac{1}{{a}_{1} + {b}_{n}} \\ \vdots... | Proof: The theorem is true if \( n = 2 \) . This follows from some computations. Suppose it is true for \( n - 1, n \geq 3 \) . \n\nContinuing to use the multilinear properties of determinants, this equals ![3f4063cc-9... | Yes |
Lemma 5.6.1 Consider the following product.\n\n\[ \left( \begin{array}{l} 0 \\ I \\ 0 \end{array}\right) \left( \begin{array}{lll} 0 & I & 0 \end{array}\right) \]\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 and there are \( ... | Proof: From the definition of the way you multiply matrices, the product is\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( \begin{array}{l} \mathbf{0} \\... | Yes |
Theorem 5.6.2 Let \( B \) be a \( q \times p \) block matrix as in 5.6.19 and let \( A \) be a \( p \times n \) block matrix as in 5.6.20 such that \( {B}_{is} \) is conformable with \( {A}_{sj} \) and each product, \( {B}_{is}{A}_{sj} \) for \( s = 1,\cdots, p \) is of the same size so they can be added. Then \( {BA} ... | Proof: From 5.6.18\n\n\[ {B}_{is}{A}_{sj} = \left( \begin{matrix} \mathbf{0} & {I}_{{r}_{i} \times {r}_{i}} & \mathbf{0} \end{matrix}\right) B\left( \begin{matrix} \mathbf{0} \\ {I}_{{p}_{s} \times {p}_{s}} \\ \mathbf{0} \end{matrix}\right) \left( \begin{matrix} \mathbf{0} & {I}_{{p}_{s} \times {p}_{s}} & \mathbf{0} \e... | Yes |
Example 5.6.3 Let an \( n \times n \) matrix have the form\n\n\[ A = \left( \begin{array}{ll} a & \mathbf{b} \\ \mathbf{c} & P \end{array}\right) \]\n\nwhere \( P \) is \( n - 1 \times n - 1 \) . Multiply it by\n\n\[ B = \left( \begin{array}{ll} p & \mathbf{q} \\ \mathbf{r} & Q \end{array}\right) \]\n\nwhere \( B \) is... | You use block multiplication\n\n\[ \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} \en... | Yes |
Theorem 5.6.4 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 5.8.3 Let \( A \) be an \( n \times n \) matrix. Then there exists a unitary matrix, \( U \) such that\n\n\[ {U}^{ * }{AU} = T \]\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 equation. | Proof: Let \( {\mathbf{v}}_{1} \) be a unit eigenvector for \( A \) . Then there exists \( {\lambda }_{1} \) such that\n\n\[ A{\mathbf{v}}_{1} = {\lambda }_{1}{\mathbf{v}}_{1},\left| {\mathbf{v}}_{1}\right| = 1 \]\n\nExtend \( \left\{ {\mathbf{v}}_{1}\right\} \) to a basis and then use Lemma 5.8.1 to obtain \( \left\{ ... | Yes |
Lemma 5.8.6 If \( T \) is upper triangular and normal, then \( T \) is a diagonal matrix. | Proof: Since \( T \) is normal, \( {T}^{ * }T = T{T}^{ * } \) . Writing this in terms of components and using the description of the adjoint as the transpose of the conjugate, yields the following for the \( i{k}^{th} \) entry of \( {T}^{ * }T = T{T}^{ * } \) .\n\n\[ \mathop{\sum }\limits_{j}{t}_{ij}{t}_{jk}^{ * } = \m... | Yes |
Theorem 5.8.7 Let \( A \) be a normal matrix. Then there exists a unitary matrix, \( U \) such that \( {U}^{ * }{AU} \) is a diagonal matrix. | Proof: From Theorem 5.8.3 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 5.8.8 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 5.8.9 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. | Proof: This follows from Theorem 5.8.4 and Corollary 5.8.8. | No |
Lemma 5.9.1 Let \( A \) be a Hermitian matrix such that all its eigenvalues are nonnegative. Then there exists a Hermitian matrix, \( {A}^{1/2} \) such that \( {A}^{1/2} \) has all nonnegative eigenvalues and \( {\left( {A}^{1/2}\right) }^{2} = A \) . | Proof: Since \( A \) is Hermitian, there exists a diagonal matrix \( D \) having all real nonnegative entries and a unitary matrix \( U \) such that \( A = {U}^{ * }{DU} \) . Then denote by \( {D}^{1/2} \) the matrix which is obtained by replacing each diagonal entry of \( D \) with its square root. Thus \( {D}^{1/2}{D... | Yes |
Lemma 5.9.2 Suppose \( \left\{ {{\mathbf{x}}_{1},{\mathbf{x}}_{2},\cdots ,{\mathbf{x}}_{r}}\right\} \) is an orthonormal set of vectors. Then if \( {c}_{1},\cdots ,{c}_{r} \) are scalars,\n\n\[{\left| \mathop{\sum }\limits_{{k = 1}}^{r}{c}_{k}{\mathbf{x}}_{k}\right| }^{2} = \mathop{\sum }\limits_{{k = 1}}^{r}{\left| {c... | Proof: This follows from the definition. From the properties of the dot product and using the fact that the given set of vectors is orthonormal,\n\n\[{\left| \mathop{\sum }\limits_{{k = 1}}^{r}{c}_{k}{\mathbf{x}}_{k}\right| }^{2} = \left( {\mathop{\sum }\limits_{{k = 1}}^{r}{c}_{k}{\mathbf{x}}_{k},\mathop{\sum }\limits... | Yes |
Lemma 5.9.3 Suppose \( \left\{ {{\mathbf{w}}_{1},\cdots ,{\mathbf{w}}_{r},{\mathbf{v}}_{r + 1},\cdots ,{\mathbf{v}}_{p}}\right\} \) is a linearly independent set of vectors such that \( \left\{ {{\mathbf{w}}_{1},\cdots ,{\mathbf{w}}_{r}}\right\} \) is an orthonormal set of vectors. Then when the Gram Schmidt process is... | Proof: Let \( \left\{ {{\mathbf{u}}_{1},\cdots ,{\mathbf{u}}_{p}}\right\} \) be the orthonormal set delivered by the Gram Schmidt process. Then \( {\mathbf{u}}_{1} = {\mathbf{w}}_{1} \) because by definition, \( {\mathbf{u}}_{1} \equiv {\mathbf{w}}_{1}/\left| {\mathbf{w}}_{1}\right| = {\mathbf{w}}_{1} \) . Now suppose ... | Yes |
Lemma 5.9.4 Let \( V \) be a subspace of dimension \( p \) and let \( \left\{ {{\mathbf{w}}_{1},\cdots ,{\mathbf{w}}_{r}}\right\} \) be an orthonormal set of vectors in \( V \) . Then this orthonormal set of vectors may be extended to an orthonormal basis for \( V \), \[ \left\{ {{\mathbf{w}}_{1},\cdots ,{\mathbf{w}}_{... | Proof: First extend the given linearly independent set \( \left\{ {{\mathbf{w}}_{1},\cdots ,{\mathbf{w}}_{r}}\right\} \) to a basis for \( V \) and then apply the Gram Schmidt theorem to the resulting basis. Since \( \left\{ {{\mathbf{w}}_{1},\cdots ,{\mathbf{w}}_{r}}\right\} \) is orthonormal it follows from Lemma 5.9... | Yes |
Lemma 5.9.5 Suppose \( R \) is an \( m \times n \) matrix with \( m > n \) 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 dot product,\n\n\[ \n{\left| \mathbf{x}\right| }^{2} + {\left| \mathbf{y}\right| }^{2} + \left( {\mathbf{x},\mathbf{y}}\right) + \left( {\mathbf{y},\mathbf... | Yes |
Lemma 5.9.7 Suppose \( \det \left( A\right) = 0 \) . Then for all sufficiently small nonzero \( \varepsilon \) , \( \det \left( {A + {\varepsilon I}}\right) \neq 0. \) | Proof: First suppose \( A \) is a \( p \times p \) matrix. Suppose also that \( \det \left( A\right) = 0 \) . Thus, the constant term of \( \det \left( {{\lambda I} - A}\right) \) is 0 . Consider \( {\varepsilon I} + A \equiv {A}_{\varepsilon } \) for small real \( \varepsilon \) . The characteristic polynomial of \( {... | Yes |
Theorem 6.2.3 Let \( \mathbf{x} \in {\mathbb{F}}^{n} \) and let \( r \geq 0 \) . Then \( B\left( {\mathbf{x}, r}\right) \) is an open set. Also,\n\n\[ D\left( {\mathbf{x}, r}\right) \equiv \left\{ {\mathbf{y} \in {\mathbb{F}}^{n} : \left| {\mathbf{y} - \mathbf{x}}\right| \leq r}\right\} \]\n\nis a closed set. | Proof: Suppose \( \mathbf{y} \in B\left( {\mathbf{x}, r}\right) \) . It is necessary to show there exists \( {r}_{1} > 0 \) such that \( B\left( {\mathbf{y},{r}_{1}}\right) \subseteq B\left( {\mathbf{x}, r}\right) \) . Define \( {r}_{1} \equiv r - \left| {\mathbf{x} - \mathbf{y}}\right| \) . Then if \( \left| {\mathbf{... | Yes |
The function, \( a\mathbf{f} + b\mathbf{g} \) is continuous at \( \mathbf{x} \) whenever \( \mathbf{f},\mathbf{g} \) are continuous at \( \mathbf{x} \) \( \in D\left( \mathbf{f}\right) \cap D\left( \mathbf{g}\right) \) and \( a, b \in \mathbb{F} \) . | For example the first claim says that \( \left( {a\mathbf{f} + b\mathbf{g}}\right) \left( \mathbf{y}\right) \) is close to \( \left( {a\mathbf{f} + b\mathbf{g}}\right) \left( \mathbf{x}\right) \) when \( \mathbf{y} \) is close to \( \mathbf{x} \) provided the same can be said about \( \mathbf{f} \) and \( \mathbf{g} \)... | No |
Theorem 6.5.3 If \( \mathop{\lim }\limits_{{\mathbf{y} \rightarrow \mathbf{x}}}\mathbf{f}\left( \mathbf{y}\right) = \mathbf{L} \) and \( \mathop{\lim }\limits_{{y \rightarrow x}}\mathbf{f}\left( \mathbf{y}\right) = {\mathbf{L}}_{1} \), then \( \mathbf{L} = {\mathbf{L}}_{1} \) . | Proof: Let \( \varepsilon > 0 \) be given. There exists \( \delta > 0 \) such that if \( 0 < \left| {\mathbf{y} - \mathbf{x}}\right| < \delta \) and \( \mathbf{y} \in D\left( \mathbf{f}\right) \), then\n\n\[ \left| {\mathbf{f}\left( \mathbf{y}\right) - \mathbf{L}}\right| < \varepsilon ,\left| {\mathbf{f}\left( \mathbf{... | Yes |
Theorem 6.5.5 Suppose \( \mathop{\lim }\limits_{{\mathbf{y} \rightarrow \mathbf{x}}}\mathbf{f}\left( \mathbf{y}\right) = \mathbf{L} \) and \( \mathop{\lim }\limits_{{\mathbf{y} \rightarrow \mathbf{x}}}\mathbf{g}\left( \mathbf{y}\right) = \mathbf{K} \) where \( \mathbf{K},\mathbf{L} \in \) \( {\mathbb{F}}^{q} \) . Then ... | Proof: The proof of 6.5.1 is left for you. It is like a corresponding theorem for continuous functions. Now 6.5.2is to be verified. Let \( \varepsilon > 0 \) be given. Then by the triangle inequality, \[ \left| {\mathbf{f} \cdot \mathbf{g}\left( \mathbf{y}\right) - \mathbf{L} \cdot \mathbf{K}}\right| \leq \left| {\math... | No |
Theorem 6.5.6 For \( \mathbf{f} : D\left( \mathbf{f}\right) \rightarrow {\mathbb{F}}^{q} \) and \( \mathbf{x} \in D\left( \mathbf{f}\right) \) a limit point of \( D\left( \mathbf{f}\right) ,\mathbf{f} \) is continuous at \( \mathbf{x} \) if and only if\n\n\[ \mathop{\lim }\limits_{{\mathbf{y} \rightarrow \mathbf{x}}}\m... | Proof: First suppose \( \mathbf{f} \) is continuous at \( \mathbf{x} \) a limit point of \( D\left( \mathbf{f}\right) \) . Then for every \( \varepsilon > 0 \) there exists \( \delta > 0 \) such that if \( \left| {\mathbf{y} - \mathbf{x}}\right| < \delta \) and \( \mathbf{y} \in D\left( \mathbf{f}\right) \), then \( \l... | Yes |
Theorem 6.5.7 Suppose \( \mathbf{f} : D\left( \mathbf{f}\right) \rightarrow {\mathbb{F}}^{q} \) . Then for \( \mathbf{x} \) a limit point of \( D\left( \mathbf{f}\right) \) , \[ \mathop{\lim }\limits_{{\mathbf{y} \rightarrow \mathbf{x}}}\mathbf{f}\left( \mathbf{y}\right) = \mathbf{L} \] if and only if \[ \mathop{\lim }... | Proof: Suppose 6.5.6. Then letting \( \varepsilon > 0 \) be given there exists \( \delta > 0 \) such that if \( 0 < \left| {\mathbf{y} - \mathbf{x}}\right| < \delta \), it follows \[ \left| {{f}_{k}\left( \mathbf{y}\right) - {L}_{k}}\right| \leq \left| {\mathbf{f}\left( \mathbf{y}\right) - \mathbf{L}}\right| < \varepsi... | Yes |
Example 6.5.8 Find \( \mathop{\lim }\limits_{{\left( {x, y}\right) \rightarrow \left( {3,1}\right) }}\left( {\frac{{x}^{2} - 9}{x - 3}, y}\right) \) . | It is clear that \( \mathop{\lim }\limits_{{\left( {x, y}\right) \rightarrow \left( {3,1}\right) }}\frac{{x}^{2} - 9}{x - 3} = 6 \) and \( \mathop{\lim }\limits_{{\left( {x, y}\right) \rightarrow \left( {3,1}\right) }}y = 1 \) . Therefore, this limit equals \( \left( {6,1}\right) \) . | Yes |
Example 6.5.9 Find \( \mathop{\lim }\limits_{{\left( {x, y}\right) \rightarrow \left( {0,0}\right) }}\frac{xy}{{x}^{2} + {y}^{2}} \) . | First of all observe the domain of the function is \( {\mathbb{F}}^{2} \smallsetminus \{ \left( {0,0}\right) \} \), every point in \( {\mathbb{F}}^{2} \) except the origin. Therefore, \( \left( {0,0}\right) \) is a limit point of the domain of the function so it might make sense to take a limit. However, just as in the... | Yes |
Theorem 6.7.2 If \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{\mathbf{a}}_{n} = \mathbf{a} \) and \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{\mathbf{a}}_{n} = {\mathbf{a}}_{1} \) then \( {\mathbf{a}}_{1} = \mathbf{a} \) . | Proof: Suppose \( {\mathbf{a}}_{1} \neq \mathbf{a} \) . Then let \( 0 < \varepsilon < \left| {{\mathbf{a}}_{1} - \mathbf{a}}\right| /2 \) in the definition of the limit. It follows there exists \( {n}_{\varepsilon } \) such that if \( n \geq {n}_{\varepsilon } \), then \( \left| {{\mathbf{a}}_{n} - \mathbf{a}}\right| <... | Yes |
Theorem 6.7.3 Let \( {\mathbf{a}}_{n} = \left( {{a}_{1}^{n},\cdots ,{a}_{p}^{n}}\right) \in {\mathbb{F}}^{p} \) . Then \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{\mathbf{a}}_{n} = \mathbf{a} \equiv \left( {{a}_{1},\cdots ,{a}_{p}}\right) \) if and only if for each \( k = 1,\cdots, p \) , \[ \mathop{\lim }\limit... | Proof: First suppose \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{\mathbf{a}}_{n} = \mathbf{a} \) . Then given \( \varepsilon > 0 \) there exists \( {n}_{\varepsilon } \) such that if \( n > {n}_{\varepsilon } \), then \[ \left| {{a}_{k}^{n} - {a}_{k}}\right| \leq \left| {{\mathbf{a}}_{n} - \mathbf{a}}\right| < \... | Yes |
Example 6.7.4 Let \( {\mathbf{a}}_{n} = \left( {\frac{1}{{n}^{2} + 1},\frac{1}{n}\sin \left( n\right) ,\frac{{n}^{2} + 3}{3{n}^{2} + {5n}}}\right) \) . | It suffices to consider the limits of the components according to the following theorem. Thus the limit is \( \left( {0,0,1/3}\right) \) . | No |
Theorem 6.7.5 Suppose \( \\left\\{ {\\mathbf{a}}_{n}\\right\\} \) and \( \\left\\{ {\\mathbf{b}}_{n}\\right\\} \) are sequences and that\n\n\[ \n\\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}{\\mathbf{a}}_{n} = \\mathbf{a}\\text{ and }\\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}{\\mathbf{b}}_{n} = \\mathb... | Proof: The first of these claims is left for you to do. To do the second, let \( \\varepsilon > 0 \) be given and choose \( {n}_{1} \) such that if \( n \\geq {n}_{1} \) then\n\n\[ \n\\left| {{\\mathbf{a}}_{n} - \\mathbf{a}}\\right| < 1\\text{.}\n\]\n\nThen for such \( n \), the triangle inequality and Cauchy Schwarz i... | No |
Theorem 6.7.7 Let \( {\left\{ {\mathbf{a}}_{n}\right\} }_{n = 1}^{\infty } \) be a Cauchy sequence in \( {\mathbb{F}}^{p} \) . Then there exists a unique \( \mathbf{a} \in {\mathbb{F}}^{p} \) such that \( {\mathbf{a}}_{n} \rightarrow \mathbf{a} \) . | Proof: Let \( {\mathbf{a}}_{n} = \left( {{a}_{1}^{n},\cdots ,{a}_{p}^{n}}\right) \) . Then\n\n\[ \left| {{a}_{k}^{n} - {a}_{k}^{m}}\right| \leq \left| {{\mathbf{a}}_{n} - {\mathbf{a}}_{m}}\right| \]\n\nwhich shows for each \( k = 1,\cdots, p \), it follows \( {\left\{ {a}_{k}^{n}\right\} }_{n = 1}^{\infty } \) is a Cau... | Yes |
Theorem 6.7.8 The set of terms in a Cauchy sequence in \( {\mathbb{F}}^{p} \) is bounded in the sense that for all \( n,\left| {\mathbf{a}}_{n}\right| < M \) for some \( M < \infty \) . | Proof: Let \( \varepsilon = 1 \) in the definition of a Cauchy sequence and let \( n > {n}_{1} \) . Then from the definition,\n\n\[ \left| {{\mathbf{a}}_{n} - {\mathbf{a}}_{{n}_{1}}}\right| < 1 \]\n\nIt follows that for all \( n > {n}_{1} \) ,\n\n\[ \left| {\mathbf{a}}_{n}\right| < 1 + \left| {\mathbf{a}}_{{n}_{1}}\rig... | Yes |
Theorem 6.7.9 If a sequence \( \left\{ {\mathbf{a}}_{n}\right\} \) in \( {\mathbb{F}}^{p} \) converges, then the sequence is a Cauchy sequence. | Proof: Let \( \varepsilon > 0 \) be given and suppose \( {\mathbf{a}}_{n} \rightarrow \mathbf{a} \) . Then from the definition of convergence, there exists \( {n}_{\varepsilon } \) such that if \( n > {n}_{\varepsilon } \), it follows that\n\n\[ \left| {{\mathbf{a}}_{n} - \mathbf{a}}\right| < \frac{\varepsilon }{2} \]\... | Yes |
Theorem 6.7.10 A function \( \mathbf{f} : D\left( \mathbf{f}\right) \rightarrow {\mathbb{F}}^{q} \) is continuous at \( \mathbf{x} \in D\left( \mathbf{f}\right) \) if and only if, whenever \( {\mathbf{x}}_{n} \rightarrow \mathbf{x} \) with \( {\mathbf{x}}_{n} \in D\left( \mathbf{f}\right) \), it follows \( \mathbf{f}\l... | Proof: Suppose first that \( \mathbf{f} \) is continuous at \( \mathbf{x} \) and let \( {\mathbf{x}}_{n} \rightarrow \mathbf{x} \) . Let \( \varepsilon > 0 \) be given. By continuity, there exists \( \delta > 0 \) such that if \( \left| {\mathbf{y} - \mathbf{x}}\right| < \delta \), then \( \left| {\mathbf{f}\left( \mat... | Yes |
The function, \( a\mathbf{f} + b\mathbf{g} \) is continuous at \( \mathbf{x} \) when \( \mathbf{f},\mathbf{g} \) are continuous at \( \mathbf{x} \in \) \( D\left( \mathbf{f}\right) \cap D\left( \mathbf{g}\right) \) and \( a, b \in \mathbb{F} \) . | Proof: Begin with 1.) Let \( \varepsilon > 0 \) be given. By assumption, there exist \( {\delta }_{1} > 0 \) such that whenever \( \left| {\mathbf{x} - \mathbf{y}}\right| < {\delta }_{1} \), it follows \( \left| {\mathbf{f}\left( \mathbf{x}\right) - \mathbf{f}\left( \mathbf{y}\right) }\right| < \frac{\varepsilon }{2\le... | Yes |
Theorem 6.10.3 Let \( C \subseteq {\mathbb{F}}^{p} \) be closed and bounded. Then \( C \) is sequentially compact. | Proof: Let \( \left\{ {\mathbf{a}}_{n}\right\} \subseteq C \) . Then let \( {\mathbf{a}}_{n} = \left( {{a}_{1}^{n},\cdots ,{a}_{p}^{n}}\right) \) . It follows the real and imaginary parts of the terms of the sequence, \( {\left\{ {a}_{j}^{n}\right\} }_{n = 1}^{\infty } \) are each contained in some sufficiently large c... | Yes |
Theorem 6.10.4 Let \( C \) be closed and bounded and let \( f : C \rightarrow \mathbb{R} \) be continuous. Then \( f \) achieves its maximum and its minimum on \( C \) . This means there exist, \( {\mathbf{x}}_{1},{\mathbf{x}}_{2} \in C \) such that for all \( \mathbf{x} \in C \) ,\n\n\[ f\left( {\mathbf{x}}_{1}\right)... | Proof: Let \( M = \sup \{ f\left( \mathbf{x}\right) : \mathbf{x} \in C\} \) . Recall this means \( + \infty \) if \( f \) is not bounded above and it equals the least upper bound of these values of \( f \) if \( f \) is bounded above. Then there exists a sequence, \( \left\{ {\mathbf{x}}_{n}\right\} \) such that \( f\l... | Yes |
Theorem 6.10.6 Let \( \mathbf{f} : C \rightarrow {\mathbb{F}}^{q} \) be continuous where \( C \) is a closed and bounded set in \( {\mathbb{F}}^{p} \) . Then \( \mathbf{f} \) is uniformly continuous on \( C \) . | Proof: If this is not so, there exists \( \varepsilon > 0 \) and pairs of points, \( {\mathbf{x}}_{n} \) and \( {\mathbf{y}}_{n} \) satisfying \( \left| {{\mathbf{x}}_{n} - {\mathbf{y}}_{n}}\right| < 1/n \) but \( \left| {\mathbf{f}\left( {\mathbf{x}}_{n}\right) - \mathbf{f}\left( {\mathbf{y}}_{n}\right) }\right| \geq ... | Yes |
Proposition 6.11.4 Let \( A\left( \mathbf{x}\right) \in \mathcal{L}\left( {{\mathbb{F}}^{n},{\mathbb{F}}^{m}}\right) \) for each \( \mathbf{x} \in U \subseteq {\mathbb{F}}^{p} \) . Then letting \( \left( {{A}_{ij}\left( \mathbf{x}\right) }\right) \) denote the matrix of \( A\left( \mathbf{x}\right) \) with respect to t... | Proof: Suppose first the second condition holds. Then from the material on linear transformations,\n\n\[ \left| {{A}_{ij}\left( \mathbf{x}\right) - {A}_{ij}\left( \mathbf{y}\right) }\right| = \left| {{\mathbf{e}}_{i} \cdot \left( {A\left( \mathbf{x}\right) - A\left( \mathbf{y}\right) }\right) {\mathbf{e}}_{j}}\right| \... | Yes |
Lemma 6.12.3 Let \( \mathbf{f} \) be differentiable at \( \mathbf{x} \) . Then \( \mathbf{f} \) is continuous at \( \mathbf{x} \) and in fact, there exists \( K > 0 \) such that whenever \( \left| \mathbf{v}\right| \) is small enough, \[ \left| {\mathbf{f}\left( {\mathbf{x} + \mathbf{v}}\right) - \mathbf{f}\left( \math... | Proof: From the definition of the derivative, \( \mathbf{f}\left( {\mathbf{x} + \mathbf{v}}\right) - \mathbf{f}\left( \mathbf{x}\right) = D\mathbf{f}\left( \mathbf{x}\right) \mathbf{v} + o\left( \mathbf{v}\right) \) . Let \( \left| \mathbf{v}\right| \) be small enough that \( \frac{o\left( \left| \mathbf{v}\right| \rig... | Yes |
Theorem 6.12.4 (The chain rule) Let \( U \) and \( V \) be open sets, \( U \subseteq {\mathbb{F}}^{n} \) and \( V \subseteq \) \( {\mathbb{F}}^{m} \) . Suppose \( \mathbf{f} : U \rightarrow V \) is differentiable at \( \mathbf{x} \in U \) and suppose \( \mathbf{g} : V \rightarrow {\mathbb{F}}^{q} \) is differentiable a... | Proof: This follows from a computation. Let \( B\left( {\mathbf{x}, r}\right) \subseteq U \) and let \( r \) also be small enough that for \( \left| \mathbf{v}\right| \leq r \), it follows that \( \mathbf{f}\left( {\mathbf{x} + \mathbf{v}}\right) \in V \) . Such an \( r \) exists because \( \mathbf{f} \) is continuous ... | Yes |
Corollary 6.13.8 Let \( U \) be an open subset of \( {\mathbb{F}}^{n} \) and let \( \mathbf{f} : U \rightarrow {\mathbb{F}}^{m} \) be \( {C}^{1} \) in the sense that all the partial derivatives of \( \mathbf{f} \) exist and are continuous. Then \( \mathbf{f} \) is differentiable and | \[ \mathbf{f}\left( {\mathbf{x} + \mathbf{v}}\right) = \mathbf{f}\left( \mathbf{x}\right) + \mathop{\sum }\limits_{{k = 1}}^{n}\frac{\partial \mathbf{f}}{\partial {x}_{k}}\left( \mathbf{x}\right) {v}_{k} + \mathbf{o}\left( \mathbf{v}\right) . \] | Yes |
Theorem 6.15.1 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}\... | 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( {x +... | Yes |
Example 6.15.5 Let\n\n\[ f\left( {x, y}\right) = \left\{ \begin{array}{l} \frac{{xy}\left( {{x}^{2} - {y}^{2}}\right) }{{x}^{2} + {y}^{2}}\text{ if }\left( {x, y}\right) \neq \left( {0,0}\right) \\ 0\text{ if }\left( {x, y}\right) = \left( {0,0}\right) \end{array}\right. \]\n\nFrom the definition of partial derivatives... | Now\n\n\[ {f}_{xy}\left( {0,0}\right) \equiv \mathop{\lim }\limits_{{y \rightarrow 0}}\frac{{f}_{x}\left( {0, y}\right) - {f}_{x}\left( {0,0}\right) }{y} \]\n\n\[ = \mathop{\lim }\limits_{{y \rightarrow 0}}\frac{-{y}^{4}}{{\left( {y}^{2}\right) }^{2}} = - 1 \]\n\nwhile\n\n\[ {f}_{yx}\left( {0,0}\right) \equiv \mathop{\... | Yes |
Theorem 6.16.1 (implicit function theorem) Suppose \( U \) is an open set in \( {\mathbb{R}}^{n} \times {\mathbb{R}}^{m} \) . Let \( \mathbf{f} : U \rightarrow {\mathbb{R}}^{n} \) be in \( {C}^{1}\left( U\right) \) and suppose\n\n\[ \mathbf{f}\left( {{\mathbf{x}}_{0},{\mathbf{y}}_{0}}\right) = \mathbf{0},{D}_{1}\mathbf... | Proof: Let\n\[ \mathbf{f}\left( {\mathbf{x},\mathbf{y}}\right) = \left( \begin{matrix} {f}_{1}\left( {\mathbf{x},\mathbf{y}}\right) \\ {f}_{2}\left( {\mathbf{x},\mathbf{y}}\right) \\ \vdots \\ {f}_{n}\left( {\mathbf{x},\mathbf{y}}\right) \end{matrix}\right) \]\n\nDefine for \( \left( {{\mathbf{x}}^{1},\cdots ,{\mathbf{... | Yes |
Theorem 6.16.3 (inverse function theorem) Let \( {\mathbf{x}}_{0} \in U \subseteq {\mathbb{F}}^{n} \) and let \( \mathbf{f} : U \rightarrow {\mathbb{F}}^{n} \). Suppose \[ \mathbf{f}\text{is}{C}^{1}\left( U\right) \text{, and}D\mathbf{f}{\left( {\mathbf{x}}_{0}\right) }^{-1} \in \mathcal{L}\left( {{\mathbb{F}}^{n},{\ma... | Proof: Apply the implicit function theorem to the function \[ \mathbf{F}\left( {\mathbf{x},\mathbf{y}}\right) \equiv \mathbf{f}\left( \mathbf{x}\right) - \mathbf{y} \] where \( {\mathbf{y}}_{0} \equiv \mathbf{f}\left( {\mathbf{x}}_{0}\right) \) . Thus the function \( \mathbf{y} \rightarrow \mathbf{x}\left( \mathbf{y}\r... | Yes |
Theorem 6.16.4 (implicit function theorem) Suppose \( U \) is an open set in \( {\mathbb{F}}^{n} \times {\mathbb{F}}^{m} \) . Let \( \mathbf{f} : U \rightarrow {\mathbb{F}}^{n} \) be in \( {C}^{k}\left( U\right) \) and suppose\n\n\[ \mathbf{f}\left( {{\mathbf{x}}_{0},{\mathbf{y}}_{0}}\right) = \mathbf{0},{D}_{1}\mathbf... | Proof: From Corollary 6.16.2 \( \mathbf{y} \rightarrow \mathbf{x}\left( \mathbf{y}\right) \) is \( {C}^{1} \) . It remains to show it is \( {C}^{k} \) for \( k > 1 \) assuming that \( \mathbf{f} \) is \( {C}^{k} \) . From 6.16.40\n\n\[ \frac{\partial \mathbf{x}}{\partial {y}^{l}} = - {D}_{1}{\left( \mathbf{x},\mathbf{y... | Yes |
Lemma 7.1.3 In a metric space, \( X \) every ball, \( B\left( {x, r}\right) \) is open. A set is closed if and only if it contains all its limit points. If \( p \) is a limit point of \( S \), then there exists a sequence of distinct points of \( S,\left\{ {x}_{n}\right\} \) such that \( \mathop{\lim }\limits_{{n \righ... | Proof: Let \( z \in B\left( {x, r}\right) \) . Let \( \delta = r - d\left( {x, z}\right) \) . Then if \( w \in B\left( {z,\delta }\right) \) ,\n\n\[ d\left( {w, x}\right) \leq d\left( {x, z}\right) + d\left( {z, w}\right) < d\left( {x, z}\right) + r - d\left( {x, z}\right) = r. \]\n\nTherefore, \( B\left( {z,\delta }\r... | No |
Theorem 7.1.4 Suppose \( \\left( {X, d}\\right) \) is a metric space. Then the sets \( \\{ B\\left( {x, r}\\right) : r > 0, x \\in X\\} \) satisfy\n\n\[ \n\\cup \\{ B\\left( {x, r}\\right) : r > 0, x \\in X\\} = X \n\]\n\n(7.1.1)\n\nIf \( p \\in B\\left( {x,{r}_{1}}\\right) \\cap B\\left( {z,{r}_{2}}\\right) \), there ... | Proof: Observe that the union of these balls includes the whole space, \( X \) so 7.1.1 is obvious. Consider 7.1.2. Let \( p \\in B\\left( {x,{r}_{1}}\\right) \\cap B\\left( {z,{r}_{2}}\\right) \). Consider\n\n\[ \nr \\equiv \\min \\left( {{r}_{1} - d\\left( {x, p}\\right) ,{r}_{2} - d\\left( {z, p}\\right) }\\right)\n... | Yes |
Lemma 7.1.5 If \( \left\{ {x}_{n}\right\} \) is a Cauchy sequence in a metric space, \( X \) and if some subsequence, \( \left\{ {x}_{{n}_{k}}\right\} \) converges to \( x \), then \( \left\{ {x}_{n}\right\} \) converges to \( x \) . Also if a sequence converges, then it is a Cauchy sequence. | Proof: Note first that \( {n}_{k} \geq k \) because in a subsequence, the indices, \( {n}_{1},{n}_{2},\cdots \) are strictly increasing. Let \( \varepsilon > 0 \) be given and let \( N \) be such that for \( k > \) \( N, d\left( {x,{x}_{{n}_{k}}}\right) < \varepsilon /2 \) and for \( m, n \geq N, d\left( {{x}_{m},{x}_{... | Yes |
Lemma 7.1.7 The function, \( x \rightarrow \operatorname{dist}\left( {x, S}\right) \) is continuous and in fact satisfies\n\n\[ \left| {\operatorname{dist}\left( {x, S}\right) - \operatorname{dist}\left( {y, S}\right) }\right| \leq d\left( {x, y}\right) . \] | Proof: Suppose \( \operatorname{dist}\left( {x, y}\right) \) is as least as large as \( \operatorname{dist}\left( {y, S}\right) \) . Then pick \( z \in S \) such that \( d\left( {y, z}\right) \leq \operatorname{dist}\left( {y, S}\right) + \varepsilon \) . Then\n\n\[ \left| {\operatorname{dist}\left( {x, S}\right) - \op... | Yes |
Proposition 7.2.3 An open ball is an open set. | Proof: Suppose \( y \in B\left( {x, r}\right) \) . We need to verify that \( y \) is an interior point of \( B\left( {x, r}\right) \) . Let \( \delta = r - d\left( {x, y}\right) \) . Then if \( z \in B\left( {y,\delta }\right) \), it follows that\n\n\[ d\left( {z, x}\right) \leq d\left( {z, y}\right) + d\left( {y, x}\r... | Yes |
Proposition 7.2.5 A point \( x \) is a limit point of the nonempty set \( A \) if and only if every \( B\left( {x, r}\right) \) contains infinitely many points of \( A \) . | Proof: \( \Leftarrow \) is obvious. Consider \( \Rightarrow \) . Let \( x \) be a limit point. Let \( {r}_{1} = 1 \) . Then \( B\left( {x,{r}_{1}}\right) \) contains \( {a}_{1} \neq x \) . If \( \left\{ {{a}_{1},\cdots ,{a}_{n}}\right\} \) have been chosen none equal to \( x \) and with no repeats in the list, let \( 0... | Yes |
Proposition 7.2.7 The limit is well defined. That is, if \( x,{x}^{\prime } \) are both limits of a sequence, then \( x = {x}^{\prime } \) . | Proof: From the definition, there exist \( N,{N}^{\prime } \) such that if \( n \geq N \), then \( d\left( {x,{x}_{n}}\right) < \) \( \varepsilon /2 \) and if \( n \geq {N}^{\prime } \), then \( d\left( {x,{x}_{n}}\right) < \varepsilon /2 \) . Then let \( M \geq \max \left( {N,{N}^{\prime }}\right) \) . Let \( n > M \)... | Yes |
Theorem 7.2.8 Let \( S \neq \varnothing \) . Then \( p \) is a limit point of \( S \) if and only if there exists a sequence of distinct points of \( S,\left\{ {x}_{n}\right\} \) none of which equal \( p \) such that \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = \) \( p \) . | Proof: \( \Rightarrow \) Suppose \( p \) is a limit point. Why does there exist the promissed convergent sequence? Let \( {x}_{1} \in B\left( {p,1}\right) \cap S \) such that \( {x}_{1} \neq p \) . If \( {x}_{1},\cdots ,{x}_{n} \) have been chosen, let \( {x}_{n + 1} \neq p \) be in \( B\left( {p,{\delta }_{n + 1}}\rig... | Yes |
Theorem 7.2.10 A set \( H \) is closed if and only if it contains all of its limit points. | Proof: \( \Rightarrow \) Let \( H \) be closed and let \( p \) be a limit point. We need to verify that \( p \in H \) . If it is not, then since \( H \) is closed, its complement is open and so there exists \( \delta > 0 \) such that \( B\left( {p,\delta }\right) \cap H = \varnothing \) . However, this prevents \( p \)... | Yes |
Corollary 7.2.11 A set \( H \) is closed if and only if whenever \( \left\{ {h}_{n}\right\} \) is a sequence of points of \( H \) which converges to a point \( x \), it follows that \( x \in H \) . | Proof: \( \Rightarrow \) Suppose \( H \) is closed and \( {h}_{n} \rightarrow x \) . If \( x \in H \) there is nothing left to show. If \( x \notin H \), then from the definition of limit, it is a limit point of \( H \) because none of the \( {h}_{n} \) are equal to \( x \) . Hence \( x \in H \) after all.\n\n\( \Lefta... | Yes |
Theorem 7.2.13 Let \( \\left\\{ {x}_{{n}_{k}}\\right\\} \) be a subsequence of a convergent sequence \( \\left\\{ {x}_{n}\\right\\} \) where \( {x}_{n} \\rightarrow x \) . Then | Proof: Let \( \\varepsilon > 0 \) be given. Then there exists \( N \) such that\n\n\[ d\\left( {{x}_{n}, x}\\right) < \\varepsilon \\text{ if }n \\geq N. \]\n\nIt follows that if \( k \\geq N \), then \( {n}_{k} \\geq N \) and so\n\n\[ d\\left( {{x}_{{n}_{k}}, x}\\right) < \\varepsilon \\text{ if }k \\geq N. \]\n\nThis... | Yes |
Theorem 7.3.2 Let \( \\left\\{ {x}_{n}\\right\\} \) be a Cauchy sequence. Then it converges if and only if any subsequence converges. | Proof: \( \\Rightarrow \) This was just done above. \( \\Leftarrow \) Suppose now that \( \\left\\{ {x}_{n}\\right\\} \) is a Cauchy sequence and \( \\mathop{\\lim }\\limits_{{k \\rightarrow \\infty }}{x}_{{n}_{k}} = x \) . Then there exists \( {N}_{1} \) such that if \( k > {N}_{1} \), then \( d\\left( {{x}_{{n}_{k}},... | Yes |
Lemma 7.3.4 If \( {x}_{n} \rightarrow x \), then \( \left\{ {x}_{n}\right\} \) is a Cauchy sequence. | Proof: Let \( \varepsilon > 0 \) . Then there exists \( {n}_{\varepsilon } \) such that if \( m \geq {n}_{\varepsilon } \), then \( d\left( {x,{x}_{m}}\right) < \) \( \varepsilon /2 \) . If \( m, k \geq {n}_{\varepsilon } \), then by the triangle inequality,\n\n\[ d\left( {{x}_{m},{x}_{k}}\right) \leq d\left( {{x}_{m},... | Yes |
Proposition 7.3.5 If \( \left\{ {x}_{n}\right\} \) is a sequence and if \( p \) is a limit point of the set \( S = \) \( { \cup }_{n = 1}^{\infty }\left\{ {x}_{n}\right\} \) then there is a subsequence \( \left\{ {x}_{{n}_{k}}\right\} \) such that \( \mathop{\lim }\limits_{{k \rightarrow \infty }}{x}_{{n}_{k}} = x \) . | Proof: By Theorem 7.2.8, there exists a sequence of distinct points of \( S \) denoted as \( \left\{ {y}_{k}\right\} \) such that none of them equal \( p \) and \( \mathop{\lim }\limits_{{k \rightarrow \infty }}{y}_{k} = p \) . Thus \( B\left( {p, r}\right) \) contains infinitely many different points of the set \( D \... | Yes |
Lemma 7.3.6 Suppose \( {x}_{n} \rightarrow x \) and \( {y}_{n} \rightarrow y \) . Then \( d\left( {{x}_{n},{y}_{n}}\right) \rightarrow d\left( {x, y}\right) \) . | Proof: Consider the following.\n\n\[ d\left( {x, y}\right) \leq d\left( {x,{x}_{n}}\right) + d\left( {{x}_{n}, y}\right) \leq d\left( {x,{x}_{n}}\right) + d\left( {{x}_{n},{y}_{n}}\right) + d\left( {{y}_{n}, y}\right) \]\n\nso\n\n\[ d\left( {x, y}\right) - d\left( {{x}_{n},{y}_{n}}\right) \leq d\left( {x,{x}_{n}}\right... | Yes |
Lemma 7.4.2 Let \( A \) be a nonempty set in \( \left( {X, d}\right) \) . Then \( \bar{A} \) is a closed set and\n\n\[ \bar{A} = A \cup {A}^{\prime } \]\n\nwhere \( {A}^{\prime } \) denotes the set of limit points of \( A \) . | Proof: First of all, denote by \( \mathcal{C} \) the set of closed sets which contain \( A \) . Then\n\n\[ \bar{A} = \cap \mathcal{C} \]\n\nand this will be closed if its complement is open. However,\n\n\[ {\bar{A}}^{C} = \cup \left\{ {{H}^{C} : H \in \mathcal{C}}\right\} \]\n\nEach \( {H}^{C} \) is open and so the uni... | Yes |
Theorem 7.5.2 A metric space is separable if and only if it is completely separable. | Proof: \( \Leftarrow \) Let \( \mathcal{B} \) be the special countable collection of open sets and for each \( B \in \mathcal{B} \), let \( {p}_{B} \) be a point of \( B \) . Then let \( \mathcal{P} \equiv \left\{ {{p}_{B} : B \in \mathcal{B}}\right\} \) . If \( B\left( {x, r}\right) \) is any ball, then it is the unio... | Yes |
Theorem 7.5.5 Every separable metric space has the Lindeloff property. | Proof: Let \( \mathcal{C} \) be an open cover of a set \( S \) . Let \( \mathcal{B} \) be a countable basis. Such exists by Theorem 7.5.2. Let \( \widehat{\mathcal{B}} \) denote those sets of \( \mathcal{B} \) which are contained in some set of \( \mathcal{C} \) . Thus \( \widehat{\mathcal{B}} \) is a countable open co... | Yes |
Example 7.6.2 Let \( X \) be any infinite set and define \( d\left( {x, y}\right) = 1 \) if \( x \neq y \) while \( d\left( {x, y}\right) = 0 \) if \( x = y \). | You should verify the details that this is a metric space because it satisfies the axioms of a metric. The set \( X \) is closed and bounded because its complement is \( \varnothing \) which is clearly open because every point of \( \varnothing \) is an interior point. (There are none.) Also \( X \) is bounded because ... | No |
Lemma 7.6.4 Let \( X \) be a metric space and suppose \( D \) is a countable dense subset of \( X \) . In other words, it is being assumed \( X \) is a separable metric space. Consider the open sets of the form \( B\left( {d, r}\right) \) where \( r \) is a positive rational number and \( d \in D \) . Denote this count... | Proof: Let \( U \) be an open set and let \( x \in U \) . Let \( B\left( {x,\delta }\right) \subseteq U \) . Then by density of \( D \), there exists \( d \in D \cap B\left( {x,\delta /4}\right) \) . Now pick \( r \in \mathbb{Q} \cap \left( {\delta /4,{3\delta }/4}\right) \) and consider \( B\left( {d, r}\right) \) . C... | Yes |
Lemma 7.6.6 A subset of \( {\mathbb{R}}^{n} \) is totally bounded if and only if it is bounded. | Proof: Let \( A \) be totally bounded. Is it bounded? Let \( {\mathbf{x}}_{1},\cdots ,{\mathbf{x}}_{p} \) be a 1 net for \( A \) . Now consider the ball \( B\left( {\mathbf{0}, r + 1}\right) \) where \( r > \max \left( {\left| {\mathbf{x}}_{i}\right| : i = 1,\cdots, p}\right) \) . If \( \mathbf{z} \in A \) , then \( \m... | Yes |
Theorem 7.6.7 A subset of \( {\mathbb{R}}^{n} \) is compact if and only if it is closed and bounded. | Proof: Since a set in \( {\mathbb{R}}^{n} \) is totally bounded if and only if it is bounded, this theorem follows from Proposition 7.6.5 and the observation that a subset of \( {\mathbb{R}}^{n} \) is closed if and only if it is complete. This proves the theorem. | No |
Proposition 7.6.8 If \( K \) is a closed, nonempty subset of a nonempty compact set \( H \), then \( K \) is compact. | Proof: Let \( \mathcal{C} \) be an open cover for \( K \) . Then \( \mathcal{C} \cup \left\{ {K}^{C}\right\} \) is an open cover for \( H \) . Thus there are finitely many sets from this last collection of open sets, \( {U}_{1},\cdots ,{U}_{m} \) which covers \( H \) . Include only those which are in \( \mathcal{C} \) ... | Yes |
Theorem 7.7.1 Let \( X \) be a compact metric space and let \( f : X \rightarrow \mathbb{R} \) be continuous. Then \( \max \{ f\left( x\right) : x \in X\} \) and \( \min \{ f\left( x\right) : x \in X\} \) both exist. | Proof: First it is shown \( f\left( X\right) \) is compact. Suppose \( \mathcal{C} \) is a set of open sets whose union contains \( f\left( X\right) \) . Then since \( f \) is continuous \( {f}^{-1}\left( U\right) \) is open for all \( U \in \mathcal{C} \) . Therefore, \( \left\{ {{f}^{-1}\left( U\right) : U \in \mathc... | Yes |
Theorem 7.7.3 Suppose \( f : X \rightarrow Y \) is continuous and \( X \) is compact. Then \( f \) is uniformly continuous. | Proof: Suppose this is not true and that \( f \) is continuous but not uniformly continuous. Then there exists \( \varepsilon > 0 \) such that for all \( \delta > 0 \) there exist points, \( {p}_{\delta } \) and \( {q}_{\delta } \) such that \( d\left( {{p}_{\delta },{q}_{\delta }}\right) < \delta \) and yet \( d\left(... | Yes |
Theorem 7.7.5 Suppose \( \mathcal{F} \) is a collection of compact sets in a metric space, \( X \) which has the finite intersection property. Then there exists a point in their intersection. \( \left( {\cap \mathcal{F} \neq \varnothing }\right) \) . | Proof: First I show each compact set is closed. Let \( K \) be a nonempty compact set and suppose \( p \notin K \) . Then for each \( x \in K \), let \( {V}_{x} = B\left( {x, d\left( {p, x}\right) /3}\right) \) and \( {U}_{x} = B\left( {p, d\left( {p, x}\right) /3}\right) \) so that \( {U}_{x} \) and \( {V}_{x} \) have... | Yes |
Theorem 7.7.6 Let \( {X}_{i} \) be a compact metric space with metric \( {d}_{i} \) . Then \( \mathop{\prod }\limits_{{i = 1}}^{m}{X}_{i} \) is also a compact metric space with respect to the metric, \( d\left( {\mathbf{x},\mathbf{y}}\right) \equiv \mathop{\max }\limits_{i}\left( {{d}_{i}\left( {{x}_{i},{y}_{i}}\right)... | Proof: This is most easily seen from sequential compactness. Let \( {\left\{ {\mathbf{x}}^{k}\right\} }_{k = 1}^{\infty } \) be a sequence of points in \( \mathop{\prod }\limits_{{i = 1}}^{m}{X}_{i} \) . Consider the \( {i}^{th} \) component of \( {\mathbf{x}}^{k},{x}_{i}^{k} \) . It follows \( \left\{ {x}_{i}^{k}\righ... | Yes |
Theorem 7.8.4 Suppose \( K \) is a nonempty compact subset of \( {\mathbb{R}}^{n} \) and \( A \subseteq C\left( {K, X}\right) \) is uniformly bounded and uniformly equicontinuous. Then if \( \left\{ {f}_{k}\right\} \subseteq A \), there exists a function, \( f \in C\left( {K, X}\right) \) and a subsequence, \( {f}_{{k}... | To give a proof of this theorem, I will first prove some lemmas. | No |
Lemma 7.8.5 If \( K \) is a compact subset of \( {\mathbb{R}}^{n} \), then there exists \( D \equiv {\left\{ {\mathbf{x}}_{k}\right\} }_{k = 1}^{\infty } \subseteq K \) such that \( D \) is dense in \( K \) . Also, for every \( \varepsilon > 0 \) there exists a finite set of points, \( \left\{ {{\mathbf{x}}_{1},\cdots ... | Proof: For \( m \in \mathbb{N} \), pick \( {x}_{1}^{m} \in K \) . If every point of \( K \) is within \( 1/m \) of \( {x}_{1}^{m} \), stop. Otherwise, pick\n\n\[ \n{x}_{2}^{m} \in K \smallsetminus B\left( {{x}_{1}^{m},1/m}\right) .\n\]\n\nIf every point of \( K \) contained in \( B\left( {{x}_{1}^{m},1/m}\right) \cup B... | Yes |
Lemma 7.9.3 Let \( S \) be a totally bounded subset of \( \left( {X, d}\right) \) a metric space. Then \( \bar{S} \) is also totally bounded. | Proof: Suppose not. Then there exists a sequence \( \left\{ {p}_{n}\right\} \subseteq \bar{S} \) such that \( d\left( {{p}_{m},{p}_{n}}\right) \geq \) \( \varepsilon \) for all \( m \neq n \) . Now let \( {q}_{n} \in B\left( {{p}_{n},\frac{\varepsilon }{8}}\right) \cap S \) . Then it follows that\n\n\[ \frac{\varepsilo... | Yes |
Theorem 7.9.5 Let \( \left( {X,{d}_{X}}\right) \) be a compact metric space and let \( \left( {Y,{d}_{Y}}\right) \) be a complete metric space. Thus \( \left( {C\left( {X, Y}\right) ,\rho }\right) \) is a complete metric space. Let \( \mathcal{A} \subseteq C\left( {X, Y}\right) \) be pointwise compact and equicontinuou... | Proof: The more useful direction is that the two conditions imply compactness of \( \overline{\mathcal{A}} \) . I prove this first. Since \( \overline{\mathcal{A}} \) is a closed subset of a complete space, it follows that \( \overline{\mathcal{A}} \) will be compact if it is totally bounded. In showing this, it follow... | Yes |
Lemma 7.10.2 Let \( \\left( {X, d}\\right) \) be a metric space and let \( S \\subseteq X \) be a nonempty subset. Define\n\n\[ \n\\operatorname{dist}\\left( {x, S}\\right) \\equiv \\inf \\{ d\\left( {x, y}\\right) : y \\in S\\} .\n\]\n\nThen \( x \\rightarrow \\operatorname{dist}\\left( {x, S}\\right) \) is a continuo... | Proof: The continuity of \( x \\rightarrow \\operatorname{dist}\\left( {x, S}\\right) \) is obvious if the inequality 7.10.10 is established. So let \( x, y \\in X \) . Without loss of generality, assume \( \\operatorname{dist}\\left( {x, S}\\right) \\geq \) \( \\operatorname{dist}\\left( {y, S}\\right) \) and pick \( ... | Yes |
Lemma 7.10.3 Let \( H, K \) be two nonempty disjoint closed subsets of a metric space, \( \left( {X, d}\right) \). Then there exists a continuous function, \( g : X \rightarrow \left\lbrack {-1,1}\right\rbrack \) such that \( g\left( H\right) = - 1/3, g\left( K\right) = 1/3, g\left( X\right) \subseteq \left\lbrack {-1/... | Proof: Let\n\n\[ f\left( x\right) \equiv \frac{\operatorname{dist}\left( {x, H}\right) }{\operatorname{dist}\left( {x, H}\right) + \operatorname{dist}\left( {x, K}\right) }.\]\n\nThe denominator is never equal to zero because if \( \operatorname{dist}\left( {x, H}\right) = 0 \), then \( x \in H \) becasue \( H \) is cl... | Yes |
Lemma 7.10.5 Suppose \( M \) is a closed set in \( X \) where \( \left( {X, d}\right) \) is a metric space and suppose \( f : M \rightarrow \left\lbrack {-1,1}\right\rbrack \) is continuous at every point of \( M \) . Then there exists a function, \( g \) which is defined and continuous on all of \( X \) such that \( \... | Proof: Let \( H = {f}^{-1}\left( \left\lbrack {-1, - 1/3}\right\rbrack \right), K = {f}^{-1}\left( \left\lbrack {1/3,1}\right\rbrack \right) \) . Thus \( H \) and \( K \) are disjoint closed subsets of \( M \) . Suppose first \( H, K \) are both nonempty. Then by Lemma 7.10.3 there exists \( g \) such that \( g \) is a... | Yes |
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