Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
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Property 1.16.3 If \( {A}^{ij} \) and \( {B}^{pq} \) are skew-symmetric tensors, then outer product is symmetric tensor. | Proof: Since \( {A}^{ij} \) and \( {B}^{pq} \) are skew-symmetric, so by definition,\n\n\[ \n{A}^{ji} = - {A}^{ij}\text{ and }{B}^{qp} = - {B}^{pq}.\n\]\n\nIf the outer product of \( {A}^{ij} \) and \( {B}^{pq} \) be \( {C}^{ijpq} \), then,\n\n\[ \n{C}^{ijpq} = {A}^{ij}{B}^{pq} \n\]\n\nThus\n\n\[ \n{C}^{ijpq} = \left( ... | Yes |
Theorem 2.1.2 The line element \( {g}_{ij}d{x}^{i}d{x}^{j} \) is an invariant. | Proof: Let us consider a co-ordinate transformation from \( {x}^{i} \) to \( {\bar{x}}^{i} \) given by\n\n\[ \n{x}^{i} = {x}^{i}\left( {{\bar{x}}^{1},{\bar{x}}^{2},\ldots ,{\bar{x}}^{N}}\right) ;i = 1,2,\ldots, N.\n\]\n\nSince \( {g}_{ij} \) is a covariant tensor of rank 2, we have\n\n\[ \n{\bar{g}}_{ij} = {g}_{pq}\fra... | Yes |
Property 2.1.1 The properties of reciprocal tensor \( {g}^{ij} \) are (i) \( {g}_{ij}{g}^{kj} = {\delta }_{i}^{k} \) ,(ii) \( {g}^{ij}{g}_{ij} = \) \( N \) and (iii) \( {g}^{ij} \) is also a symmetric contravariant tensor of rank 2 . | Proof: (i) Let the cofactor of \( {g}_{ij} \) in \( g \) be denoted by \( \xi \left( {i, j}\right) \) . From properties of determinants, we have\n\n\[ \n{g}_{ij}\xi \left( {i, j}\right) = g;\;g = \left| {g}_{ij}\right| \n\] \n\nor \n\[ \n{g}_{ij}\frac{\xi \left( {i, j}\right) }{g} = 1 \Rightarrow {g}_{ij}{g}^{ij} = 1;\... | Yes |
Theorem 2.1.3 Formula (2.9) for arc length does not depend on the particular parameterisation of the curve. | Proof: Given a curve \( \mathcal{C} : {x}^{i} = {x}^{i}\left( t\right) ;a \leq t \leq b \), suppose that \( \mathcal{C} : {x}^{i} = {x}^{i}\left( \bar{t}\right) ;\bar{a} \leq \bar{t} \leq \bar{b} \) is a different parameterisation, where \( \bar{t} = \phi \left( t\right) \), with \( {\phi }^{\prime }\left( t\right) > 0... | Yes |
Theorem 2.3.1 The necessary and sufficient condition for the existence of an \( N \) ply orthogonal system of co-ordinate hypersurfaces is that the fundamental form must be of the form\n\n\[ d{s}^{2} = {g}_{11}{\left( d{x}^{1}\right) }^{2} + {g}_{22}{\left( d{x}^{2}\right) }^{2} + \cdots + {g}_{NN}{\left( d{x}^{N}\righ... | Proof: Condition necessary: Let us suppose that co-ordinate hypersurfaces form an \( N \) ply orthogonal system of hypersurfaces. Since the co-ordinate hypersurfaces form an \( N \) ply orthogonal system of hypersurfaces, we have\n\n\[ {g}^{ij} = 0;\text{ for every }i, j = 1,2,\ldots, N\text{ and }i \neq j. \]\n\n(2.36... | Yes |
Property 3.1.1 The Christoffel symbols of First and Second kind defined in Eqs. (3.1) and (3.2) are symmetric with respect to the indices \( i \) and \( j \) . | Proof: In the definition of Christoffel symbol (3.1) of first kind, interchanging of \( i \) and \( j \), provides\n\n\[ \left\lbrack {{ji}, k}\right\rbrack = \frac{1}{2}\left( {\frac{\partial {g}_{jk}}{\partial {x}^{i}} + \frac{\partial {g}_{ik}}{\partial {x}^{j}} - \frac{\partial {g}_{ji}}{\partial {x}^{k}}}\right) \... | Yes |
Property 3.1.2 The necessary and sufficient condition that all the Christoffel symbols vanish at a point is that \( {g}_{ij} \) are constants. | Proof: Let \( {g}_{ij} \) be constants, at a point \( P\left( {x}^{i}\right) \), then\n\n\[ \frac{\partial {g}_{ij}}{\partial {x}^{k}} = 0,\frac{\partial {g}_{ik}}{\partial {x}^{j}} = 0\;\text{ and }\;\frac{\partial {g}_{jk}}{\partial {x}^{i}} = 0. \]\n\nUsing definition (3.1) of Christoffel symbol of first kind, we ge... | Yes |
Property 3.1.3 To establish \( \left\lbrack {{ij}, m}\right\rbrack = {g}_{km}\left\{ \begin{matrix} k \\ i\;j \end{matrix}\right\} \) . | Proof: We see from the defining formula (3.2) that we can pass from the symbol of the first kind \( \left\lbrack {{ij}, m}\right\rbrack \) to the symbol \( \left\{ \begin{matrix} k \\ i\;j \end{matrix}\right\} \) by forming the inner product \( {g}^{km}\left\lbrack {{ij}, m}\right\rbrack \) . Therefore, the inner multi... | Yes |
Property 3.1.4 To establish \( \left\lbrack {{ij}, k}\right\rbrack + \left\lbrack {{jk}, i}\right\rbrack = \frac{\partial {g}_{ik}}{\partial {x}^{j}} \) . | Proof: Using definition (3.1) of Christoffel symbol of first kind, we get\n\n\[ \left\lbrack {{ij}, k}\right\rbrack + \left\lbrack {{jk}, i}\right\rbrack = \frac{1}{2}\left( {\frac{\partial {g}_{jk}}{\partial {x}^{i}} + \frac{\partial {g}_{ik}}{\partial {x}^{j}} - \frac{\partial {g}_{ij}}{\partial {x}^{k}}}\right) + \f... | Yes |
Property 3.1.5 To establish \( {\partial }_{j}{g}_{ik} = \frac{\partial {g}^{ik}}{\partial {x}^{j}} = - {g}^{hk}\left\{ \begin{matrix} i \\ {hj} \end{matrix}\right\} - {g}^{hi}\left\{ \begin{matrix} k \\ {hj} \end{matrix}\right\} \) . | Proof: The formula for the partial derivatives of the contravariant tensor \( {g}^{ik} \) can be obtained by differentiating the identity \( {g}_{pm}{g}^{mi} = {\delta }_{p}^{i} \), where \( {\delta }_{p}^{i} \) is the Kronecker delta, with respect to \( {x}^{j} \), we get\n\n\[ \frac{\partial }{\partial {x}^{j}}\left(... | Yes |
Property 3.1.6 To establish \( \left\{ \begin{matrix} i \\ i - j \end{matrix}\right\} = \left\{ \begin{matrix} i \\ j - i \end{matrix}\right\} = \frac{\partial }{\partial {x}^{j}}\left( {\log \sqrt{g}}\right) \), where \( g = \left| {g}_{ij}\right| \neq 0 \) . | Proof: According to the definition of reciprocal tensor, we have\n\n\[ \n{g}^{ij} = \frac{\text{ cofactor of }{g}_{ij}\text{ in }\left| {g}_{ij}\right| }{\left| {g}_{ij}\right| } = \frac{{G}^{ij}}{g},\n\]\n\nwhere \( {G}^{ij} \) denotes the cofactor of \( {g}_{ij} \) in \( g = \left| {g}_{ij}\right| \) . Since, \( g = ... | Yes |
Theorem 3.3.1 A necessary and sufficient condition that the curl of a vector field vanishes is that the vector field be gradient. | Proof: Let \( {A}_{i} \) be a covariant vector. Let the curl of the vector \( {A}_{i} \) vanish, so that\n\n\[ \operatorname{curl}{A}_{i} = {A}_{i, j} - {A}_{j, i} = \frac{\partial {A}_{i}}{\partial {x}^{j}} - \frac{\partial {A}_{j}}{\partial {x}^{i}} = 0. \]\n\nWe have to show that \( {A}_{i} = \nabla \phi \), where \... | Yes |
Property 4.1.1 The necessary and sufficient condition that the covariant differentiation of all vectors be commutative is that the Riemann tensor \( {R}_{ijk}^{\alpha } \) vanishes identically. | Proof: If the left-hand side of Eq. (4.8) is to vanish, i.e. the order of covariant differentiation is to be immaterial, then \( {R}_{ijk}^{\alpha } = 0 \) . Since \( {B}_{i} \) is arbitrary, in general \( {R}_{ijk}^{\alpha } \neq 0 \) , so that the order of covariant differentiation is no immaterial. Thus, it is clear... | No |
Property 4.1.2 The Riemann-Christoffel curvature tensor \( {R}_{ijk}^{\alpha } \) is skew-symmetric with respect to the indices \( j \) and \( k \) . | Proof: The Riemann-Christoffel curvature tensor \( {R}_{ijk}^{\alpha } \) is given by Eq. (4.6). Interchanging \( j \) and \( k \), in the expression for \( {R}_{ijk}^{\alpha } \), we get\n\n\[ \n{R}_{ikj}^{\alpha } = - \frac{\partial }{\partial {x}^{j}}\left\{ \begin{matrix} \alpha \\ i\;k \end{matrix}\right\} + \frac... | Yes |
Property 4.1.3 The curvature tensor \( {R}_{ijk}^{\alpha } \) satisfies cyclic property, i.e.\n\n\[ {R}_{ijk}^{\alpha } + {R}_{jki}^{\alpha } + {R}_{kij}^{\alpha } = 0\text{; i.e.}{R}_{\left\lbrack ijk\right\rbrack }^{\alpha } = 0\text{.} \] | Proof: The Riemannian curvature tensor \( {R}_{ijk}^{\alpha } \) is given by Eq. (4.6). Taking the sum of this and two similar equations obtained by cyclic permutation of \( i, j, k \) we obtain\n\n\[ {R}_{\left\lbrack ijk\right\rbrack }^{\alpha } = {R}_{ijk}^{\alpha } + {R}_{jki}^{\alpha } + {R}_{kij}^{\alpha } \]\n\n... | Yes |
Property 4.1.6 The covariant curvature tensor \( {R}_{hijk} \) satisfies cyclic property, i.e.\n\n\[ \n{R}_{hijk} + {R}_{hjki} + {R}_{hkji} = 0. \n\] | Proof: The covariant curvature tensor \( {R}_{hijk} \) is given by Eq. (4.16,) from which we get,\n\n\[ \n{R}_{hjki} = \frac{1}{2}\left( {\frac{{\partial }^{2}{g}_{jk}}{\partial {x}^{h}\partial {x}^{i}} + \frac{{\partial }^{2}{g}_{hi}}{\partial {x}^{j}\partial {x}^{k}} - \frac{{\partial }^{2}{g}_{ji}}{\partial {x}^{h}\... | Yes |
Property 4.1.7 The curvature tensor \( {R}_{hijk} \) satisfies the differential property\n\n\[ {R}_{{ijk}, m}^{\alpha } + {R}_{{ikm}, j}^{\alpha } + {R}_{{imj}, k}^{\alpha } = 0\text{; i.e.}{R}_{{hijk}, l} + {R}_{{hikl}, j} + {R}_{{hilj}, k} = 0\text{.} \] | Proof: Let us choose a system of geodesic co-ordinates with the pole at \( {P}_{0} \), then at \( {P}_{0} \) Christoffel symbols vanish and first covariant derivative reduce to corresponding ordinary partial derivatives, i.e. at \( {P}_{0} \) ,\n\n\[ \left\lbrack {{ki}, j}\right\rbrack = 0;\;\left\{ \begin{matrix} j \\... | Yes |
Theorem 4.3.1 A space of constant curvature is an Einstein space. | Proof: Let the Riemannian curvature \( \kappa \) at \( P \) of \( {V}_{N} \) for the orientation determined by \( {p}^{i} \) and \( {q}^{i} \), is given in Eq. (4.34) as\n\n\[ \kappa = \frac{{p}^{h}{q}^{i}{p}^{j}{q}^{k}{R}_{hijk}}{{p}^{h}{q}^{i}{p}^{j}{q}^{k}\left\lbrack {{g}_{hj}{g}_{ik} - {g}_{ij}{g}_{kh}}\right\rbra... | Yes |
Theorem 6.7.1 Prove the \( \kappa = 0 \) is the necessary and sufficient condition for a surface to be a developable. | Proof: The Gaussian curvature of the total curvature \( \kappa \) of the surface \( \mathcal{S} \) is given by\n\n\[ \kappa = \frac{{R}_{\alpha \beta \gamma \delta }}{a};\;a = \left| {a}_{\alpha \beta }\right| ,\]\n\nwhere the tensor \( {R}_{\alpha \beta \gamma \delta } \) is skew symmetric in the first two and last tw... | Yes |
Theorem 6.8.1 The necessary and sufficient condition for a curve on a surface to be a geodesic is that its geodesic curvature is zero. | Proof: The differential equation of the geodesic is\n\n\[ \frac{{d}^{2}{u}^{\alpha }}{d{s}^{2}} + \left\{ \begin{matrix} \alpha \\ \beta /\gamma \end{matrix}\right\} \frac{d{u}^{\beta }}{ds}\frac{d{u}^{\gamma }}{ds} = 0 \]\n\nor\n\[ \frac{d}{ds}\left( \frac{d{u}^{\alpha }}{ds}\right) + \left\{ \begin{matrix} \alpha \\ ... | Yes |
Theorem 7.4.1 (Theorema Egregium of Gauss): The Gaussian curvature \( \kappa \) is an intrinsic property of a surface, depending only on the first fundamental form and their derivatives but is independent of the second fundamental form. | Proof: The total curvature \( \kappa \) is given by\n\n\[ \kappa = \frac{b}{a} = \frac{{b}_{11}{b}_{22} - {b}_{12}^{2}}{{a}_{11}{a}_{22} - {a}_{12}^{2}}. \]\n\nThus, the total curvature, as obtained, depends upon the first fundamental form as well as the second fundamental form. But we have the equations of Gauss,\n\n\... | Yes |
Theorem 8.3.1 (Rodrigue's formula): A line of curvature is characterised by\n\n\\[ \n\\frac{\\partial {\\xi }^{i}}{\\partial s} + {\\chi }_{\\left( n\\right) }\\frac{d{x}^{i}}{ds} = 0 \n\\]\n\nwhere \\( {\\chi }_{\\left( n\\right) } \\) is the principal curvature of the surface. | Proof: From Weingarten formula Eq. (7.49), we have\n\n\\[ \n{\\xi }_{,\\alpha }^{i} = - {a}^{\\beta \\gamma }{b}_{\\beta \\gamma }{x}_{\\gamma }^{i} \n\\]\n\nor\n\\[ \n{\\xi }_{,\\alpha }^{i}\\frac{d{u}^{\\alpha }}{ds} = - {a}^{\\beta \\gamma }{b}_{\\beta \\gamma }{x}_{\\gamma }^{i}\\frac{d{u}^{\\alpha }}{ds} \n\\]\n\n... | Yes |
Theorem 8.3.2 (Euler's theorem on normal curvature): If the lines of curvature are not indeterminant at a given point \( P \) on the surface and if \( \theta \) is the angle between a given direction and a principal direction at \( P \), then the normal curvature is given by the formula\n\n\[ \n{\chi }_{\left( n\right)... | Proof: We assume that \( P \) is not an umbilic. If the parametric curves are taken as the lines of curvature, then the principal curvatures are given by\n\n\[ \n{\chi }_{\left( 1\right) } = \frac{{b}_{11}}{{a}_{11}}\text{ and }{\chi }_{\left( 2\right) } = \frac{{b}_{22}}{{a}_{22}}.\n\]\n\nLet \( \theta \) be the angle... | Yes |
Theorem 8.3.4 The parametric curves are asymptotic lines if and only if\n\n\[ \n{b}_{11} = {b}_{22} = 0 \n\] | Proof: First, let the parametric curves \( {u}^{1} = \) constant and \( {u}^{2} = \) constant be asymptotic lines. Then from Eq. (8.39), we get\n\n\[ \n{b}_{\alpha \beta }d{u}^{\alpha }d{u}^{\beta } = {b}_{\alpha \beta }{\lambda }^{\alpha }{\lambda }^{\beta } = 0. \n\]\n\nFor \( {u}^{1} \) curve, i.e. \( d{u}^{2} = 0 \... | Yes |
Theorem 8.3.5 The torsion of an asymptotic line equals to \( \pm \sqrt{-\kappa } \), where \( \kappa \) is the Gaussian curvature of the surface. | Proof: From Eq. (8.6), we have, for an asymptotic curve,\n\n\[ \chi {\mu }^{i} = {b}_{\alpha \beta }{\lambda }^{\alpha }{\lambda }^{\beta }{\xi }^{i} + {\chi }_{g}{\zeta }^{i}, = {\chi }_{g}{\zeta }^{i};\;\text{ as }{b}_{\alpha \beta }{\lambda }^{\alpha }{\lambda }^{\beta } = 0, \]\n\n(8.42)\n\nwhere \( {\mu }^{i} \) i... | Yes |
Theorem 9.1.1 A necessary and sufficient condition that a force field \( {F}_{i} \), defined in a simply connected region, be conservative is that \( {F}_{i, j} = {F}_{j, i} \) . | Proof: First, let the force field \( {F}_{i} \) be conservative then \( {F}_{i} = - \frac{\partial V}{\partial {x}^{i}} \) . Now,\n\n\[ \n{F}_{i, j} = \frac{\partial {F}_{i}}{\partial {x}^{j}} - \left\{ \begin{matrix} k \\ i\;j \end{matrix}\right\} {F}_{k} = \frac{\partial }{\partial {x}^{j}}\left( {-\frac{\partial V}{... | Yes |
Theorem 9.1.2 (Integral of energy): The motion of a particle in a conservative field of force is such that the sum of its kinetic and potential energies is a constant. | Proof: Consider a particle moving on the curve\n\n\[ \mathcal{C} : {x}^{i} = {x}^{i}\left( t\right) ;{t}_{1} \leq t \leq {t}_{2} \]\n\nwhere \( t \) denotes the time. The kinetic energy \( T \) of the particle is given by\n\n\[ T\left( {{q}^{i},{\dot{q}}^{i}}\right) = \frac{1}{2}m{g}_{ij}{\dot{q}}^{i}{\dot{q}}^{j} = \f... | Yes |
Theorem 1.4. Let \( V \) be a vector space with basis \( {\mathbf{a}}_{1},\ldots ,{\mathbf{a}}_{m} \) . Let \( W \) be a vector space. Given any \( m \) vectors \( {\mathbf{b}}_{1},\ldots ,{\mathbf{b}}_{m} \) in \( W \), there is exactly one linear transformation \( T : V \rightarrow W \) such that, for all \( i, T\lef... | In fact, every linear transformation of \( {\mathbb{R}}^{m} \) to \( {\mathbb{R}}^{n} \) has this form. The proof is easy. Given \( T \), let \( {\mathbf{b}}_{1},\ldots ,{\mathbf{b}}_{m} \) be the vectors of \( {\mathbb{R}}^{n} \) such that \( T\left( {\mathbf{e}}_{j}\right) = {\mathbf{b}}_{j} \) . Then let \( A \) be ... | No |
Theorem 1.6. If \( B \) is the matrix obtained by applying an elementary row operation to \( A \), then \(\operatorname{rank}B = \operatorname{rank}A\text{.}\) | Gauss-Jordan reduction is the process of applying elementary operations to \( A \) to reduce it to a special form called echelon form (or stairstep form), for which the rank is obvious. An example of a matrix in this form is the following:\n\n\[ B = \left\lbrack \begin{matrix} \oplus & * & * & * & * & * \\ 0 & 0 & * & ... | No |
Theorem 2.1. Let \( A \) be an \( n \) by \( m \) matrix. Any elementary row operation on A may be carried out by premultiplying A by the corresponding elementary matrix. | Proof. One proceeds by direct computation. The effect of multiplying \( A \) on the left by the matrix \( E \) is to interchange rows \( {i}_{1} \) and \( {i}_{2} \) of \( A \) . Similarly, multiplying \( A \) by \( {E}^{\prime } \) has the effect of replacing row \( {i}_{1} \) by itself plus \( c \) times row \( {i}_{... | Yes |
Theorem 2.2. If \( A \) has both a left inverse \( B \) and a right inverse \( C \), then they are unique and equal. | Proof. Equality follows from the computation\n\n\[ C = {I}_{m} \cdot C = \left( {B \cdot A}\right) \cdot C = B \cdot \left( {A \cdot C}\right) = B \cdot {I}_{n} = B\text{.} \]\n\nIf \( {B}_{1} \) is another left inverse for \( A \), we apply this same computation with \( {B}_{1} \) replacing \( B \) . We conclude that ... | Yes |
Theorem 2.3. Let \( A \) be a matrix of size \( n \) by \( m \) . If \( A \) is invertible, then\n\n\[ n = m = \operatorname{rank}A\text{.} \] | Proof. Step 1. We show that for any \( k \) by \( n \) matrix \( D \) ,\n\n\[ \text{rank}\left( {D \cdot A}\right) \leq \text{rank}A\text{.} \]\n\nThe proof is easy. If \( R \) is a row matrix of size 1 by \( n \), then \( R \cdot A \) is a row matrix that equals a linear combination of the rows of \( A \), so it is an... | Yes |
Theorem 2.4. Let \( A \) be a matrix of size \( n \) by \( m \) . Suppose\n\n\[ n = m = \operatorname{rank}A\text{.} \]\n\nThen \( A \) is invertible; and furthermore, \( A \) equals a product of elementary matrices. | Proof. Step 1. We note first that every elementary matrix is invertible, and that its inverse is an elementary matrix. This follows from the fact that elementary operations are invertible. Alternatively, you can check directly that the matrix \( E \) corresponding to an operation of the first type is its own inverse, t... | Yes |
Theorem 2.5. If \( A \) is a square matrix and if \( B \) is a left inverse for \( A \) , then \( B \) is also a right inverse for \( A \) . | Proof. Since \( A \) has a left inverse, Step 2 of the proof of Theorem 2.3 implies that the rank of \( A \) equals the number of columns of \( A \) . Since \( A \) is square, this is the same as the number of rows of \( A \), so the preceding theorem implies that \( A \) has an inverse. By Theorem 2.2, this inverse mu... | Yes |
Theorem 2.6. Let \( A \) be an \( n \) by \( n \) matrix.\n\n(a) If \( E \) is the elementary matrix corresponding to the operation that exchanges rows \( {i}_{1} \) and \( {i}_{2} \), then \( \det \left( {E \cdot A}\right) = - \det A \) . | Proof. Property (a) is a restatement of Axiom 1, and (d) is a restatement of Axiom 3. Property (c) follows directly from linearity (Axiom 2); it states merely that\n\n\[ \text{det}{A}_{i}\left( {\lambda \mathrm{x}}\right) = \lambda \text{(det}{A}_{i}\left( \mathbf{x}\right) \text{).} \]\n\nNow we verify (b). Note first... | No |
Theorem 2.7. Let \( A \) be a square matrix. If the rows of \( A \) are independent, then det \( A \neq 0 \) ; if the rows are dependent, then det \( A = 0 \) . Thus an \( n \) by \( n \) matrix \( A \) has rank \( n \) if and only if \( \det A \neq 0 \) . | Proof. First, we note that if the \( {i}^{\text{th }} \) row of \( A \) is the zero row, then \( \det A = 0 \) . For multiplying row \( i \) by 2 leaves \( A \) unchanged; on the other hand, it must multiply the value of the determinant by 2 .\n\nSecond, we note that applying one of the elementary row operations to \( ... | Yes |
Theorem 2.8. Given a square matrix \( A \), let us reduce it to echelon form \( B \) by elementary row operations of types (1) and (2). If \( B \) has a zero row, then \( \det A = 0 \) . Otherwise, let \( k \) be the number of row exchanges involved in the reduction process. Then det \( A \) equals \( {\left( -1\right)... | Proof. If \( B \) has a zero row, then rank \( A < n \) and \( \det A = 0 \) . So suppose that \( B \) has no zero row. We know from (a) and (b) of Theorem 2.6 that \( \det A = ( - \) \( 1{)}^{k} \) det \( B \) . Furthermore, \( B \) must have the form\n\n\[ B = \left\lbrack \begin{matrix} {b}_{11} & * & \ldots & * \\ ... | Yes |
Corollary 2.9. The determinant function is uniquely characterized by its three axioms. It is also characterized by the four properties listed in Theorem 2.6. | Proof. The calculation of det \( A \) just given uses only properties (a)-(d) of Theorem 2.6. These in turn follow from the three axioms. | Yes |
Theorem 2.10. Let \( A \) and \( B \) be \( n \) by \( n \) matrices. Then\n\n\[ \det \left( {A \cdot B}\right) = \left( {\det A}\right) \cdot \left( {\det B}\right) \text{.} \] | Proof. Step 1. The theorem holds when \( A \) is an elementary matrix. Indeed:\n\n\[ \det \left( {E \cdot B}\right) = - \det B = \left( {\det E}\right) \left( {\det B}\right) \text{,} \]\n\n\[ \det \left( {{E}^{\prime } \cdot B}\right) = \det B = \left( {\det {E}^{\prime }}\right) \left( {\det B}\right) , \]\n\n\[ \tex... | Yes |
Theorem 2.11. \( \\det {A}^{\\mathrm{{tr}}} = \\det A \) . | Proof. Step 1. We show the theorem holds when \( A \) is an elementary matrix.\n\nLet \( E,{E}^{\\prime } \), and \( {E}^{\\prime \\prime } \) be elementary matrices of the three basic types. Direct inspection shows that \( {E}^{\\mathrm{{tr}}} = E \) and \( {\\left( {E}^{\\prime \\prime }\\right) }^{\\mathrm{{tr}}} = ... | Yes |
Theorem 2.13 (Cramer’s rule). Let \( A \) be an \( n \) by \( n \) matrix with successive columns \( {\mathbf{a}}_{1},\ldots ,{\mathbf{a}}_{n} \) . Let\n\n\[ \mathbf{x} = \left\lbrack \begin{matrix} {x}_{1} \\ \vdots \\ {x}_{n} \end{matrix}\right\rbrack \;\text{ and }\;\mathbf{c} = \left\lbrack \begin{matrix} {c}_{1} \... | Proof. Let \( {\mathbf{e}}_{1},\ldots ,{\mathbf{e}}_{n} \) be the standard basis for \( {\mathbb{R}}^{n} \), where each \( {\mathbf{e}}_{i} \) is written as a column matrix. Let \( C \) be the matrix\n\n\[ C = \left\lbrack {{\mathbf{e}}_{1}\ldots {\mathbf{e}}_{i - 1}\mathrm{x}{\mathbf{e}}_{i + 1}\ldots {e}_{n}}\right\r... | Yes |
Theorem 2.14. Let \( A \) be an \( n \) by \( n \) matrix of rank \( n \) ; let \( B = {A}^{-1} \) . Then\n\n\[ \n{b}_{ij} = \frac{{\left( -1\right) }^{j + i}\det {A}_{ji}}{\det A}.\n\] | Proof. Let \( j \) be fixed throughout this argument. Let\n\n\[ \n\mathbf{x} = \left\lbrack \begin{matrix} {x}_{1} \\ \vdots \\ {x}_{n} \end{matrix}\right\rbrack\n\]\n\ndenote the \( {j}^{\text{th }} \) column of the matrix \( B \) . The fact that \( A \cdot B = {I}_{n} \) implies in particular that \( A \cdot \mathbf{... | Yes |
Theorem 3.2. Let \( X \) be a metric space; let \( Y \) be a subspace. A subset A of \( Y \) is open in \( Y \) if and only if it has the form\n\n\[ A = U \cap Y \]\n\nwhere \( U \) is open in \( X \) . Similarly, a subset \( A \) of \( Y \) is closed in \( Y \) if and only if it has the form\n\n\[ A = C \cap Y \]\n\nw... | It follows that if \( A \) is open in \( Y \) and \( Y \) is open in \( X \), then \( A \) is open in \( X \) . Similarly, if \( A \) is closed in \( Y \) and \( Y \) is closed in \( X \), then \( A \) is closed in \( X \) . | No |
Theorem 4.1. A subspace \( X \) of \( {\mathbb{R}}^{n} \) is compact if and only if for every collection of sets open in \( X \) whose union is \( X \), there is a finite subcollection whose union equals \( X \) . | Proof. Suppose \( X \) is compact. Let \( \left\{ {A}_{\alpha }\right\} \) be a collection of sets open in \( X \) whose union is \( X \) . Choose, for each \( \alpha \), an open set \( {U}_{\alpha } \) of \( {\mathbb{R}}^{n} \) such that \( {A}_{\alpha } = {U}_{\alpha } \) \( \cap X \) . Since \( X \) is compact, some... | Yes |
Theorem 4.3. If \( X \) is a compact subspace of \( {\mathbb{R}}^{n} \), then \( X \) is closed and bounded. | Proof. Step 1. We show that \( X \) is bounded. For each positive integer \( N \), let \( {U}_{N} \) denote the open cube \( {U}_{N} = C\left( {\mathbf{0};N}\right) \) . Then \( {U}_{N} \) is an open set; and \( {U}_{1} \subset {U}_{2} \) \( \subset \cdots \) ; and the sets \( {U}_{N} \) cover all of \( {\mathbb{R}}^{n... | Yes |
Corollary 4.4. Let \( X \) be a compact subspace of \( \mathbb{R} \) . Then \( X \) has a largest element and a smallest element. | Proof. Since \( X \) is bounded, it has a greatest lower bound and a least upper bound. Since \( X \) is closed, these elements must belong to \( X \) . | Yes |
Theorem 4.5 (Extreme-value theorem). Let \( X \) be a compact subspace of \( {\mathbb{R}}^{m} \) . If \( f : X \rightarrow {\mathbb{R}}^{n} \) is continuous, then \( f\left( X\right) \) is a compact subspace of \( {\mathbb{R}}^{n} \) . | Proof. Let \( \left\{ {V}_{\alpha }\right\} \) be a collection of open sets of \( {\mathbb{R}}^{n} \) that covers \( f\left( X\right) \) . The sets \( {f}^{-1}\left( {V}_{\alpha }\right) \) form an open covering of \( X \) . Hence some finitely many of them cover \( X \), say for \( \alpha = {\alpha }_{1},\ldots ,{\alp... | Yes |
Theorem 4.6 (The \( \in \) -neighborhood theorem). Let \( X \) be a compact subspace of \( {\mathbb{R}}^{n} \) ; let \( U \) be an open set of \( {\mathbb{R}}^{n} \) containing \( X \) . Then there is an \( \in > \) \( 0 \) such that the \( \in \) -neighborhood of \( X \) (in either metric) is contained in \( U \) . | Proof. The \( \in \) -neighborhood of \( X \) in the euclidean metric is contained in the \( \in \) -neighborhood of \( X \) in the sup metric. Therefore it suffices to deal only with the latter case.\n\nStep 1. Let \( C \) be a fixed subset of \( {\mathbb{R}}^{n} \) . For each \( \mathbf{x} \in {\mathbb{R}}^{n} \), we... | Yes |
Theorem 4.7 (Uniform continuity). Let \( X \) be a compact subspace of \( {\mathbb{R}}^{\mathrm{m}} \) ; let \( f : X \rightarrow {\mathbb{R}}^{n} \) be continuous. Given \( \in > 0 \), there is a \( \delta > 0 \) such that whenever \( x, y \in X \)\n\n\[ \left| {\mathrm{x} - \mathrm{y}}\right| < \delta \;\text{ implie... | Proof. Consider the subspace \( X \times X \) of \( {\mathbb{R}}^{m} \times {\mathbb{R}}^{m} \) ; and within this, consider the space\n\n\[ \Delta = \{ \left( {\mathrm{x},\mathrm{x}}\right) \mid \mathrm{x} \in X\} , \]\n\nwhich is called the diagonal of \( X \times X \) . The diagonal is a compact subspace of \( {\math... | Yes |
Lemma 4.8. The rectangle\n\n\[ Q = \left\lbrack {{a}_{1},{b}_{1}}\right\rbrack \times \cdots \times \left\lbrack {{a}_{n},{b}_{n}}\right\rbrack \]\nin \( {\mathbb{R}}^{n} \) is compact. | Proof. We proceed by induction on \( n \) . The lemma is true for \( n = 1 \) ; we suppose it true for \( n - 1 \) and prove it true for \( n \) . We can write\n\n\[ Q = X \times \left\lbrack {{a}_{n},{b}_{n}}\right\rbrack \]\n\nwhere \( X \) is a rectangle in \( {\mathbb{R}}^{n - 1} \) . Then \( X \) is compact by the... | Yes |
Theorem 4.9. If \( X \) is a closed and bounded subspace of \( {\mathbb{R}}^{n} \), then \( X \) is compact. | Proof. Let \( \mathbf{A} \) be a collection of open sets that covers \( X \) . Let us adjoin to this collection the single set \( {\mathbb{R}}^{n} - X \), which is open in \( {\mathbb{R}}^{n} \) because \( X \) is closed. Then we have an open covering of all of \( {\mathbb{R}}^{n} \) . Because \( X \) is bounded, we ca... | Yes |
Theorem 4.11 (Intermediate-value theorem). Let \( X \) be connected. If \( f \) : \( X \) \( \rightarrow Y \) is continuous, then \( f\left( X\right) \) is a connected subspace of \( Y \). In particular, if \( f : X \rightarrow \mathbb{R} \) is continuous and if \( f\left( {x}_{0}\right) < r < f\left( {x}_{1}\right) \)... | Proof. Suppose \( f\left( X\right) = A \cup B \), where \( A \) and \( B \) are disjoint sets open in \( f\left( X\right) \). Then \( {f}^{-1}\left( A\right) \) and \( {f}^{-1}\left( B\right) \) are disjoint sets whose union is \( X \), and each is open in \( X \) because \( f \) is continuous. This contradicts connect... | Yes |
Theorem 5.1. Let \( A \subset {\mathbb{R}}^{m} \) ; let \( f : A \rightarrow {\mathbb{R}}^{n} \) . If \( f \) is differentiable at \( \mathbf{a} \), then all the directional derivatives of \( f \) at \( \mathbf{a} \) exist, and \[ {f}^{\prime }\left( {\mathrm{a};\mathrm{u}}\right) = {Df}\left( \mathrm{a}\right) \cdot \... | Proof. Let \( B = {Df}\left( \mathbf{a}\right) \) . Set \( \mathbf{h} = t\mathbf{u} \) in the definition of differentiability, where \( t \neq 0 \) . Then by hypothesis, \[ \frac{f\left( {\mathrm{a} + t\mathrm{u}}\right) - f\left( \mathrm{a}\right) - B \cdot t\mathrm{u}}{\left| t\mathrm{u}\right| } \rightarrow 0 \] \( ... | Yes |
Theorem 5.2. Let \( A \subset {\mathbb{R}}^{m} \) ; let \( f : A \rightarrow {\mathbb{R}}^{n} \) . If \( f \) is differentiable at \( \mathbf{a} \), then \( f \) is continuous at \( \mathbf{a} \) . | Proof. Let \( B = {Df}\left( \mathbf{a}\right) \) . For \( \mathbf{h} \) near \( \mathbf{0} \) but different from \( \mathbf{0} \), write\n\n\[ f\left( {\mathrm{a} + \mathrm{h}}\right) - f\left( \mathrm{a}\right) = \left| \mathrm{h}\right| \left\lbrack \frac{f\left( {\mathrm{a} + \mathrm{h}}\right) - f\left( \mathrm{a}... | Yes |
Theorem 5.3. Let \( A \subset {\mathbb{R}}^{m} \) ; let \( f : A \rightarrow \mathbb{R} \) . If \( f \) is differentiable at \( \mathbf{a} \), then\n\n\[ \n{Df}\left( \mathrm{a}\right) = \left\lbrack \begin{array}{llll} {D}_{1}f\left( \mathrm{a}\right) & {D}_{2}f\left( \mathrm{a}\right) & \ldots & {D}_{m}f\left( \mathr... | Proof. By hypothesis, \( {Df}\left( \mathbf{a}\right) \) exists and is a matrix of size 1 by \( m \) . Let\n\n\[ \n{Df}\left( \mathrm{a}\right) = \left\lbrack \begin{array}{llll} {\lambda }_{1} & {\lambda }_{2} & \ldots & {\lambda }_{m} \end{array}\right\rbrack \text{.}\n\]\n\nIt follows (using Theorem 5.1) that\n\n\[ ... | Yes |
Theorem 5.4. Let \( A \subset {\mathbb{R}}^{m} \) ; let \( f : A \rightarrow {\mathbb{R}}^{n} \) . Suppose \( A \) contains a neighborhood of \( \mathbf{a} \) . Let \( {f}_{i} : A \rightarrow \mathbb{R} \) be the \( {i}^{\text{th }} \) component function of \( f \), so that\n\n\[ f\left( \mathrm{x}\right) = \left\lbrac... | Proof. Let \( B \) be an arbitrary \( n \) by \( m \) matrix. Consider the function\n\n\[ F\left( \mathbf{h}\right) = \frac{f\left( {\mathrm{a} + \mathbf{h}}\right) - f\left( \mathrm{a}\right) - B \cdot \mathbf{h}}{\left| \mathbf{h}\right| },\]\n\nwhich is defined for \( 0 < \left| \mathbf{h}\right| < \in \) (for some ... | Yes |
Theorem 6.1 (Mean-value theorem). If \( \phi : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is continuous at each point of the closed interval \( \left\lbrack {a, b}\right\rbrack \), and differentiable at each point of the open interval \( \left( {a, b}\right) \), then there exists a point \( c \) of \( \... | \[ \varphi \left( b\right) - \varphi \left( a\right) = {\varphi }^{\prime }\left( c\right) \left( {b - a}\right) . \] | No |
Theorem 6.3. Let \( A \) be open in \( {\mathbb{R}}^{m} \) ; let \( f : A \rightarrow \mathbb{R} \) be a function of class \( {C}^{2} \) . Then for each \( \mathbf{a} \in A \) ,\n\n\[ \n{D}_{k}{D}_{j}f\left( \mathrm{a}\right) = {D}_{j}{D}_{k}f\left( \mathrm{a}\right) .\n\] | Proof. Since one calculates the partial derivatives in question by letting all variables other than \( {x}_{k} \) and \( {x}_{j} \) remain constant, it suffices to consider the case where \( f \) is a function merely of two variables. So we assume that \( A \) is open in \( {\mathbb{R}}^{2} \), and that \( f : A \right... | No |
Theorem 7.1. Let \( A \subset {\mathbb{R}}^{m} \) ; let \( B \subset {\mathbb{R}}^{n} \) . Let\n\n\[ f : A \rightarrow {\mathbb{R}}^{n}\;\text{ and }\;g : B \rightarrow {\mathbb{R}}^{p}, \]\n\nwith \( f\left( A\right) \subset B \) . Suppose \( f\left( \mathbf{a}\right) = \mathbf{b} \) . If \( f \) is differentiable at ... | Proof. For convenience, let \( \mathbf{x} \) denote the general point of \( {\mathbb{R}}^{m} \), and let \( \mathbf{y} \) denote the general point of \( {\mathbb{R}}^{n} \).\n\nBy hypothesis, \( g \) is defined in a neighborhood of \( \mathbf{b} \) ; choose \( \in \) so that \( g\left( \mathbf{y}\right) \) is defined f... | Yes |
Corollary 7.2. Let \( A \) be open in \( {\mathbb{R}}^{m} \) ; let \( B \) be open in \( {\mathbb{R}}^{n} \) . Let\n\n\[ f : A \rightarrow {\mathbb{R}}^{n}\;\text{ and }\;g : B \rightarrow {\mathbb{R}}^{p},\]\n\nwith \( f\left( A\right) \subset B \) . If \( f \) and \( g \) are of class \( {C}^{r} \), so is the composi... | Proof. The chain rule gives us the formula\n\n\[ D\left( {g \circ f}\right) \left( \mathrm{x}\right) = {Dg}\left( {f\left( \mathrm{x}\right) }\right) \cdot {Df}\left( \mathrm{x}\right) ,\]\n\nwhich holds for \( \mathbf{x} \in A \) .\n\nSuppose first that \( f \) and \( g \) are of class \( {C}^{1} \) . Then the entries... | Yes |
Theorem 7.3 (Mean-value theorem). Let \( A \) be open in \( {\mathbb{R}}^{m} \) ; let \( f : A \rightarrow \mathbb{R} \) be differentiable on A. If A contains the line segment with end points \( \mathbf{a} \) and \( \mathbf{a} \) + \( \mathbf{h} \), then there is a point \( \mathbf{c} = \mathbf{a} + {t}_{0}\mathbf{h} \... | Proof. Set \( \phi \left( t\right) = f\left( {\mathbf{a} + t\mathbf{h}}\right) \) ; then \( \phi \) is defined for \( t \) in an open interval about \( \left\lbrack {0,1}\right\rbrack \) . Being the composite of differentiable functions, \( \phi \) is differentiable; its derivative is given by the formula\n\n\[ \dot{\v... | Yes |
Theorem 7.4. Let \( A \) be open in \( {\mathbb{R}}^{n} \) ; let \( f : A \rightarrow {\mathbb{R}}^{n} \) ; let \( f\left( \mathbf{a}\right) = \mathbf{b} \) . Suppose that \( g \) maps a neighborhood of \( \mathbf{b} \) into \( {\mathbb{R}}^{n} \), that \( g\left( \mathbf{b}\right) = \mathbf{a} \), and\n\n\[ g\left( {f... | Proof. Let \( i : {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{n} \) be the identity function; its derivative is the identity matrix \( {I}_{n} \) . We are given that\n\n\[ g\left( {f\left( \mathrm{x}\right) }\right) = i\left( \mathrm{x}\right) \]\n\nfor all \( \mathbf{x} \) in a neighborhood of \( \mathbf{a} \) . The ch... | Yes |
Lemma 8.1. Let \( A \) be open in \( {\mathbb{R}}^{n} \) ; let \( f : A \rightarrow {\mathbb{R}}^{n} \) be of class \( {C}^{1} \) . If \( {Df}\left( \mathbf{a}\right) \) is non-singular, then there exists an \( \alpha > 0 \) such that the inequality\n\n\[ \left| {f\left( {\mathbf{x}}_{0}\right) - f\left( {\mathbf{x}}_{... | Proof. Let \( E = {Df}\left( \mathbf{a}\right) \) ; then \( E \) is non-singular. We first consider the linear transformation that maps \( \mathbf{x} \) to \( E \cdot \mathbf{x} \) . We compute\n\n\[ \left| {{\mathbf{x}}_{0} - {\mathbf{x}}_{1}}\right| = \left| {{E}^{-1} \cdot \left( {E \cdot {\mathbf{x}}_{0} - E \cdot ... | Yes |
Theorem 8.3 (The inverse function theorem). Let \( A \) be open in \( {\mathbb{R}}^{n} \) ; let \( f \) : \( A \rightarrow {\mathbb{R}}^{n} \) be of class \( {C}^{r} \) . If \( {Df}\left( \mathbf{x}\right) \) is non-singular at the point \( \mathbf{a} \) of \( A \), there is a neighborhood \( U \) of the point \( \math... | Proof. By Lemma 8.1, there is a neighborhood \( {U}_{0} \) of a on which \( f \) is one-to-one. Because \( \det {Df}\left( \mathbf{x}\right) \) is a continuous function of \( \mathbf{x} \), and \( \det {Df}\left( \mathbf{a}\right) \neq 0 \) , there is a neighborhood \( {U}_{1} \) of a such that \( \det {Df}\left( \math... | Yes |
Theorem 9.1. Let \( A \) be open in \( {\mathbb{R}}^{k + n} \) ; let \( f : A \rightarrow {\mathbb{R}}^{n} \) be differentiable. Write \( f \) in the form \( f\left( {\mathbf{x},\mathbf{y}}\right) \), for \( \mathbf{x} \in {\mathbb{R}}^{k} \) and \( \mathbf{y} \in {\mathbb{R}}^{n} \) ; then \( D \) f has the form\n\n\[... | Proof. Given \( g \), let us define \( h : B \rightarrow {\mathbb{R}}^{k + n} \) by the equation\n\n\[ \nh\left( x\right) = \left( {x, g\left( x\right) }\right) .\n\]\n\nThe hypotheses of the theorem imply that the composite function\n\n\[ \nH\left( \mathbf{x}\right) = f\left( {h\left( \mathbf{x}\right) }\right) = f\le... | Yes |
Theorem 9.2 (Implicit function theorem). Let \( A \) be open in \( {\mathbb{R}}^{k + n} \) ; let \( f : A \) \( \rightarrow {\mathbb{R}}^{n} \) be of class \( {C}^{r} \) . Write \( f \) in the form \( f\left( {\mathbf{x},\mathbf{y}}\right) \), for \( \mathbf{x} \in {\mathbb{R}}^{k} \) and \( \mathbf{y} \in {\mathbb{R}}... | Proof. We construct a function \( F \) to which we can apply the inverse function theorem. Define \( F : A \rightarrow {\mathbb{R}}^{k + n} \) by the equation \[ F\left( {\mathbf{x},\mathbf{y}}\right) = \left( {\mathbf{x}, f\left( {\mathbf{x},\mathbf{y}}\right) }\right) . \] Then \( F \) maps the open set \( A \) of \(... | Yes |
Lemma 10.1. Let \( P \) be a partition of the rectangle \( Q \) ; let \( f : Q \rightarrow \mathbb{R} \) be a bounded function. If \( {P}^{\prime \prime } \) is a refinement of \( P \), then\n\n\[ L\left( {f, P}\right) \leq L\left( {f,{P}^{\prime \prime }}\right) \text{ and }U\left( {f,{P}^{\prime \prime }}\right) \leq... | Proof. Let \( Q \) be the rectangle\n\n\[ Q = \left\lbrack {{a}_{1},{b}_{1}}\right\rbrack \times \cdots \times \left\lbrack {{a}_{n},{b}_{n}}\right\rbrack . \]\n\nIt suffices to prove the lemma when \( {P}^{\prime \prime } \) is obtained by adjoining a single additional point to the partition of one of the component in... | Yes |
Lemma 10.2. Let \( Q \) be a rectangle; let \( f : Q \rightarrow \mathbb{R} \) be a bounded function. If \( P \) and \( {P}^{\prime } \) are any two partitions of \( Q \), then\n\n\[ L\left( {f, P}\right) \leq U\left( {f,{P}^{\prime }}\right) . \] | Proof. In the case where \( P = {P}^{\prime } \), the result is obvious: For any subrectangle \( R \) determined by \( P \), we have \( {m}_{R}\left( f\right) \leq {M}_{R}\left( f\right) \) . Multiplying by \( v\left( R\right) \) and summing gives the desired inequality.\n\nIn general, given partitions \( P \) and \( {... | Yes |
Theorem 10.3 (The Riemann condition). Let \( Q \) be a rectangle; let \( f \) : \( Q \) \( \rightarrow \) R be a bounded function. Then\n\n\[ \n{\int }_{Q}f \leq {\int }_{Q}f \n\]\n\nequality holds if and only if given \( \in > 0 \), there exists a corresponding partition \( P \) of \( Q \) for which\n\n\[ \nU\left( {f... | Proof. Let \( {P}^{\prime } \) be a fixed partition of \( Q \) . It follows from the fact that \( L\left( {f, P}\right) \leq \) \( U\left( {f,{P}^{\prime }}\right) \) for every partition \( P \) of \( Q \), that\n\n\[ \n{\int }_{\underline{Q}}f \leq U\left( {f,{P}^{\prime }}\right) .\n\]\n\nNow we use the fact that \( ... | Yes |
Theorem 10.4. Every constant function \( f\left( x\right) = c \) is integrable. Indeed, if \( Q \) is a rectangle and if \( P \) is a partition of \( Q \), then\n\n\[{\int }_{Q}c = c \cdot v\left( Q\right) = c\mathop{\sum }\limits_{R}v\left( R\right) ,\]\n\nwhere the summation extends over all subrectangles determined ... | Proof. If \( R \) is a subrectangle determined by \( P \), then \( {m}_{R}\left( f\right) = c = {M}_{R}\left( f\right) \) . It follows that\n\n\[L\left( {f, P}\right) = c\mathop{\sum }\limits_{R}v\left( R\right) = U\left( {f, P}\right) ,\]\n\nso the Riemann condition holds trivially. Thus \( {\int }_{Q}c \) exists; sin... | Yes |
Corollary 10.5. Let \( Q \) be a rectangle in \( {\mathbb{R}}^{n} \) ; let \( \left\{ {{Q}_{1},\ldots ,{Q}_{k}}\right\} \) be a finite collection of rectangles that covers \( Q \) . Then\n\n\[ v\left( Q\right) \leq \mathop{\sum }\limits_{{i = 1}}^{k}v\left( {Q}_{i}\right) \] | Proof. Choose a rectangle \( {Q}^{\prime } \) containing all the rectangles \( {Q}_{1},\ldots ,{Q}_{k} \) . Use\nthe end points of the component intervals of the rectangles \( Q,{Q}_{1},\ldots ,{Q}_{k} \) to define a partition \( P \) of \( {Q}^{\prime } \) . Then each of the rectangles \( Q,{Q}_{1},\ldots ,{Q}_{k} \) ... | Yes |
Theorem 11.1. (a) If B \( \subset \) A and A has measure zero in \( {\mathbb{R}}^{n} \) , then so does B. (b) Let \( A \) be the union of the countable collection of sets \( {A}_{1},{A}_{2},\ldots \) If each \( {A}_{i} \) has measure zero in \( {\mathbb{R}}^{n} \) ; so does \( A \) . | Proof. (a) is immediate. To prove (b), cover the set \( {A}_{j} \) by countably many rectangles\n\n\[ {Q}_{1j},{Q}_{2j},\;{Q}_{3j},\;\ldots \]\n\nof total volume less than \( \in /{2}^{j} \) Do this for each \( j \) . Then the collection of rectangles \( \left\{ {Q}_{ij}\right\} \) is countable, it covers \( A \), and ... | Yes |
Theorem 11.2. Let \( Q \) be a rectangle in \( {\mathbb{R}}^{n} \); let \( f : Q \rightarrow \mathbb{R} \) be a bounded function. Let \( D \) be the set of points of \( Q \) at which \( f \) fails to be continuous. Then \( {\int }_{Q}f \) exists if and only if \( D \) has measure zero in \( {\mathbb{R}}^{n} \). | Proof. Choose \( M \) so that \( \left| {f\left( \mathbf{x}\right) }\right| \leq M \) for \( \mathbf{x} \in Q \). Step 1. We prove the \ | No |
Theorem 11.3. Let \( Q \) be a rectangle in \( {\mathbb{R}}^{n} \); let \( f : Q \rightarrow \mathbb{R} \); assume \( f \) is integrable over \( Q \). (a) If \( f \) vanishes except on a set of measure zero, then \( {\int }_{Q}f = 0 \). | Proof. (a) Suppose \( f \) vanishes except on a set \( E \) of measure zero. Let \( P \) be a partition of \( Q \). If \( R \) is a subrectangle determined by \( P \), then \( R \) is not contained in \( E \), so that \( f \) vanishes at some point of \( R \). Then \( {m}_{R}\left( f\right) \leq 0 \) and \( {M}_{R}\lef... | Yes |
Corollary 12.3. Let \( Q = A \times B \), where \( A \) is a rectangle in \( {\mathbb{R}}^{k} \) and \( B \) is a rectangle in \( {\mathbb{R}}^{n} \) . Let \( f : Q \rightarrow \mathbb{R} \) be a bounded function. If \( {\int }_{Q}f \) exists, and if \( {\int }_{y \in B}f\left( {\mathbf{x},\mathbf{y}}\right) \) exists ... | \[ {\int }_{Q}f = {\int }_{\mathrm{x} \in A}{\int }_{\mathrm{y} \in B}f\left( {\mathrm{x},\mathrm{y}}\right) . \] | Yes |
Lemma 13.1. Let \( Q \) and \( {Q}^{\prime } \) be two rectangles in \( {\mathbb{R}}^{n} \) . If \( f : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) is a bounded function that vanishes outside \( Q \cap {Q}^{\prime } \), then\n\n\[ \n{\int }_{Q}f = {\int }_{{Q}^{\prime }}f \n\]\n\none integral exists if and only if the o... | Proof. We consider first the case where \( Q \subset {Q}^{\prime } \) . Let \( E \) be the set of points of Int \( Q \) at which \( f \) fails to be continuous. Then both the maps \( f : Q \rightarrow \mathbb{R} \) and \( f \) \( : {Q}^{\prime } \rightarrow \mathbb{R} \) are continuous except at points of \( E \) and p... | Yes |
Lemma 13.2. Let \( S \) be a subset of \( {\mathbb{R}}^{n} \) ; let \( f, g : S \rightarrow {\mathbb{R}}^{n} \) . Let \( F, G : S \rightarrow {\mathbb{R}}^{n} \) be defined by the equations\n\n\[ F\left( \mathbf{x}\right) = \max \{ f\left( \mathbf{x}\right), g\left( \mathbf{x}\right) \} \;\text{ and }\;G\left( \mathbf{... | Proof. (a) Suppose \( f \) and \( g \) are continuous at \( {\mathbf{x}}_{0} \) . Consider first the case in which \( f\left( {\mathbf{x}}_{0}\right) = g\left( {\mathbf{x}}_{0}\right) = r \) . Then \( F\left( {\mathbf{x}}_{0}\right) = G\left( {\mathbf{x}}_{0}\right) = r \) . By continuity, given \( \in > 0 \) , we can ... | Yes |
(a) (Linearity). If \( f \) and \( g \) are integrable over \( S \), so is \( af + bg \), and\n\n\[ \n{\int }_{S}\left( {{af} + {bg}}\right) = a{\int }_{S}f + b{\int }_{S}g.\n\] | Proof. (a) It suffices to prove this result for the integral over a rectangle, since\n\n\[ \n{\left( af + bg\right) }_{S} = a{f}_{S} + b{g}_{S}.\n\]\n\nSo suppose \( f \) and \( g \) are integrable over \( Q \). Then \( f \) and \( g \) are continuous except on sets \( D, E \), respectively, of measure zero. It follows... | Yes |
Corollary 13.4. Let \( {S}_{1},\ldots ,{S}_{k} \) be bounded sets in \( {\mathbb{R}}^{n} \) ; assume \( {S}_{i} \cap {S}_{j} \) has measure zero whenever \( i \neq j \) . Let \( S = {S}_{1} \cup \cdots \cup {S}_{k} \) . If \( f : S \rightarrow \mathbb{R} \) is integrable over each set \( {S}_{i} \), then \( f \) is int... | Proof. The case \( k = 2 \) follows from additivity, since the integral of \( f \) over \( {S}_{1} \cap {S}_{2} \) vanishes by Theorem 11.3. The general case follows by induction. | No |
Theorem 13.5. Let \( S \) be a bounded set in \( {\mathbb{R}}^{n} \) ; let \( f : S \rightarrow \mathbb{R} \) be a bounded continuous function. Let \( E \) be the set of points \( {\mathbf{x}}_{0} \) of Bd \( S \) for which the condition\n\n\[ \mathop{\lim }\limits_{{\mathrm{x} \rightarrow {\mathrm{x}}_{0}}}f\left( \ma... | Proof. Let \( {\mathbf{x}}_{0} \) be a point of \( {\mathbb{R}}^{n} \) not in \( E \) . We show that the function \( {f}_{S} \) is continuous at \( {\mathbf{x}}_{0} \) ; the theorem follows.\n\nIf \( {\mathbf{x}}_{0} \in \) Int \( S \), then the functions \( f \) and \( {f}_{S} \) agree in a neighborhood of \( {\mathbf... | Yes |
Theorem 13.6. Let \( S \) be a bounded set in \( {\mathbb{R}}^{n} \) ; let \( f : S \rightarrow \mathbb{R} \) be a bounded continuous function; let \( A = \operatorname{Int}S \) . If \( f \) is integrable over \( S \), then \( f \) is integrable over \( A \), and \( {\int }_{S}f = {\int }_{A}f \) . | Proof. Step 1. We show that if \( {f}_{S} \) is continuous at \( {\mathbf{x}}_{0} \), then \( {f}_{A} \) is continuous, and agrees with \( {f}_{S} \), at \( {\mathbf{x}}_{0} \) . The proof is easy. If \( {\mathbf{x}}_{0} \in \) Int \( S \) or \( {\mathbf{x}}_{0} \in \) Ext \( S \), then \( {f}_{S} \) and \( {f}_{A} \) ... | Yes |
Theorem 14.1. A subset \( S \) of \( {\mathbb{R}}^{n} \) is rectifiable if and only if \( S \) is bounded and \( \mathrm{{Bd}}S \) has measure zero. | Proof. The function \( {1}_{S} \) that equals 1 on \( S \) and 0 outside \( S \) is continuous on the open sets Ext \( S \) and Int \( S \) . It fails to be continuous at each point of Bd \( S \) . By Theorem 11.2, the function \( {1}_{S} \) is integrable over a rectangle \( Q \) containing \( S \) if and only if \( \m... | Yes |
Theorem 14.2. (a) (Positivity). If \( S \) is rectifiable, \( v\left( S\right) \geq 0 \) . | Proof. Parts (a), (b), and (c) follow from Theorem 13.3. Part (d) follows by applying Theorem 11.3 to the non-negative function \( {1}_{S} \) . Part (e) follows from Theorem 13.6, and (f) from Theorem 13.5. | No |
Lemma 15.1. Let \( A \) be an open set in \( {\mathbb{R}}^{n} \). Then there exists a sequence \( {C}_{1} \), \( {C}_{2},\ldots \) of compact rectifiable subsets of \( A \) whose union is \( A \), such that \( {C}_{N} \subset \) Int \( {C}_{N + 1} \) for each \( N \). | Proof. Let \( d \) denote the sup metric \( d\left( {\mathbf{x},\mathbf{y}}\right) = \left| {\mathbf{x} - \mathbf{y}}\right| \) on \( {\mathbb{R}}^{n} \). If \( B \subset {\mathbb{R}}^{n} \), let \( d\left( {\mathbf{x}, B}\right) \) denote the distance from \( \mathbf{x} \) to \( B \), as usual. (See \( §4 \).)\n\nNow ... | Yes |
Theorem 15.2. Let \( A \) be open in \( {\mathbb{R}}^{n} \) ; let \( f : A \rightarrow \mathbb{R} \) be continuous. Choose a sequence \( {C}_{N} \) of compact rectifiable subsets of \( A \) whose union is \( A \) such that \( {C}_{N} \subset \) Int \( {C}_{N + 1} \) for each \( N \) . Then \( f \) is integrable over \(... | Proof. Step 1. We prove the theorem first in the case where \( f \) is nonnegative. Here \( f = \left| f\right| \) . Since the sequence \( {\int }_{{C}_{N}}f \) is increasing (by monotonicity), it converges if and only if it is bounded.\n\nSuppose first that \( f \) is integrable over \( A \) . If we let \( D \) range ... | Yes |
Theorem 15.3. Let \( A \) be an open set in \( {\mathbb{R}}^{n} \). Let \( f, g : A \rightarrow \mathbb{R} \) be continuous functions.\n\n(a) (Linearity). If \( f \) and \( g \) are integrable over \( A \), so is \( af + bg \); and\n\n\[ \n{\int }_{A}\left( {{af} + {bg}}\right) = a{\int }_{A}f + b{\int }_{A}g.\n\] | Proof. Let \( {C}_{N} \) be a sequence of compact rectifiable sets whose union is \( A \), such that \( {C}_{N} \subset \) Int \( {C}_{N + 1} \) for all \( N \).\n\n(a) We have\n\n\[ \n{\int }_{{C}_{N}}\left| {{af} + {bg}}\right| \leq \left| a\right| {\int }_{{C}_{N}}\left| f\right| + \left| b\right| {\int }_{{C}_{N}}\... | Yes |
Theorem 15.4. Let \( A \) be a bounded open set in \( {\mathbb{R}}^{n} \) ; let \( f : A \rightarrow \mathbb{R} \) be a bounded continuous function. Then the extended integral \( {\int }_{A}f \) exists. If the ordinary integral \( {\int }_{A}f \) also exists, then these two integrals are equal. | Proof. Let \( Q \) be a rectangle containing \( A \) .\n\nStep 1. We show the extended integral of \( f \) exists. Choose \( M \) so that \( \left| {f\left( x\right) }\right| \leq \) \( M \) for \( \mathbf{x} \in A \) . Then for any compact rectifiable subset \( D \) of \( A \) ,\n\n\[{\int }_{D}\left| f\right| \leq {\... | Yes |
Corollary 15.5. Let \( S \) be a bounded set in \( {\mathbb{R}}^{n} \) ; let \( f : S \rightarrow \mathbb{R} \) be a bounded continuous function. If \( f \) is integrable over \( S \) in the ordinary sense, then\n\n\[ \n\text{(ordinary)}{\int }_{S}f = \text{(extended)}{\int }_{\operatorname{Int}S}f\text{.\n\] | Proof. One applies Theorems 13.6 and 15.4. | No |
Lemma 16.1. Let \( Q \) be a rectangle in \( {\mathbb{R}}^{n} \). There is a \( {C}^{\infty } \) function \( \phi \) : \( {\mathbb{R}}^{n} \rightarrow \) R such that \( \phi \left( \mathbf{x}\right) > 0 \) for \( \mathbf{x} \in \) Int \( Q \) and \( \phi \left( \mathbf{x}\right) = 0 \) otherwise. | Proof. Let \( f : \mathbb{R} \rightarrow \mathbb{R} \) be defined by the equation\n\n\[ f\left( x\right) = \left\{ \begin{matrix} {e}^{-1/x} & \text{ if }x > 0, \\ 0 & \text{ otherwise }. \end{matrix}\right. \]\n\nThen \( f\left( x\right) > 0 \) for \( x > 0 \). It is a standard result of single-variable analysis that ... | No |
Lemma 16.2. Let \( \mathbf{A} \) be a collection of open sets in \( {\mathbb{R}}^{n} \) ; let \( A \) be their union. Then there exists a countable collection \( {Q}_{1},{Q}_{2},\ldots \) of rectangles contained in A such that:\n\n(1) The sets Int \( {Q}_{i} \) cover \( A \) .\n\n(2) Each \( {Q}_{i} \) is contained in ... | Proof. It is not difficult to find rectangles \( {Q}_{i} \) satisfying (1) and (2). Choosing them so they also satisfy (3), the so-called \ | No |
Theorem 16.3 (Existence of a partition of unity). Let A be a collection of open sets in \( {\mathbb{R}}^{n} \) ; let \( A \) be their union. There exists a sequence \( {\phi }_{1},{\phi }_{2},\ldots \) of continuous functions \( {\phi }_{i} : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) such that:\n\n1. \( {\phi }_{i}\le... | Proof. Given \( A \) and \( A \), let \( {Q}_{1},{Q}_{2},\ldots \) be a sequence of rectangles in \( A \) satisfying the conditions stated in Lemma 16.2. For each \( i \), let \( {\psi }_{i} : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) be a \( {C}^{\infty } \) function that is positive on Int \( {Q}_{i} \) and zero els... | Yes |
Lemma 16.4. Let \( A \) be open in \( {\mathbb{R}}^{n} \); let \( f : A \rightarrow \mathbb{R} \) be continuous. If \( f \) vanishes outside the compact subset \( C \) of \( A \), then the integrals \( {\int }_{A}f \) and \( {\int }_{C}f \) exist and are equal. | Proof. The integral \( {\int }_{C}f \) exists because \( C \) is bounded, and the function \( {f}_{c} \) , which equals \( f \) on \( A \) and vanishes outside \( C \), is continuous and bounded on all of \( {\mathbb{R}}^{n} \). Let \( {C}_{i} \) be a sequence of compact rectifiable sets whose union is \( A \), such th... | Yes |
Theorem 17.2 (Change of variables theorem). Let \( g : A \rightarrow B \) be a diffeomorphism of open sets in \( {\mathbb{R}}^{n} \) . Let \( f : B \rightarrow \mathbb{R} \) be a continuous function. Then \( f \) is integrable over \( B \) if and only if the function \( \left( {f \circ g}\right) \mid \det {Dg} \mid \) ... | \[ {\int }_{B}f = {\int }_{A}\left( {f \circ g}\right) \left| {\det {Dg}}\right| . \] | No |
Theorem 18.2. Let \( g : A \rightarrow B \) be a diffeomorphism of class \( {C}^{r} \), where \( A \) and \( B \) are open sets in \( {\mathbb{R}}^{n} \). Let \( D \) be a compact subset of \( A \), and let \( E = g\left( D\right) \). (a) We have \( g\left( {\operatorname{Int}D}\right) = \operatorname{Int}E \) and \( g... | Proof. (a) The map \( {g}^{-1} \) is continuous. Therefore, for any open set \( U \) contained in \( A \), the set \( g\left( U\right) \) is an open set contained in \( B \). In particular, \( g( \) Int \( D) \) is an open set in \( {\mathbb{R}}^{n} \) contained in the set \( g\left( D\right) = E \). Thus \[ g\left( {\... | Yes |
Lemma 19.2. Let \( g : A \rightarrow B \) be a diffeomorphism of open sets in \( {\mathbb{R}}^{n} \) ; let \( f \) : \( B \rightarrow \mathbb{R} \) be continuous. If \( \left( {f \circ g}\right) \left| {\det {Dg}}\right| \) is integrable over \( A \), then \( f \) is integrable over B. | Proof. We apply the lemma just proved to the diffeomorphism \( {g}^{-1} : B \rightarrow \) \( A \) . The function \( F = \left( {f \circ g}\right) \left| {\det {Dg}}\right| \) is continuous on \( A \), and is integrable over A by hypothesis. It follows from Lemma 19.1 that the function\n\n\[ \left( {F \circ {g}^{-1}}\r... | Yes |
Theorem 20.1. Let \( A \) be an \( n \) by \( n \) matrix. Let \( h : {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{n} \) be the linear transformation \( h\left( \mathbf{x}\right) = A \cdot \mathbf{x} \) . Let \( S \) be a rectifiable set in \( {\mathbb{R}}^{n} \), and let \( T = h\left( S\right) \) . Then \[ v\left( T\ri... | Proof. Consider first the case where \( A \) is non-singular. Then \( h \) is a diffeomorphism of \( {\mathbb{R}}^{n} \) with itself; \( h \) carries Int \( S \) onto Int \( T \) ; and \( T \) is rectifiable. We have \[ v\left( T\right) = v\left( {\text{ Int }T}\right) = {\int }_{\text{Int }T}1 = {\int }_{\text{Int }S}... | Yes |
Theorem 20.2. Let \( {\mathbf{a}}_{1},\ldots ,{\mathbf{a}}_{n} \) be \( n \) independent vectors in \( {\mathbb{R}}^{n} \) . Let \( A = \) \( \left\lbrack {{\mathbf{a}}_{1}\ldots {\mathbf{a}}_{n}}\right\rbrack \) be the \( n \) by \( n \) matrix with columns \( {\mathbf{a}}_{1},\ldots ,{\mathbf{a}}_{n} \) . Then\n\n\[ ... | Proof. Consider the linear transformation \( h : {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{n} \) given by \( h\left( \mathbf{x}\right) = A \) . \( \mathbf{x} \) . Then \( h \) carries the unit basis vectors \( {\mathbf{e}}_{1},\ldots ,{\mathbf{e}}_{n} \) to the vectors \( {\mathbf{a}}_{1},\ldots ,{\mathbf{a}}_{n} \), ... | Yes |
Theorem 20.3. Let \( C \) be a non-singular \( n \) by \( n \) matrix. Let \( h : {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{n} \) be the linear transformation \( h\left( \mathbf{x}\right) = C \cdot \mathbf{x} \) . Let \( \left( {{\mathbf{a}}_{1}\ldots ,{\mathbf{a}}_{n}}\right) \) be a frame in \( {\mathbb{R}}^{n} \) .... | Proof. Let \( {\mathbf{b}}_{i} = h\left( {\mathbf{a}}_{i}\right) \) for each \( i \) . Then\n\n\[ C \cdot \left\lbrack {{\mathbf{a}}_{1}\ldots {\mathbf{a}}_{n}}\right\rbrack = \left\lbrack {{\mathbf{b}}_{1}\ldots {\mathbf{b}}_{n}}\right\rbrack \]\n\nso that\n\n\[ \text{(det}C\text{) .}\det \left\lbrack {{\mathbf{a}}_{1... | Yes |
Theorem 20.4. Let A, B, C be orthogonal n by n matrices. Then:\n\n(a) \( A \cdot B \) is orthogonal\n\n(b) \( A.\left( {B.C}\right) = \left( {A.B}\right) .C \) .\n\n(c) There is an orthogonal matrix \( {I}_{n} \) such that \( A \cdot {I}_{n} = {I}_{n} \cdot A = A \) for all orthogonal A.\n\n(d) Given \( A \), there is ... | Proof. To check (a), we compute\n\n\[ \n{\left( A \cdot B\right) }^{\mathrm{{tr}}} \cdot \left( {A \cdot B}\right) = \left( {{B}^{\mathrm{{tr}}} \cdot {A}^{\mathrm{{tr}}}}\right) \cdot \left( {A \cdot B}\right) \n\]\n\n\[ \n= {B}^{\mathrm{{tr}}} \cdot B = {I}_{n}\text{.} \n\]\n\nCondition (b) is immediate and (c) follo... | Yes |
Theorem 20.5. Let \( h : {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{n} \) be a map such that \( h\left( 0\right) = 0 \). (a) The map \( h \) is an isometry if and only if it preserves dot products. | Proof. (a) Given \( \mathbf{x} \) and \( \mathbf{y} \), we compute:\n\n\[ \parallel h\left( \mathrm{x}\right) - h\left( \mathrm{y}\right) {\parallel }^{2} = h\left( \mathrm{x}\right), h\left( \mathrm{x}\right) - {2h}\left( \mathrm{x}\right), h\left( \mathrm{y}\right) + h\left( \mathrm{y}\right), h\left( \mathrm{y}\righ... | Yes |
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