Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Corollary 3.2 For every \( X, Y \in \mathcal{W} \) and every \( h \in H \), one has\n\n\[ \mathbf{E}\left( {Y\langle \mathcal{D}X, h\rangle }\right) = \mathbf{E}\left( {{XYW}\left( h\right) - X\langle \mathcal{D}Y, h\rangle }\right) . \] | Proof. Note that \( {XY} \in \mathcal{W} \) and that Leibniz’s rule holds. | No |
Proposition 3.3 The operator \( \mathcal{D} \) is closable. In other words if, for some sequence \( {X}_{n} \in \mathcal{W} \), one has \( {X}_{n} \rightarrow 0 \) in \( {L}^{2}\left( {\Omega ,\mathbf{P}}\right) \) and \( \mathcal{D}{X}_{n} \rightarrow Y \) in \( {L}^{2}\left( {\Omega ,\mathbf{P}, H}\right) \), then \(... | Proof. Let \( {X}_{n} \) be as in the statement of the proposition and let \( Z \in \mathcal{W} \), so that in particular both \( Z \) and \( \mathcal{D}Z \) have moments of all orders. It then immediately follows from 3.5 that on has\n\n\[ \mathbf{E}\left( {Z\langle Y, h\rangle }\right) = \mathop{\lim }\limits_{{n \ri... | Yes |
Theorem 3.5 The space \( {L}_{a}^{2}\left( {\Omega ,\mathbf{P}, H}\right) \) is included in the domain of \( \delta \) and, on it, \( \delta \) coincides with the Itô integration operator. | Proof. Let \( u \) be an elementary adapted process of the form 3.6 with each \( {Y}_{k}^{\left( i\right) } \) in \( \mathcal{W} \) . For \( X \in \mathcal{W} \) one then has, as a consequence of (3.5), \[ \mathbf{E}\left( {\langle u,\mathcal{D}X\rangle }\right) = \mathop{\sum }\limits_{{k = 1}}^{N}\mathbf{E}\left( {{Y... | Yes |
Theorem 3.6 The space \( {\mathcal{W}}^{1,2}\left( H\right) \) is included in the domain of \( \delta \) and, on it, the identity\n\n\[ \mathbf{E}{\left| \delta u\right| }^{2} = \mathbf{E}{\int }_{0}^{\infty }{\left| u\left( t\right) \right| }^{2}{dt} + \mathbf{E}{\int }_{0}^{\infty }{\int }_{0}^{\infty }{\mathcal{D}}_... | Proof. Consider similarly to before \( u \) to be a process of the form \( u = \mathop{\sum }\limits_{{i = 1}}^{N}{Y}^{\left( i\right) }{h}^{\left( i\right) } \) with \( {Y}^{\left( i\right) } \in \mathcal{W} \) and \( {h}^{\left( i\right) } \in H \), but this time without any adaptedness condition on the \( Y \) ’s. I... | Yes |
Proposition 3.7 The spaces \( {\mathcal{H}}_{n} \) are invariant for \( \Delta = \delta \mathcal{D} \) and one has \( {\Delta X} = {nX} \) for every \( X \in {\mathcal{H}}_{n} \) . | Proof. Fix an orthonormal basis \( \left\{ {e}_{k}\right\} \) of \( H \) . Then, by definition, the random variables \( {\Phi }_{k} \) as in 2.6 with \( \left| k\right| = n \) are dense in \( {\mathcal{H}}_{n} \) . Recalling that \( \mathcal{D}{H}_{k}\left( {W\left( h\right) }\right) = \) \( k{H}_{k - 1}\left( {W\left(... | Yes |
Proposition 3.8 Let \( u \in {L}_{a}^{2}\left( {\Omega ,\mathbf{P}, H}\right) \) be such that \( {u}_{i}\left( t\right) \in {\mathcal{W}}^{1,2} \) for almost every \( t \) and \( {\int }_{0}^{\infty }\mathbf{E}{\begin{Vmatrix}\mathcal{D}{u}_{i}\left( t\right) \end{Vmatrix}}^{2}{dt} < \infty \) . Then 3.12) holds. | Proof. Take \( u \) of the form (3.6) with each \( {Y}_{k}^{\left( i\right) } \) in \( \mathcal{W} \), so that \( X = \sum {Y}_{k}^{\left( i\right) }W\left( {\mathbf{1}}_{\left\lbrack {s}_{k},{t}_{k}\right) }^{\left( i\right) }\right) \) . It then follows from the chain rule that\n\n\[ \mathcal{D}X = \mathop{\sum }\lim... | Yes |
Proposition 3.9 For every \( p \geq 2 \) there exist constants \( k \) and \( C \) such that, for every separable Hilbert space \( K \) and every \( u \in \mathcal{S}\left( {H \otimes K}\right) \), one has the bound\n\n\[ \mathbf{E}{\left| \delta u\right| }^{p} \leq C\mathop{\sum }\limits_{{0 \leq \ell \leq k}}{\left( ... | Proof. For \( p \in \left\lbrack {1,2}\right\rbrack \), the bound follows immediately from Theorem 3.6 and Jensen’s inequality. Take now \( p > 2 \) . Using the definition of \( \delta \) combined with the chain rule for \( \mathcal{D} \), Proposition 3.8, and Young’s inequality, we obtain the bound\n\n\[ \mathbf{E}{\l... | Yes |
Corollary 3.10 The operator \( \delta \) maps \( \mathcal{S}\left( {H \otimes K}\right) \) into \( \mathcal{S}\left( K\right) \) . | Proof. In order to estimate \( \mathbf{E}{\left| {\mathcal{D}}^{k}\delta u\right| }^{p} \), it suffices to first apply Proposition \( {3.8}\mathrm{k} \) times and then Proposition 3.9 | No |
Lemma 4.1 Let \( X \) be an \( {\mathbf{R}}^{n} \) -valued random variable for which there exist constants \( {C}_{k} \) such that \( \left| {\mathbf{E}{D}^{\left( k\right) }G\left( X\right) }\right| \leq {C}_{k}\parallel G{\parallel }_{\infty } \) for every \( G \in {\mathcal{C}}_{0}^{\infty } \) and \( k \geq 1 \) . ... | Proof. Denoting by \( \mu \) the law of \( X \), our assumption can be rewritten as\n\n\[ \left| {{\int }_{{\mathbf{R}}^{n}}{D}^{\left( k\right) }G\left( x\right) \mu \left( {dx}\right) }\right| \leq {C}_{k}\parallel G{\parallel }_{\infty }.\]\n\n(4.1)\n\nLet now \( s > n/2 \) so that \( \parallel G{\parallel }_{\infty... | Yes |
Lemma 4.4 Let \( X \) be as above and let \( Z \in \mathcal{S} \). Then, there exists \( \bar{Z} \in \mathcal{S} \) such that the identity\n\n\[ \mathbf{E}\left( {Z{\partial }_{i}G\left( X\right) }\right) = \mathbf{E}\left( {G\left( X\right) \bar{Z}}\right) \]\n\nholds for every \( G \in {\mathcal{C}}_{0}^{\infty } \). | Proof. Following the calculation above, defining the \( H \) -valued random variable \( {Y}_{i} \) by\n\n\[ {Y}_{i} = \mathop{\sum }\limits_{{j = 1}}^{n}\left( {\mathcal{D}{X}_{j}}\right) {\mathcal{M}}_{ji}^{-1}, \]\n\nwe have the identity \( {\partial }_{i}G\left( X\right) = \left\langle {\mathcal{D}G\left( X\right) ,... | Yes |
Proposition 5.1 If \( {V}_{i} \in {\mathcal{C}}_{b}^{\infty } \) for all \( i \), then \( \mathop{\sup }\limits_{{s, t \leq T}}\mathbf{E}{\left| {J}_{s, t}\right| }^{p} < \infty \) for every \( T > 0 \) and every \( p \geq 1 \) . | Proof. We write \( \left| A\right| \) for the Frobenius norm of a matrix \( A \) . A tedious application of Itô’s formula shows that for even integers \( p \geq 4 \) one has\n\n\[ d{\left| {J}_{s, t}\right| }^{p} = p{\left| {J}_{s, t}\right| }^{p - 2}\left( {\left\langle {{J}_{s, t}, D{\widetilde{V}}_{0}\left( {X}_{t}\... | Yes |
Lemma 5.3 Let \( k \geq 1 \) and let \( v \) be a stochastic process on \( {\mathbf{R}}^{k} \) with \( \mathbf{E}\parallel v{\parallel }_{{L}^{p}}^{p} < \infty \) for some \( p \geq 2 \) . Then, one has the bound\n\n\[ \mathbf{E}{\left| {\int }_{0}^{t}\cdots {\int }_{0}^{{s}_{2}}v\left( {s}_{1},\ldots ,{s}_{k}\right) d... | Proof. The proof goes by induction over \( k \) . For \( k = 1 \), it follows from the Burkholder-David-Gundy inequality followed by Hölder's inequality that\n\n\[ \mathbf{E}{\left| {\int }_{0}^{t}v\left( s\right) d{W}_{i}\left( s\right) \right| }^{p} \leq C\mathbf{E}{\left| {\int }_{0}^{t}{\left| v\left( s\right) \rig... | Yes |
Theorem 5.4 Let \( {x}_{0} \in {\mathbf{R}}^{n} \) and let \( {X}_{t} \) be the solution to (1.1). If the vector fields \( \left\{ {V}_{j}\right\} \subset {\mathcal{C}}_{b}^{\infty } \) satisfy the parabolic Hörmander condition, then the law of \( {X}_{t} \) has a smooth density with respect to Lebesgue measure. | Proof. Denote by \( {\mathcal{A}}_{0, t} \) the operator \( {\mathcal{A}}_{0, t}v = {\int }_{0}^{t}{J}_{s, t}V\left( {X}_{s}\right) v\left( s\right) {ds} \), where \( v \) is a square integrable, not necessarily adapted, \( {\mathbf{R}}^{m} \) -valued stochastic process and \( V \) is the \( n \times m \) matrix-valued... | Yes |
Lemma 6.2 Let \( M \) be a symmetric positive semidefinite \( n \times n \) matrix-valued random variable such that \( \mathbf{E}\parallel M{\parallel }^{p} < \infty \) for every \( p \geq 1 \) and such that, for every \( p \geq 1 \) there exists \( {C}_{p} \) such that\n\n\[ \mathop{\sup }\limits_{{\left| \eta \right|... | Proof. The non-trivial part of the result is that the supremum over \( \eta \) is taken outside of the probability in (6.1). For \( \varepsilon > 0 \), let \( {\left\{ {\eta }_{k}\right\} }_{k \leq N} \) be a sequence of vectors with \( \left| {\eta }_{k}\right| = 1 \) such that for every \( \eta \) with \( \left| \eta... | Yes |
Lemma 6.5 Let \( f : \left\lbrack {0,1}\right\rbrack \rightarrow \mathbf{R} \) be continuously differentiable and let \( \alpha \in (0,1\rbrack \) . Then, the bound\n\n\[ \n{\begin{Vmatrix}{\partial }_{t}f\end{Vmatrix}}_{\infty } = \parallel f{\parallel }_{1} \leq 4\parallel f{\parallel }_{\infty }\max \left\{ {1,\para... | Proof. Denote by \( {x}_{0} \) a point such that \( \left| {{\partial }_{t}f\left( {x}_{0}\right) }\right| = {\begin{Vmatrix}{\partial }_{t}f\end{Vmatrix}}_{\infty } \) . It follows from the definition of the \( \alpha \) -Hölder constant \( {\begin{Vmatrix}{\partial }_{t}f\end{Vmatrix}}_{{\mathcal{C}}^{\alpha }} \) th... | Yes |
Lemma 6.6 Let \( W \) be an \( m \) -dimensional Wiener process and let \( A \) and \( B \) be \( \mathbf{R} \) and \( {\mathbf{R}}^{m} \) -valued adapted processes such that, for \( \alpha = \frac{1}{3} \), one has \( \mathbf{E}{\left( \parallel A{\parallel }_{\alpha } + \parallel B{\parallel }_{\alpha }\right) }^{p} ... | Proof. Recall the exponential martingale inequality [RY99, p. 153], stating that if \( M \) is any continuous martingale with quadratic variation process \( \langle M\rangle \left( t\right) \), then\n\n\[ \n\mathbf{P}\left( {\mathop{\sup }\limits_{{t \leq T}}\left| {M\left( t\right) }\right| \geq x\;\& \;\langle M\rang... | Yes |
Lemma 7.2 If \( p \geq q \geq 1 \), then one has \( \parallel \parallel X{\parallel }_{{L}_{1}^{q}}{\parallel }_{{L}^{p}} \leq {\begin{Vmatrix}\parallel X{\parallel }_{{L}_{2}^{p}}\end{Vmatrix}}_{{L}^{q}} \) . | Proof. The holds for \( q = 1 \) by the triangle inequality since\n\n\[ \parallel \parallel X{\parallel }_{{L}_{1}^{1}}{\parallel }_{{L}^{p}} = {\begin{Vmatrix}\int \left| X\left( {\omega }_{1}, \cdot \right) \right| {\mathbf{P}}_{1}\left( d{\omega }_{1}\right) \end{Vmatrix}}_{{L}^{p}} \leq \int {\begin{Vmatrix}\left| ... | Yes |
Corollary 7.3 In the above setting if, for some \( p \geq q \geq 1 \) one has \( {\begin{Vmatrix}{T}_{i}X\end{Vmatrix}}_{{L}^{p}} \leq \) \( \parallel X{\parallel }_{{L}^{q}} \), then one also has \( \parallel {TX}{\parallel }_{{L}^{p}} \leq \parallel X{\parallel }_{{L}^{q}} \) . | Proof. One has\n\n\[ \parallel {TX}{\parallel }_{{L}^{p}} = {\begin{Vmatrix}{\begin{Vmatrix}{\widehat{T}}_{1}{\widehat{T}}_{2}X\end{Vmatrix}}_{{L}_{1}^{p}}\end{Vmatrix}}_{{L}^{p}} \leq {\begin{Vmatrix}{\begin{Vmatrix}{\widehat{T}}_{2}X\end{Vmatrix}}_{{L}_{1}^{q}}\end{Vmatrix}}_{{L}^{p}} \]\n\n\[ \leq \parallel \paralle... | Yes |
Theorem 7.4 The measure \( \mu \) satisfies the log-Sobolev inequality, namely\n\n\[ \int {f}^{2}\log {f}^{2}{d\mu } - \int {f}^{2}{d\mu }\log \int {f}^{2}{d\mu } \leq 2\int {\left| {\partial }_{x}f\right| }^{2}{d\mu } \]\n\nfor all \( f \in {\mathcal{W}}^{1,2} \) . | Proof. (This proof is essentially taken from [Led92].) By a simple density argument, we can assume that \( f \) is smooth and bounded. One then has \( \mathop{\lim }\limits_{{t \rightarrow \infty }}\left( {{P}_{t}{f}^{2}}\right) \left( x\right) = \) \( \int {f}^{2}{d\mu } \), uniformly over compact sets. As a consequen... | Yes |
Lemma 8.4 For \( n, m \geq 0 \), one has\n\n\[ \n{H}_{n}\left( x\right) {H}_{m}\left( x\right) = \mathop{\sum }\limits_{{p \geq 0}}C\left( {m, n;p}\right) {H}_{n + m - {2p}}\left( x\right) ,\;C\left( {m, n;p}\right) = p!\left( \begin{array}{l} m \\ p \end{array}\right) \left( \begin{array}{l} n \\ p \end{array}\right) ... | Proof. We fix \( n \) and proceed by induction on \( m \) . The case \( m = 0 \) is trivial, while the case \( m = 1 \) reads\n\n\[ \n{H}_{1} \cdot {H}_{n} = {H}_{n + 1} + n{H}_{n - 1},\n\]\n\nwhich is immediate from \( {H}_{1}\left( x\right) = x \), combined with (2.3) and (2.5). Since \( n \) is fixed, we write \( {C... | Yes |
Lemma 8.6 Let \( F \in {\mathcal{H}}_{n} \) for some \( n \geq 2 \) with \( F \neq 0 \) . Then, there exists \( c > 0 \) depending only on \( n \) such that\n\n\[ \mathbf{E}{F}^{4} - 3{\left( \mathbf{E}{F}^{2}\right) }^{2} \geq c\left( {\mathbf{E}\parallel \mathcal{D}F{\parallel }^{4} - {\left( \mathbf{E}\parallel \mat... | Proof. Since \( F \in {\mathcal{H}}_{n} \), we can represent it and its Malliavin derivative as\n\n\n\nAs a consequence of Proposition 8.2, we then have\n\n and let \( {\left\{ {F}_{k}\right\} }_{k \geq 0} \in {\mathcal{H}}_{n} \) be a sequence of random variables such that \( \mathbf{E}{F}_{k}^{2} = 1 \) (say) for all \( k \) . Then, the \( {F}_{k} \) converge in law to \( \mathcal{N}\left( {0,1}\right) \) if and only if \( \mathop{\lim }\li... | Proof. We follow the exposition of [NOL08]. Since all moments of the sequence \( {F}_{k} \) are uniformly bounded by (7.2), the necessity of \( \mathbf{E}{F}_{k}^{4} \rightarrow 3 \) is immediate. For the converse implication, we note first that by tightness we can assume that the \( {F}_{k} \) have some limit in law (... | Yes |
Theorem 3.2. Let \( f\left( T\right) \in \mathbf{Q}\left\lbrack T\right\rbrack \) be an irreducible polynomial of prime degree \( p \) with all but two roots in \( \mathbf{R} \) . The Galois group of \( f\left( T\right) \) over \( \mathbf{Q} \) is isomorphic to \( {S}_{p} \) . | Proof. Let \( L = \mathbf{Q}\left( {{r}_{1},\ldots ,{r}_{p}}\right) \) be the splitting field of \( f\left( T\right) \) over \( \mathbf{Q} \) . The permutations of the \( {r}_{i} \) ’s by \( \operatorname{Gal}\left( {L/\mathbf{Q}}\right) \) provide an embedding \( \operatorname{Gal}\left( {L/\mathbf{Q}}\right) \hookrig... | Yes |
Lemma 1. Let one eigenvalue of \( A \) be zero, WLOG we can set \( {\lambda }_{n}\left( A\right) = 0 \) . Then,\n\n\[ \mathop{\prod }\limits_{{i = 1}}^{{n - 1}}{\lambda }_{i}\left( A\right) {\left| \det \left( \begin{array}{ll} B & {v}_{n} \end{array}\right) \right| }^{2} = \det \left( {{B}^{ * }{AB}}\right) ,\] \n\nfo... | Proof. If we diagonalize \( A = {VD}{V}^{ * } \) where \( D \equiv \operatorname{diag}\left( {{\lambda }_{1}\left( A\right) ,\ldots ,{\lambda }_{n - 1}\left( A\right) ,0}\right) \) and make the replacements \( B \rightarrow {V}^{ * }B \) and \( {v}_{n} \rightarrow {V}^{ * }{v}_{n} = {e}_{n} \), we can assume that \( A ... | Yes |
Lemma 2. The norm squared of the elements of the eigenvectors are related to the eigenvalues and the submatrix eigenvalues,\n\n\[ \n{\left| {v}_{i, j}\right| }^{2}\mathop{\prod }\limits_{{k = 1;k \neq i}}^{n}\left( {{\lambda }_{i}\left( A\right) - {\lambda }_{k}\left( A\right) }\right) = \mathop{\prod }\limits_{{k = 1}... | Proof. WLOG we take \( j = 1 \) and \( i = n \) . We shift \( A \) by \( {\lambda }_{n}\left( A\right) {I}_{n} \) so that \( {\lambda }_{n}\left( A\right) = 0 \) ; this also shifts all the remaining eigenvalues of \( A \) as well as those of \( {M}_{j} \), then eq. 2\n\n---\n\nbecomes,\n\n(3)\n\n\[ \n{\left| {v}_{n,1}\... | Yes |
Corollary 3. If one element of an eigenvector vanishes, \( {v}_{i, j} = 0 \), then one of the eigenvalues of \( {M}_{j} \) must match \( {\lambda }_{i}\left( A\right) \) . | Proof. The proof follows directly from eq. 2 In addition, if \( {v}_{i, j} = 0 \), then the eigenvector equation of \( A \) collapses to an eigenvector equation of \( {M}_{j} \) . | No |
Solve \( {dy}/{dt} - {5y} = 3 \) starting from \( y\left( 0\right) = 2 \) . Here \( a = 5 \) and \( q = 3 \) . | This fits perfectly with \( {y}^{\prime } - {ay} = q \) . Equation (5) gives the solution \( y\left( t\right) \) : \n\nSolution \( y\left( t\right) = {y}_{n} + {y}_{p} = 2{e}^{5t} + \frac{3}{5}\left( {{e}^{5t} - 1}\right) \) . Set \( t = 0 \) to check that \( y\left( 0\right) = 2 \) . | Yes |
Example 2 Solve \( {dy}/{dt} = 3 - {6y} \) starting from \( y\left( 0\right) = 2 \) . | Formula (5) still gives the answer, but this \( y\left( t\right) \) is decreasing because \( \mathbf{a} = - \mathbf{6} \) is negative :\n\n\[ y\left( t\right) = 2{e}^{-{6t}} + \frac{3}{-6}\left( {{e}^{-{6t}} - 1}\right) = \frac{3}{2}{e}^{-{6t}} + \frac{1}{2}. \]\n\nWhen \( t = 0 \), that solution starts at \( y\left( 0... | Yes |
Suppose the input turns on at time \( t = 0 \) and turns off at \( t = T \) . Find \( y\left( t\right) \) . | The input is \( H\left( t\right) - H\left( {t - T}\right) \) . The output is \( y\left( t\right) = \frac{1}{a}\left( {{e}^{at} - {e}^{a\left( {t - T}\right) }}\right), t \geq T \) . | No |
The integral of all impulses for \( T \geq 0 \) is the step function \( H\left( t\right) \). | Then the integral of all impulse responses is the step response. The integral of \( {e}^{at} \) from 0 to \( t \) is \( \left( {{e}^{at} - 1}\right) /a \). Derivative of step response \( = \) impulse response as in (13). | No |
Describe the paths of the numbers \( {e}^{st} \) and \( {e}^{i\omega t} \) and \( {e}^{\left( {s + {i\omega }}\right) t} \) in the complex plane (real \( s \) and real \( \omega \) ). The time \( t \) goes from 0 to \( \infty \) . Those paths start at 1 . | Solution If \( s > 0 \), the number \( {e}^{st} \) goes from 1 out the real axis to infinity. If \( s < 0 \) , then \( {e}^{st} \) goes from 1 in to zero. All real.\n\nThe path of \( {e}^{i\omega t} \) goes around the unit circle with constant speed. At time \( T = {2\pi }/\omega \) (and also \( {2T},{3T},\ldots \) ) i... | Yes |
Example 2 Write \( q\left( t\right) = \cos {3t} + \sin {3t} \) as \( R\cos \left( {{3t} - \phi }\right) \) : the real part of \( R{e}^{i\left( {{3t} - \phi }\right) } \) . | Solution \( A = 1 \) and \( B = 1 \) so that \( R = \sqrt{2} \) . The angle \( \phi = \frac{\pi }{4} \) has \( \tan \phi = B/A = 1 \) . Then \( \cos {3t} + \sin {3t} = \sqrt{2}\cos \left( {{3t} - \frac{\pi }{4}}\right) \) . | Yes |
Write the real part of \( {e}^{i5t}/\left( {\sqrt{3} + i}\right) \) in the form \( A\cos {5t} + B\sin {5t} \) . | Solution \( \sqrt{3} + i \) is \( 2{e}^{{i\pi }/6} \) (why?) Then \( {e}^{i5t}/\left( {\sqrt{3} + i}\right) \) is \( \frac{1}{2}{e}^{i\left( {{5t} - \pi /6}\right) } \). Its real part is\n\n\[ \frac{1}{2}\cos \left( {{5t} - \frac{\pi }{6}}\right) = \frac{1}{2}\left( {\cos {5t}\cos \frac{\pi }{6} + \sin {5t}\sin \frac{\... | Yes |
Take those three steps real-complex-real to solve \( {y}^{\prime } - y = \cos t - \sin t \) . | We have to find \( R,\phi, G \), and \( \alpha \) from the numbers \( a = 1,\omega = 1, A = 1 \), and \( B = - 1 \) . Notice that \( {RG} = 1 \) .\n\n\[ R = \sqrt{{A}^{2} + {B}^{2}} = \sqrt{2}\;\tan \phi = \frac{B}{A} = - 1\text{ and }\phi = - \frac{\pi }{4}\;G = \frac{1}{\sqrt{{\omega }^{2} + {a}^{2}}} = \frac{1}{\sqr... | Yes |
The growth rate \( a\left( t\right) = {2t} \) puts the economy into serious inflation. The integral of \( a\left( t\right) \) is \( {\int }_{s}^{t}{2TdT} = {t}^{2} - {s}^{2} \) . Then \( G \) is the growth from \( s \) to \( t \) : | \[ G\left( {s, t}\right) = {e}^{{t}^{2} - {s}^{2}}\;{y}^{\prime } = {2ty} + q\left( t\right) \text{ has }{y}_{p}\left( t\right) = {\int }_{0}^{t}{e}^{{t}^{2} - {s}^{2}}q\left( s\right) {ds}. \] | Yes |
Suppose the interest rate \( a \) goes to zero. What happens to the solution formula? | The first term \( {y}_{n} \) becomes \( y\left( 0\right) \) . This deposit doesn’t grow or disappear, it stays fixed. The growth factor is \( G = 1 \) and we just add up all the inputs (they didn't grow):\n\n\[ a = 0\;{y}^{\prime } = q\left( t\right) \text{ has the particular solution }{y}_{p}\left( t\right) = {\int }_... | Yes |
The derivative of \( {ay} - b{y}^{2} \) is \( {df}/{dy} = a - {2by} \) . | At the steady state \( Y = 0,{df}/{dy} \) is \( a > 0 : \mathbf{Y} = \mathbf{0} \) is unstable.\n\nAt \( Y = a/b \), this derivative is \( a - {2b}\left( {a/b}\right) = - a.Y = a/b \) is stable.\n\nFor \( {dy}/{dt} = {ay} - b{y}^{2} \) this stability line shows which way \( y\left( t\right) \) moves from any \( y\left(... | Yes |
Example 1 \( \;\frac{dy}{dt} = \frac{t}{y}\; \) is \( \;y\;{dy} = t\;{dt}.\; \) Integrate to find \( \;\frac{1}{2}\left( {y(t{)}^{2} - y(0{)}^{2}}\right) = \frac{1}{2}{t}^{2}. \) | Solve this implicit equation to find \( y\left( t\right) \) explicitly : Solution \( y\left( t\right) = \sqrt{y{\left( 0\right) }^{2} + {t}^{2}} \) . Then \( \frac{dy}{dt} = \frac{t}{\sqrt{y{\left( 0\right) }^{2} + {t}^{2}}} = \frac{t}{y} \) | Yes |
Example 2 \( {dy}/{dt} = {2ty}\; \) has \( \;g\left( t\right) = {2t}\; \) divided by \( \;f\left( y\right) = 1/y \) . | Solution Separate \( 1/y \) from \( {2t} \) and integrate to get \( F = \ln y - \ln y\left( 0\right) \) and \( G = {t}^{2} \) :\n\n\[ \frac{dy}{y} = {2tdt}\;\text{ leads to }\;{\int }_{y\left( 0\right) }^{y}\frac{du}{u} = \ln y - \ln y\left( 0\right) \;\text{ and }\;{\int }_{0}^{t}{2xdx} = {t}^{2} \]\n\nIn this example... | Yes |
Our favorite equation \( \frac{dy}{dt} = {ay} + q \) is separable when \( a \) and \( q \) are constant. | Move \( y + \frac{q}{a} \) to the left side below \( {dy} \) . Keep \( {adt} \) on the right side. Then integrate both sides, and you have solved this equation once more !\n\n\[ \frac{dy}{y + \frac{q}{a}} = {adt}\;\text{ gives }\;\ln \left( {y + \frac{q}{a}}\right) = {at} + C. \]\n\nTake exponentials to find \( y \), a... | Yes |
Example 4 (Logistic equation )\n\n\\[ \n\\frac{dy}{dt} = {ay} - b{y}^{2}\\;{\\int }_{y\\left( 0\\right) }^{y}\\frac{du}{{au} - b{u}^{2}} = {\\int }_{t\\left( 0\\right) }^{t}{dx} \n\\] | The right side is certainly \\( G\\left( t\\right) = t - t\\left( 0\\right) \\) . I am including \\( t\\left( 0\\right) \\) to show how the system allows any starting value for \\( t \\) as well as \\( y \\) . We don’t know a perfect starting time for the Earth’s population, so we pick a year like \\( t\\left( 0\\right... | Yes |
The equation \( \frac{dy}{dt} = \frac{{2yt} - 1}{{y}^{2} - {t}^{2}} \) has \( g = {2yt} - 1 \) and \( f = {y}^{2} - {t}^{2} \) . | Step 1 Integrate \( {fdy} = \left( {{y}^{2} - {t}^{2}}\right) {dy} \) to find \( F\left( {y, t}\right) = \frac{1}{3}{y}^{3} - y{t}^{2} \) . Then \( \frac{\partial F}{\partial t} = - {2ty} \).\n\nStep 2 Solve equation (11) for \( C\left( t\right) \) . For our particular \( f \) and \( g \), this is possible :\n\n\[ - {2... | Yes |
Example 6 Steps \( 1,2,3 \) must be possible because this non-separable equation is exact :\n\n\[ \frac{dy}{dt} = \frac{t - y}{t + y} = \frac{g\left( {y, t}\right) }{f\left( {y, t}\right) }\;\text{ has }\;\frac{\partial f}{\partial t} = - \frac{\partial g}{\partial y} = 1. \] | Step 1 Integrate \( \int {fdy} = \int \left( {t + y}\right) {dy} \) to find \( F = {ty} + \frac{1}{2}{y}^{2} \).\n\nStep 2 Write out \( \frac{\partial }{\partial t}\left( {F + C}\right) = - g = y - t \) to find \( C\left( t\right) = - \frac{1}{2}{t}^{2} \)\n\nStep 3 The example is solved by \( F + C = {ty} + \frac{1}{2... | Yes |
Solve \( {y}^{\prime \prime } - 3{y}^{\prime } + {2y} = 0 \) . | Solution Substitute \( y = {e}^{st} \) as before. Negative damping gives positive \( s \) . \[ {s}^{2} - {3s} + 2 = 0\;\left( {s - 1}\right) \left( {s - 2}\right) = 0\;{s}_{1} = 2\text{ and }{s}_{2} = 1. \] The complete solution is now \( y\left( t\right) = {c}_{1}{e}^{2t} + {c}_{2}{e}^{t} \) . Exponential growth \( = ... | Yes |
Solve \( {y}^{\prime \prime } - 2{y}^{\prime } + y = 0 \) . | Solution Substitute \( y = {e}^{st} \) as usual. The root \( s = 1 \) is repeated: two equal roots.\n\n\[ \n{s}^{2} - {2s} + 1 = 0\;{\left( s - 1\right) }^{2} = 0\;{s}_{1} = 1 = {s}_{2} \]\n\nWith that root, \( y = {e}^{t} \) solves the equation : easy to check. A second solution is needed ! We now confirm that \( y = ... | Yes |
Example 4 Solve \( {y}^{\prime \prime } = 0 \) . The coefficients \( 1,0,0 \) have \( {B}^{2} = {4AC} \) . | Solution Substitute \( y = {e}^{st} \) to find \( {s}^{2}{e}^{st} = 0 \) and \( {s}^{2} = 0 \) . The double root is \( s = 0 \) . The usual solution \( y = {e}^{st} = {e}^{0t} = 1 \) does have \( {y}^{\prime \prime } = 0 \) . We need a second solution.\n\nThe rule \( y = t{e}^{st} \) still applies when \( s = 0 \) . Th... | Yes |
Solve \( {y}^{\prime \prime } + 5{y}^{\prime } + {6y} = {e}^{4t} \) . | One particular solution will be \( {y}_{p} = Y{e}^{4t} \). When \( Y{e}^{4t} \) is substituted into the equation, all terms contain \( {e}^{4t} \):\n\n\[ {y}^{\prime \prime } + 5{y}^{\prime } + {6y} = {16Y}{e}^{4t} + {20Y}{e}^{4t} + {6Y}{e}^{4t} = {e}^{4t}. \]\n\nThe left side is \( {42}{\mathrm{{Ye}}}^{4t} \). This ma... | Yes |
Example 2 \( {y}^{\prime \prime } + {y}^{\prime } = {e}^{it} \) has \( s = {i\omega } = i \) . Substitute \( y = Y{e}^{it} \) and solve for \( Y \) : | \[ {i}^{2}Y{e}^{it} + {iY}{e}^{it} = {e}^{it}\;\left( {{i}^{2} + i}\right) Y = 1\;{y}_{p}\left( t\right) = \frac{1}{-1 + i}{e}^{it}. \] | Yes |
Solve \( {y}^{\prime \prime } + {y}^{\prime } = \cos t \) . | Here \( {y}_{p}\left( t\right) \) in Example 3 is the real part of \( {y}_{p}\left( t\right) \) in Example 2. Please use this idea:\n\nThe real part of the input \( {e}^{i\omega t} \) produces the real part of the output \( Y{e}^{i\omega t} \) .\n\nStep 1 Write \( Y = \frac{1}{-1 + i} = \frac{1}{-1 + i}\left( \frac{-1 ... | No |
Solve \( {y}^{\prime \prime } + {y}^{\prime } + {2y} = \cos t \) in rectangular form and also in polar form. | The equation has \( A = 1, B = 1, C = 2 \), and \( \omega = 1 \) . We are finding a particular solution. Let me use the formulas directly and then comment briefly. The numbers give \( C - A{\omega }^{2} = 1 \) and \( {B\omega } = 1 \), so \( D = {1}^{2} + {1}^{2} = 2 \) .\n\nTherefore the solution has \( G = \sqrt{1/2}... | Yes |
Example 1 Suppose the RLC circuit has resistance \( R = {10} \) ohms and inductance \( L = {0.1} \) henry and capacitance \( C = {10}^{-4} \) farad. The units of \( R \) and \( {\omega L} \) and \( 1/{\omega C} \) must agree. Since frequency \( \omega \) is measured in inverse seconds, all three units can be given in t... | R Ohm \( \Omega = V/A\; = 1 \) volt per amp\n\nL Henry \( H = V \cdot \sec /A = 1 \) volt-second per amp\n\nC Farad \( F = A \cdot \sec /V = 1 \) amp-second per volt | Yes |
Example 2 Find the impedance \( Z \), its magnitude \( \left| Z\right| \), and the phase angle \( \alpha \) for an RLC loop when the frequency is \( \omega = {60} \) cycles/second \( = {60}\mathrm{\;{Hz}} = {120\pi } \) radians/second. | The impedance of this loop is \( \;Z = R + i\left( {{\omega L} - \frac{1}{\omega C}}\right) = \left| Z\right| {e}^{-{i\alpha }} \). | No |
To tune a radio to a station with frequency \( \omega \), what should be the capacitance \( C \) (which you adjust)? Suppose \( R \) and \( L \) are fixed and known. | The goal of tuning is to achieve \( {\omega L} = 1/{\omega C} \) . Then the imaginary part of \( Z \) is zero: inductance cancels capacitance. Tuning achieves \( Z = R \), that real part \( R \) is fixed.\n\n\[{\omega L} = \frac{1}{\omega C}\;{\omega }^{2} = \frac{1}{LC}\;C = \frac{1}{L{\omega }^{2}}\] | Yes |
Example 4 Suppose the network contains two RLC branches in parallel. Find the total impedance \( {Z}_{12} \) from the impedances \( {Z}_{1} \) and \( {Z}_{2} \) of the two separate branches. | \[ \frac{1}{{Z}_{12}} = \frac{1}{{Z}_{1}} + \frac{1}{{Z}_{2}} = \frac{{Z}_{1} + {Z}_{2}}{{Z}_{1}{Z}_{2}} \] | Yes |
Find a particular solution to \( {y}^{\prime \prime } + y = t{e}^{st} = \) polynomial times \( {e}^{st} \) . | The good form to assume for \( y\left( t\right) \) is \( \left( {{at} + b}\right) {e}^{st} \) . Please notice that \( b{e}^{st} \) is included. Even though \( f \) doesn’t have \( {e}^{st} \) by itself, that will appear in the derivatives of \( t{e}^{st} \) . To be sure we capture every derivative, \( {at} + b \) must ... | Yes |
Start from \( y\left( 0\right) = 0 \) and \( {y}^{\prime }\left( 0\right) = 0 \) . With those initial conditions, the transform of \( {y}^{\prime } \) is \( {sY} \) and the transform of \( {y}^{\prime \prime } \) is \( {s}^{2}Y \) . We can transform a whole equation:\n\nStep \( 1{y}^{\prime \prime } - 4{y}^{\prime } + ... | Step 2 The transform of \( y\left( t\right) \) is \( Y\left( s\right) = \frac{1}{\left( {{s}^{2} - {4s} + 3}\right) \left( {s - a}\right) } = \frac{1}{\left( {s - 3}\right) \left( {s - 1}\right) \left( {s - a}\right) }\)\n\nStep 3 The inverse Laplace transform of \( Y\left( s\right) \) is \( \mathbf{y}\left( t\right) =... | Yes |
Example 2 Change from \( f = {e}^{at} \) to \( f = \delta \left( t\right) = \) impulse. Keep \( y\left( 0\right) = {y}^{\prime }\left( 0\right) = 0 \) . | Step \( 1{y}^{\prime \prime } + B{y}^{\prime } + {Cy} = \delta \left( t\right) \) transforms to \( \left( {{s}^{2} + {Bs} + C}\right) Y\left( s\right) = 1 \) . \n\nStep 2 The transform of \( y\left( t\right) \) is \( Y\left( s\right) = \frac{1}{{s}^{2} + {Bs} + C} = \) transfer function. \n\nStep 3 The inverse transfor... | Yes |
Example 1 \( \;{dy}/{dt} = 2 - y\; \) | Solution \( y\left( t\right) = 2 + C{e}^{-t}\;y\left( \infty \right) = 2 \) | No |
Example 2 \( \;\frac{dy}{dt} = y - {y}^{2}\; \) Solutions \( y\left( t\right) = \frac{1}{1 + C{e}^{-t}}\;y\left( t\right) \rightarrow 1 \) or \( - \infty \) | The slope of every small arrow is \( y - {y}^{2} \) . In the range \( 0 < y < 1, y \) will be larger than \( {y}^{2} \) . The arrows have positive slope \( y - {y}^{2} \) in this range (small slope near \( y = 0 \) , small slope near \( y = 1 \), all up and to the right). The other two ranges are above \( y = 1 \) and ... | Yes |
Linearize \( {y}^{\prime } = \sin \left( {{ay} + {bz}}\right) \) and \( {z}^{\prime } = \sin \left( {{cy} + {dz}}\right) \) at \( Y = 0, Z = 0 \) . | Solution Check first: \( f = \sin \left( {{ay} + {bz}}\right) \) and \( g = \sin \left( {{cy} + {dz}}\right) \) are zero at \( \left( {Y, Z}\right) = \left( {0,0}\right) \) . This is a critical point. The first derivatives of \( f \) and \( g \) at that point go into \( A \) .\n\n\[ \partial f/\partial y = a\cos \left(... | Yes |
Apply all three methods to \( {dy}/{dt} = y \) . The true solution \( y = {e}^{t} \) reaches \( y = e = {2.71828}\ldots \) at time \( t = 1 \) . Try \( {\Delta t} = {0.2} \) and 0.1 . | <table><thead><tr><th>\( {\Delta t} = {0.2} \)</th><th>\( {y}^{E} \)</th><th>\( {y}^{S} \)</th><th>\( {y}^{RK} \)</th><th>\( \mathbf{\Delta }t = {0.1} \)</th><th>\( {y}^{E} \)</th><th>ys</th><th>\( {y}^{RK} \)</th></tr></thead><tr><td>\( t = 0 \)</td><td>1</td><td>1</td><td>1</td><td>\( t = 0 \)</td><td>1</td><td>1</td... | No |
The column vectors \( \left( {1,2}\right) \) and \( \left( {-2,1}\right) \) have a zero dot product : | Dot product is zero\n\[ \left\lbrack \begin{array}{l} 1 \\ 2 \end{array}\right\rbrack \cdot \left\lbrack \begin{array}{r} - 2 \\ 1 \end{array}\right\rbrack = - 2 + 2 = 0. \] | Yes |
Example 4 Invertible matrix \( A \), one solution \( v \) for any right side \( b \) . | \[ A\mathbf{v} = \mathbf{b}\;\text{ is }\;\left\lbrack \begin{array}{rrr} 1 & 0 & 0 \\ - 1 & 1 & 0 \\ 0 & - 1 & 1 \end{array}\right\rbrack \left\lbrack \begin{array}{l} {v}_{1} \\ {v}_{2} \\ {v}_{3} \end{array}\right\rbrack = \left\lbrack \begin{array}{l} 1 \\ 3 \\ 5 \end{array}\right\rbrack . \] This matrix is lower t... | Yes |
\[ x - {2y} = 1 \] \[ {2x} - {4y} = 7 \] | Subtract 2 times the first equation from the second equation: \[ {2x} - {4y} = 7 \] \[ x - {2y} = 1 \] \[ \mathbf{0}y = 5 \] There is no solution to \( {0y} = 5 \). This system has no second pivot. (Zero is never allowed as a pivot!) If there is no solution, elimination discovers that fact by reaching an impossible equ... | Yes |
Example 2 Failure with infinitely many solutions. Change \( b = \left( {1,7}\right) \) to \( \left( {1,2}\right) \) . | \[ x - {2y} = 1\;\text{Subtract 2 times}\;x - {2y} = 1\;\text{Too few pivots} \] \[ {2x} - {4y} = 2\;\text{eqn. 1 from eqn. 2}\;\mathbf{0}y = \mathbf{0}\;\text{Too many solutions} \] Every \( y \) satisfies \( {0y} = 0 \) . There is really only one equation \( x - {2y} = 1 \) . The unknown \( y \) is \ | No |
Example 3 Temporary failure (zero in pivot ). A row exchange produces two pivots : | \[ \begin{array}{l} {0x} + {2y} = 4 \\ {3x} - {2y} = 5 \\ {3x} - {2y} = 5\;\text{two equations} \\ {2y} = 4\text{.} \end{array} \] The new system is already triangular. This small example is ready for back substitution. The last equation gives \( y = 2 \), and then the first equation gives \( x = 3 \) . The row picture... | Yes |
Here \( A \) has two columns and \( B \) has two rows. We can multiply \( {AB} \) . | \[ {A}_{2 \times 2}{B}_{2 \times 3} = {\left( AB\right) }_{2 \times 3}\;\left\lbrack \begin{array}{ll} a & b \\ c & d \end{array}\right\rbrack \left\lbrack \begin{array}{lll} 1 & 0 & 1 \\ 0 & 1 & 1 \end{array}\right\rbrack = \left\lbrack \begin{array}{lll} a & b & a + b \\ c & d & c + d \end{array}\right\rbrack . \] | Yes |
If \( A \) is any square matrix and \( I \) has the same size, then \( \mathbf{A}\mathbf{I} = \mathbf{I}\mathbf{A} = \mathbf{A} \). | The first column of that answer is \( A \) times the first column \( \left( {1,0,0}\right) \) of \( B = I \) . This just reproduces the first column of \( A \) . Each column of \( A \) is unchanged in \( {AI} \) . Now put the identity matrix first, as in \( {IB} \) . Multiplication gives \( {IB} = B \) for every \( B \... | No |
Another special matrix is the inverse of \( A \) . That matrix \( B \) is written \( {A}^{-1} \) : | \[ A\text{ times }{A}^{-1}\text{ is }I\;\left\lbrack \begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{array}\right\rbrack \left\lbrack \begin{array}{rrr} 2 & - 1 & 0 \\ - 1 & 2 & - 1 \\ 0 & - 1 & 1 \end{array}\right\rbrack = \left\lbrack \begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\righ... | Yes |
Example 4 Suppose \( A \) and \( C \) are 3 by 1 matrices (those are column vectors ). Suppose \( B \) is 1 by 3 (a row vector). Compute and compare \( \left( {AB}\right) C \) and \( A\left( {BC}\right) \) . | Solution \( {BC} \) is \( \left( {1 \times 3}\right) \) times \( \left( {3 \times 1}\right) = 1 \times 1 \) . One number \( d \) from one dot product :\n\n\[ A\\text{ times }{BC}\\;\\left\\lbrack \\begin{array}{l} {a}_{1} \\\\ {a}_{2} \\\\ {a}_{3} \\end{array}\\right\\rbrack \\left( {\\left\\lbrack \\begin{array}{lll} ... | Yes |
Choose the multiplier \( {\ell }_{21} = c/a \) to produce zero in \( {U}_{21} \), using \( E = {E}_{21} \): | \[ {EA} = \left\lbrack \begin{matrix} 1 & 0 \\ - c/a & 1 \end{matrix}\right\rbrack \left\lbrack \begin{array}{ll} a & b \\ c & d \end{array}\right\rbrack = \left\lbrack \begin{matrix} a & b \\ 0 & d - \left( {c/a}\right) b \end{matrix}\right\rbrack = U. \] (6) Undo this elimination by adding \( c/a \) times row 1 of \(... | Yes |
The 2 by 2 matrix \( A = \left\lbrack \begin{array}{ll} 1 & 2 \\ 1 & 2 \end{array}\right\rbrack \) is not invertible. | It fails the test in Note 5, because \( {ad} - {bc} \) equals \( 2 - 2 = 0 \) . It fails the test in Note 3, because \( {Av} = \mathbf{0} \) when \( \mathbf{v} = \left( {2, - 1}\right) \) . It fails to have two pivots as required by Note 1 . Elimination turns the second row of this matrix \( A \) into a zero row. | Yes |
Inverse of an elimination matrix. If \( E \) subtracts 5 times row 1 from row 2, then \( {E}^{-1} \) adds 5 times row 1 to row 2 : | \[ \begin{array}{l} E\text{ subtracts } \\ {E}^{-1}\text{ adds } \end{array}\;E = \left\lbrack \begin{array}{rrr} 1 & 0 & 0 \\ - \mathbf{5} & 1 & 0 \\ 0 & 0 & 1 \end{array}\right\rbrack \;\text{ and }\;{E}^{-1} = \left\lbrack \begin{array}{lll} 1 & 0 & 0 \\ \mathbf{5} & 1 & 0 \\ 0 & 0 & 1 \end{array}\right\rbrack . \] ... | Yes |
Example 3 Suppose \( F \) subtracts 4 times row 2 from row 3, and \( {F}^{-1} \) adds it back : | \[ F = \left\lbrack \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & - 4 & 1 \end{matrix}\right\rbrack \;\text{ and }\;{F}^{-1} = \left\lbrack \begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 4 & 1 \end{array}\right\rbrack \]\n\nNow multiply \( F \) by the matrix \( E \) in Example 2 to find \( {FE} \) . Also multiply \( {... | Yes |
Find \( {A}^{-1} \) by Gauss-Jordan elimination starting from \( A = \left\lbrack \begin{array}{ll} 2 & 3 \\ 4 & 7 \end{array}\right\rbrack \) . | \[ \left\lbrack \begin{array}{ll} \mathbf{A} & I \end{array}\right\rbrack = \left\lbrack \begin{array}{llll} \mathbf{2} & \mathbf{3} & 1 & 0 \\ \mathbf{4} & \mathbf{7} & 0 & 1 \end{array}\right\rbrack \rightarrow \left\lbrack \begin{array}{rrrr} 2 & 3 & 1 & 0 \\ \mathbf{0} & \mathbf{1} & - \mathbf{2} & \mathbf{1} \end{... | Yes |
Example 5 Here \( L \) is lower triangular with 1 ’s on the diagonal. Then \( {L}^{-1} \) is too. | Here \( L \) has 1’s so \( {L}^{-1} \) also has 1’s. Use the Gauss-Jordan method to construct \( {L}^{-1} \) . Start by subtracting multiples of pivot rows from rows below. Normally this gets us halfway to the inverse, but for \( L \) it gets us all the way. \( {L}^{-1} \) appears on the right when \( I \) appears on t... | Yes |
Keep only the vectors \( \left( {x, y}\right) \) whose components are positive or zero (this is a quarter-plane). | The vector \( \left( {2,3}\right) \) is included but \( \left( {-2, - 3}\right) \) is not. So rule (ii) is violated when we try to multiply by \( c = - 1 \) . The quarter-plane is not a subspace. | Yes |
Example 3 Inside the vector space \( \mathbf{M} \) of all 2 by 2 matrices, here are two subspaces :\n\n(U) All upper triangular matrices \( \left\lbrack \begin{array}{ll} a & b \\ 0 & d \end{array}\right\rbrack \; \) (D) All diagonal matrices \( \left\lbrack \begin{array}{ll} a & 0 \\ 0 & d \end{array}\right\rbrack \) ... | For a smaller subspace of diagonal matrices, we could require \( a = d \) . The matrices are multiples of the identity matrix \( I \) . These \( {aI} \) form a \ | No |
Describe the column spaces (they are subspaces of \( {\mathbf{R}}^{2} \) ) for these matrices : \[ I = \left\lbrack \begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right\rbrack \;\text{ and }\;A = \left\lbrack \begin{array}{ll} 1 & 2 \\ 2 & 4 \end{array}\right\rbrack \;\text{ and }\;B = \left\lbrack \begin{array}{lll} 1 &... | Solution The column space of \( I \) is the whole space \( {\mathbf{R}}^{2} \) . Every vector is a combination of the columns of \( I \) . In vector space language, \( \mathbf{C}\left( I\right) \) equals \( {\mathbf{R}}^{2} \) .\n\nThe column space of \( A \) is only a line. The second column \( \left( {2,4}\right) \) ... | Yes |
The columns of this \( A \) are dependent. The nonzero vector \( \mathbf{v} \) has \( A\mathbf{v} = \mathbf{0} \) . | \[ A\mathbf{v} = \left\lbrack \begin{array}{lll} 1 & 0 & 3 \\ 2 & 1 & 5 \\ 1 & 0 & 3 \end{array}\right\rbrack \left\lbrack \begin{array}{r} - 3 \\ 1 \\ 1 \end{array}\right\rbrack \;\text{ is }\; - 3\left\lbrack \begin{array}{l} 1 \\ 2 \\ 1 \end{array}\right\rbrack + 1\left\lbrack \begin{array}{l} 0 \\ 1 \\ 0 \end{array... | Yes |
The column space of \( A \) is a plane. The row space is all of \( {\mathbf{R}}^{2} \) . | \[ A = \left\lbrack \begin{array}{ll} 1 & 4 \\ 2 & 7 \\ 3 & 5 \end{array}\right\rbrack \text{ and }{A}^{\mathrm{T}} = \left\lbrack \begin{array}{lll} 1 & 2 & 3 \\ 4 & 7 & 5 \end{array}\right\rbrack \text{. Here }m = 3\text{ and }n = 2\text{. } \n\nThe row space is spanned by the three rows of \( A \) (which are columns... | Yes |
The columns of every invertible \( n \) by \( n \) matrix give a basis for \( {\mathbf{R}}^{n} \) | The only solution to \( A\mathbf{v} = \mathbf{0} \) is \( \mathbf{v} = {A}^{-1}\mathbf{0} = \mathbf{0} \) . The columns are independent. They span the whole space \( {\mathbf{R}}^{n} \) -because every vector \( \mathbf{b} \) is a combination of the columns. \( A\mathbf{v} = \mathbf{b} \) can always be solved by \( \mat... | No |
Example 8 This matrix is not invertible. Its columns are not a basis for anything ! | \[ \begin{array}{l} \text{ One pivot column } \\ \text{ One pivot row }\left( {r = 1}\right) \end{array}\;A = \left\lbrack \begin{array}{ll} 2 & 4 \\ 3 & 6 \end{array}\right\rbrack \text{ reduces to }R = \left\lbrack \begin{array}{ll} 1 & 2 \\ 0 & 0 \end{array}\right\rbrack . \]\n\nColumn 1 of \( A \) is the pivot colu... | Yes |
If \( {\mathbf{u}}_{1},\ldots ,{\mathbf{u}}_{m} \) and \( {\mathbf{w}}_{1},\ldots ,{\mathbf{w}}_{n} \) are both bases for the same vector space, then \( m = n \). | Suppose that there are more \( \mathbf{w} \) ’s than \( \mathbf{u} \) ’s. From \( n > m \) we want to reach a contradiction. The \( \mathbf{u} \) ’s are a basis, so \( {\mathbf{w}}_{1} \) must be a combination of the \( \mathbf{u} \) ’s. If \( {\mathbf{w}}_{1} \) equals \( {a}_{11}{\mathbf{u}}_{1} + \cdots + {a}_{m1}{\... | Yes |
For \( A = \left\lbrack \begin{array}{ll} 4 & 1 \\ 3 & 2 \end{array}\right\rbrack \), subtract \( \lambda \) from the diagonal and find the determinant : | \[ \det \left( {\mathbf{A} - \lambda \mathbf{I}}\right) = \det \left\lbrack \begin{matrix} 4 - \lambda & 1 \\ 3 & 2 - \lambda \end{matrix}\right\rbrack = {\lambda }^{2} - {6\lambda } + 5 = \left( {\mathbf{\lambda } - \mathbf{5}}\right) \left( {\mathbf{\lambda } - \mathbf{1}}\right) . \] | Yes |
Example 3 Find the eigenvalues and eigenvectors of \( S = \left\lbrack \begin{array}{ll} 2 & 1 \\ 1 & 2 \end{array}\right\rbrack \) . | Solution You can see that \( \mathbf{x} = \left( {1,1}\right) \) will be in the same direction as \( S\mathbf{x} = \left( {3,3}\right) \) .\n\nThen \( \mathbf{x} \) is an eigenvector of \( S \) with \( \lambda = 3 \) . We want the matrix \( S - {\lambda I} \) to be singular.\n\n\[ S = \left\lbrack \begin{array}{ll} 2 &... | No |
This real matrix has imaginary eigenvalues \( i, - i \) and complex eigenvectors : | \[ A = \left\lbrack \begin{array}{rr} 0 & - 1 \\ 1 & 0 \end{array}\right\rbrack = - {A}^{\mathrm{T}}\;\det \left( {A - {\lambda I}}\right) = \det \left\lbrack \begin{array}{rr} - \lambda & - 1 \\ 1 & - \lambda \end{array}\right\rbrack = {\mathbf{\lambda }}^{\mathbf{2}} + \mathbf{1} = 0. \] That determinant \( {\lambda ... | Yes |
Rotation comes from an orthogonal matrix \( Q \) . Then \( {\lambda }_{1} = {e}^{i\theta } \) and \( {\lambda }_{2} = {e}^{-{i\theta }} \) : | \[ Q = \left\lbrack \begin{matrix} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{matrix}\right\rbrack \;\begin{array}{ll} {\lambda }_{1} = \cos \theta + i\sin \theta & {\lambda }_{1} + {\lambda }_{2} = 2\cos \theta = \text{ trace } \\ {\lambda }_{2} = \cos \theta - i\sin \theta & {\lambda }_{1}{\lambda ... | Yes |
Example 1 Here \( A \) is triangular so the \( \lambda \) ’s are on its diagonal : \( \lambda = 1 \) and \( \lambda = 6 \) . | \[ \begin{matrix} \text{ Eigenvectors in }V & \left\lbrack \begin{array}{rr} 1 & - 1 \\ 0 & 1 \end{array}\right\rbrack \left\lbrack \begin{array}{ll} 1 & 5 \\ 0 & 6 \end{array}\right\rbrack \left\lbrack \begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right\rbrack = \left\lbrack \begin{array}{ll} 1 & 0 \\ 0 & 6 \end{array}... | Yes |
Example 2 Powers of \( A \) The Markov matrix \( A \) in the last section had \( {\lambda }_{1} = 1 \) and \( {\lambda }_{2} = {.5} \) . Here is \( A = {V\Lambda }{V}^{-1} \) with those eigenvalues in the matrix \( \Lambda \) : | \[ \left\lbrack \begin{array}{ll} {.8} & {.3} \\ {.2} & {.7} \end{array}\right\rbrack = \left\lbrack \begin{array}{rr} {.6} & 1 \\ {.4} & - 1 \end{array}\right\rbrack \left\lbrack \begin{array}{ll} 1 & 0 \\ 0 & {.5} \end{array}\right\rbrack \left\lbrack \begin{array}{rr} 1 & 1 \\ {.4} & - {.6} \end{array}\right\rbrack ... | Yes |
Example 3 Start from \( {\mathbf{u}}_{0} = \left( {1,0}\right) \) . Compute \( {A}^{k}{\mathbf{u}}_{0} \) when \( V \) and \( \Lambda \) contain these eigenvectors and eigenvalues : \[ A = \left\lbrack \begin{array}{ll} 1 & 2 \\ 1 & 0 \end{array}\right\rbrack \;\text{ has }\;{\lambda }_{1} = 2\;\text{ and }\;{\mathbf{x... | Example 3 in three steps Find \( {\mathbf{u}}_{0} = {c}_{1}{\mathbf{x}}_{1} + {c}_{2}{\mathbf{x}}_{2} \) and \( {\mathbf{u}}_{k} = {c}_{1}{\left( {\lambda }_{1}\right) }^{k}{\mathbf{x}}_{1} + {c}_{2}{\left( {\lambda }_{2}\right) }^{k}{\mathbf{x}}_{2} \) \[ \text{Step 1}\;{u}_{0} = \left\lbrack \begin{array}{l} 1 \\ 0 \... | Yes |
Find all solutions to \( {\mathbf{y}}^{\prime } = \left\lbrack \begin{array}{rr} - 2 & 1 \\ 1 & - 2 \end{array}\right\rbrack \mathbf{y} \) . Which solution has \( \mathbf{y}\left( 0\right) = \left\lbrack \begin{array}{l} 6 \\ 2 \end{array}\right\rbrack \) ? | Solution First we find \( \lambda = - 1 \) and -3 . Their eigenvectors \( {\mathbf{x}}_{1} \) and \( {\mathbf{x}}_{2} \) go into \( V \) :\n\n\[ \det \left\lbrack \begin{matrix} - 2 - \lambda & 1 \\ 1 & - 2 - \lambda \end{matrix}\right\rbrack = {\lambda }^{2} + {4\lambda } + 3\;\text{ factors into }\;\left( {\lambda + ... | Yes |
\[ {\mathbf{y}}^{\prime } = \left\lbrack \begin{array}{l} {y}_{1}^{\prime } \\ {y}_{2}^{\prime } \end{array}\right\rbrack = \left\lbrack \begin{array}{rr} 0 & 1 \\ - 1 & 0 \end{array}\right\rbrack \left\lbrack \begin{array}{l} {y}_{1} \\ {y}_{2} \end{array}\right\rbrack = \left\lbrack \begin{array}{r} {y}_{2} \\ - {y}_... | Discussion The equations are \( {y}_{1}^{\prime } = {y}_{2} \) and \( {y}_{2}^{\prime } = - {y}_{1} \) . One solution is \( {y}_{1} = \sin t \) and \( {y}_{2} = \cos t \) . A second solution is \( {y}_{1} = \cos t \) and \( {y}_{2} = - \sin t \) . We need two solutions to match two required values \( {y}_{1}\left( 0\ri... | Yes |
Example 4 \( {\left( \lambda - 2\right) }^{2} = {\lambda }^{2} - {4\lambda } + 4 = 0 \) comes from \( {y}^{\prime \prime } - 4{y}^{\prime } + {4y} = 0 \) : | \( \begin{array}{l} \text{ Companion matrix }A \\ \text{ Repeated root }\lambda = 2,2 \end{array}\;A = \left\lbrack \begin{array}{rr} 0 & 1 \\ - 4 & 4 \end{array}\right\rbrack \;\det \left( {A - {\lambda I}}\right) = {\lambda }^{2} - {4\lambda } + 4. \)\n\n\( \lambda = 2 \) must have one eigenvector, and it is \( x = \... | No |
The rotation matrix \( A = \left\lbrack \begin{array}{rr} 0 & 1 \\ - 1 & 0 \end{array}\right\rbrack \) has eigenvalues \( {\lambda }_{1} = i \) and \( {\lambda }_{2} = - i \). | \[ {e}^{At} = V{e}^{\Lambda t}{V}^{-1} = \left\lbrack \begin{array}{rr} 1 & 1 \\ i & - i \end{array}\right\rbrack \left\lbrack \begin{array}{ll} {e}^{it} & 0 \\ 0 & {e}^{-{it}} \end{array}\right\rbrack \frac{1}{2}\left\lbrack \begin{array}{rr} 1 & - i \\ 1 & i \end{array}\right\rbrack = \left\lbrack \begin{array}{rr} \... | Yes |
Example 2 Suppose \( A \) is triangular but we can’t diagonalize it (only one eigenvector):\n\n\[ \n{\mathbf{y}}^{\prime } = A\mathbf{y} = \left\lbrack \begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right\rbrack \left\lbrack \begin{array}{l} {y}_{1} \\ {y}_{2} \end{array}\right\rbrack \;\begin{array}{l} {y}_{1}^{\prime }... | Solution Since \( A \) is triangular, back substitution will solve \( {\mathbf{y}}^{\prime } = A\mathbf{y} \) . Begin by solving the last equation \( {y}_{2}{}^{\prime } = {y}_{2} \) . Then solve for \( {y}_{1} \) :\n\n\[ \n{y}_{2}\left( t\right) = {e}^{t}{y}_{2}\left( 0\right) \;\text{ Then }{y}_{1}{}^{\prime } = {y}_... | Yes |
Example 2 (using \( {e}^{At} \) ) For this triangular matrix \( A \), we can also add the series for \( {e}^{At} \) : | \[ {e}^{At} = I + {At} + \frac{1}{2}{\left( At\right) }^{2} + \frac{1}{6}{\left( At\right) }^{3} + \cdots \] \[ = \left\lbrack \begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right\rbrack + \left\lbrack \begin{array}{ll} t & t \\ 0 & t \end{array}\right\rbrack + \frac{1}{2}\left\lbrack \begin{array}{ll} {t}^{2} & 2{t}^{2}... | Yes |
The eigenvectors \( \left( {1,1}\right) \) and \( \left( {-1,1}\right) \) with \( \lambda = {16} \) and 4 give unit eigenvectors \( {\mathbf{x}}_{1} = \left( {1,1}\right) /\sqrt{2} \) and \( {\mathbf{x}}_{2} = \left( {-1,1}\right) /\sqrt{2} \) : | \[ S = \left\lbrack \begin{array}{rr} {10} & - 6 \\ - 6 & {10} \end{array}\right\rbrack \;{Q\Lambda }{Q}^{\mathrm{T}} = \frac{1}{\sqrt{2}}\left\lbrack \begin{array}{rr} 1 & - 1 \\ 1 & 1 \end{array}\right\rbrack \left\lbrack \begin{array}{ll} {16} & \\ & 4 \end{array}\right\rbrack \frac{1}{\sqrt{2}}\left\lbrack \begin{a... | Yes |
This 2 by 2 complex matrix is Hermitian (notice \( i \) and \( - i \) ) :\n\n\[ A = \left\lbrack \begin{array}{rr} 3 & i \\ - i & 3 \end{array}\right\rbrack = {A}^{ * } \] | The determinant is 8 (real). The trace is 6 (the main diagonal of a Hermitian matrix is real). The eigenvalues of this matrix are 2 and 4 (both real !). \n\nHermitian matrices \( A = {A}^{ * } \) have real eigenvalues and perpendicular eigenvectors.\n\nThe eigenvectors of \( A \) are \( {\mathbf{x}}_{1} = \left( {1, i}... | Yes |
Find the straight line \( b = C + {Dt} \) that goes through 4 points: \( b = 1,9,9,{21} \) at \( t = 0,1,3,4 \) . Those are four equations for \( C \) and \( D \), and they have no solution. | \[ \begin{array}{ll} {Av} = b\text{ has } & C + {0D} = 1 \\ \text{ no solution } & C + {1D} = 9 \\ \text{ no solution } & C + {3D} = 9 \\ & C + {4D} = {21} \end{array}\text{ is }\left\lbrack \begin{array}{ll} 1 & 0 \\ 1 & 1 \\ 1 & 3 \\ 1 & 4 \end{array}\right\rbrack \left\lbrack \begin{array}{l} C \\ D \end{array}\righ... | Yes |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.