True or False
\(e^{t\ABld} = \sum_{k=0}^{\infty}\frac{1}{k!}t^k{\ABld}^k\)
If \({\ABld} = \begin{pmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}\) and \(a_{12}\) and \(a_{21}\) are nonzero, then \(e^{t\ABld} =\) \(\begin{pmatrix}e^{ta_{11}} & e^{ta_{12}} \\ e^{ta_{21}} & e^{ta_{22}} \end{pmatrix}\)
For any \(m\times m\) matrix, \(e^{t\ABld} = N_0(t){\bf I} + N_1(t){\bf A} + \cdots + N_{m-1}(t){\bf A}^{m-1}\), where \(N_0(t), N_1(t), \cdots, N_{m-1}(t)\) is the natural fundamental set of solutions for the \(m^{th}\)-order differential equation corresponding to matrix \({\bf A}\) (with initial time \(0\)).
The natural fundamental set of solutions \(N_0(t), N_1(t), \cdots, N_{m-1}(t)\) can be obtained by solving the \(m^{th}\)-order differential equation \(p(\Dop)y = 0\), with general initial conditions of \(y(0) = y_0, y'(0) = y_1, \cdots, y^{(m-1)}(0)=y_{m-1}\), where \(p(z)\) is the characteristic polynomial of \(\ABld\)
The solution to the initial value problem, \(\xBld' = \ABld\xBld\) with \(\xBld(0) = \xBld^{0}\), is given by \(\xBld(t) = e^{t\ABld} \xBld^{0}\).
True or False (The following refers only to \(2\times 2\) matrices)
If the characteristic polynomial corresponding to the matrix \(\ABld\), has simple real roots \(\mu \pm \nu\), then \(e^{t\ABld} = e^{\mu t}\left[\cos(\nu t){\IBld} + \frac{\sin(\nu t)}{\nu}({\ABld}-\mu{\IBld})\right]\). (Note: \(\mu \pm \nu\) is obtained by completing the squares on the characteristic polynomial (ie \((z-\mu)^2 - \delta)\) with \(\nu = \sqrt{\delta}\))
If the characteristic polynomial corresponding to the matrix \(\ABld\), has double roots \(\mu\), then \(e^{t\ABld} = e^{\mu t}\left[{\IBld} + t({\ABld} - \mu{\IBld})\right]\)
If the characteristic polynomial corresponding to the matrix \(\ABld\), has complex conjugate roots \(\mu \pm i \nu\), then \(e^{t\ABld} = e^{\mu t}\left[\cosh(\nu t){\IBld} + \frac{\sinh(\nu t)}{\nu}({\ABld}-\mu{\IBld})\right]\)
Let \(\ABld\) be a \(2\times 2\) matrix. Use the Caley-Hamilton Theorem to derive the formula for the inverse of \(\ABld = \begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22}\end{pmatrix}\), \[\ABld^{-1} = \frac{1}{\det\left(\ABld\right)} \begin{pmatrix} a_{22} & -a_{12}\\ -a_{21} & a_{11}\end{pmatrix}.\]
Let \(\ABld\) be a \(2\times 2\) matrix. Use the Caley-Hamilton Theorem to prove the following formula for the determinant of \(\ABld\), \[\det\left(\ABld\right) = \frac{1}{2}\left(\text{tr}\left(\ABld\right)^2 - \text{tr}\left(\ABld^2\right)\right).\]
Consider the matrix \[\ABld = \begin{pmatrix} 0 & 1 & 1\\ 1 & 0 & 1\\ 1 & 1 & 0 \end{pmatrix}\] Given that the characteristic polynomial is \(p(z) = z^3 - 3z -2\), use the Caley-Hamilton Theorem to find the inverse of \(\ABld\).
In the text it was stated that for any constant \(n\times n\) matrix \(\ABld\) and any \(t\) and \(s\), the following property holds \[e^{(t+s)\ABld} = e^{t\ABld}e^{s\ABld}.\] As was outlined in the text, show this by showing that both sides of the equation satisfy the same initial value problem.
For 7–9 verify that the matrix \(\PhiBld(t)\) is a matrix exponential \(e^{t\ABld}\) for some \(\ABld\) by checking that \(\PhiBld(t)\) satisfies the properties of a matrix exponential, i.e. \(\PhiBld(0) = \IBld\) and \(\PhiBld(s)\PhiBld(t) = \PhiBld(t+s)\) for any \(t,s\). Use this to find the inverse \(\PhiBld(t)^{-1}\).
\( \PhiBld(t) = \begin{pmatrix} \cos{(2t)} & \frac{1}{2}\sin{(2t)}\\ -2\sin{(2t)} & \cos{(2t)} \end{pmatrix} \)
\( \PhiBld(t) = e^{t} \begin{pmatrix} 1& t\\ 0 & 1 \end{pmatrix} \)
\( \PhiBld(t) = e^t\begin{pmatrix} \cosh{(2t)} & 2\sinh{(2t)}\\ \frac{1}{2}\sinh{(2t)} & \cosh{(2t)} \end{pmatrix} \)
For 10-19 compute \(e^{t\ABld}\).
\(\ABld = \begin{pmatrix} 3 & 2 \\ 2 & 3 \end{pmatrix}\)
\({\bf A} = \begin{pmatrix} -1 & -4 \\ 1 & -1 \end{pmatrix}\)
\({\bf A} = \begin{pmatrix}2 & -1 \\ 3 & -2 \end{pmatrix}\)
\({\bf A} = \begin{pmatrix}1 & \frac{-1}{2} \\ 2 & -1 \end{pmatrix}\)
\({\bf A} = \begin{pmatrix}1 & -1 \\ 5 & -3 \end{pmatrix}\)
\({\bf A} = \begin{pmatrix}7 & -9 \\ 1 & 1 \end{pmatrix}\)
\({\bf A} = \begin{pmatrix}1 & 5 \\ -1 & 3 \end{pmatrix}\)
\({\bf A} = \begin{pmatrix}5 & 3 \\ 3 & 5 \end{pmatrix}\)
\({\bf A} = \begin{pmatrix}1 & -4 \\ 4 & -7 \end{pmatrix}\)
\(\ABld = \begin{pmatrix} -3 & 2 & 0\\ -1 & 0 & 0\\ -3 & 3 & 1 \end{pmatrix}\)
For 20-25, find the solution to the initial value problem using \(e^{t\ABld}\).
\({\bf x}' = \begin{pmatrix} -2 & 1 \\ -5 & 4 \end{pmatrix} {\bf x}, \hspace{.5 in} {\bf x}(0) = \begin{pmatrix} 2 \\ 1 \end{pmatrix} \)
\({\bf x}' = \begin{pmatrix} -\frac{5}{2} & \frac{3}{2} \\ -\frac{3}{2} & \frac{1}{2} \end{pmatrix} {\bf x}, \hspace{.5 in} {\bf x}(0) = \begin{pmatrix} -1 \\ 1 \end{pmatrix} \)
\({\bf x}' = \begin{pmatrix} 1 & -5 \\ 1 & -3 \end{pmatrix} {\bf x}, \hspace{.5 in} {\bf x}(0) = \begin{pmatrix} 1 \\ 4 \end{pmatrix} \)
\({\bf x}' = \begin{pmatrix} 2 & \frac{-5}{2} \\ \frac{9}{5} & -1 \end{pmatrix} {\bf x}, \hspace{.5 in} {\bf x}(0) = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \)
\({\bf x}' = \begin{pmatrix} 1 & 0 & 0 \\ -4 & 1 & 0 \\ 3 & 6 & 2 \end{pmatrix} {\bf x}, \hspace{.5 in} {\bf x}(0) = \begin{pmatrix} -1 \\ 3 \\ 1 \end{pmatrix} \)
\({\bf x}' = \begin{pmatrix} 1 & -1 & 4 \\ 0 & 2 & -1 \\ 0 & 0 & -1 \end{pmatrix} {\bf x}, \hspace{.5 in} {\bf x}(0) = \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} \)
Solve the general initial value problems to find the natural fundamental solutions given in equations (4.18a), (4.18b), (4.18c) in the text.
Let \(\ABld\) be a \(3{\times}3\) matrix of the form \[\ABld = \begin{pmatrix} 0 & a_{12} & a_{13} \\ 0 & 0 & a_{23} \\ 0 & 0 & 0 \end{pmatrix}.\] Compute the matrix exponential in terms of \(\ABld\). How does this compare to the series representation of the the matrix exponential? Suppose that \(\ABld\) is more generally given by \[\ABld = \begin{pmatrix} 0 & a_{12} & a_{13} & \cdots & a_{1n} \\ 0 & 0 & a_{23} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 & a_{(n-1),n} \\ 0 & 0 & \cdots & 0 & 0 \end{pmatrix} \,.\] Can you guess the form of the matrix exponential in this case? (see exercise 30 in the supplement on matricies and vectors for a hint on the annihilator for \(\ABld\)).
Derive \(\frac{\dee}{\dt}e^{t\ABld} = {\ABld}e^{t\ABld}\) using the series representation of \(e^{t\ABld}\). (You may pull derivatives inside infinite sums without justification).
Compute the natural fundamental sets of \(y'' - 6y' + 5y = 0\) using Green functions. Confirm that these are the same natural sets obtained from Problem 10.
Compute the natural fundamental sets of \(-y''' + 4y'' - 5y' + 2y = 0\) using Green functions. (The characteristic polynomial of this equation has roots \(z = 1, 1, 2\).) Confirm that these are the same natural sets obtained from Problem 24.
Calculate \(e^{t\ABld}\) using Hermite interpolation methods.
\({\bf A} = \begin{pmatrix}2 & -1 \\ 3 & -2 \end{pmatrix}\)
\({\bf A} = \begin{pmatrix}1 & -4 \\ 4 & -7 \end{pmatrix}\)