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\textbf{Exercises on differential equations and $e^{At}$}
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\begin{problem} (6.3 \#14.a  {\em Introduction to Linear Algebra:} Strang) The matrix in this question is skew-symmetric $(A^T = -A):$
$$\frac{d\vb u}{dt} =\threebythree 0 c {-b} {-c} 0 a b {-a} 0 \vb u \hspace{8pt}\text{ or } \hspace{8pt}
\begin{array}{lll} 
{u_1}' = cu_2 - bu_3 \\
{{u_2}' = au_3 - cu_1} \\
{{u_3}' =bu_1 - au_2}. \end{array} $$
Find the derivative of $|| \vb u (t)||^2$ using the definition: $$|| \vb u (t)||^2 = {u_1}^2 + {u_2}^2 + {u_3}^2.$$ What does this tell you about the rate of change of the length of $\vb u$? What does this tell you about the range of values of $\vb u(t)$?
\end{problem}

\solution{
\begin{align*}
\frac{d|| \vb u (t)||^2}{dt} &= \frac{d({u_1}^2 + {u_2}^2 + {u_3}^2)}{dt} \\
&= 2u_1 {u_1}' + 2u_2 {u_2}' + 2u_3 {u_3}'  \\
&= 2u_1 (cu_2 - bu_3 ) + 2u_2 (au_3 - cu_1) + 2u_3 (bu_1 - au_2)\\ &=0.
\end{align*}
This means $||\vb u (t)||^2 $ stays equal to $||\vb u (0) ||^2$.  Because $\vb u(t)$ never changes length, it is always on the circumference of a circle of radius $||\vb u(0)||$. 
}

\begin{problem} (6.3 \#24.) Write $A = \small \twobytwo 1 1 0 3 $
as $S \Lambda S^{-1}$. Multiply $S e^{\Lambda t} S^{-1}$ to
find the matrix exponential $e^{At}.$ Check your work by evaluating
$e^{A t}$ and the derivative of $e^{At}$ when $t=0.$
\end{problem}

\solution{ The eigenvalues of $A$ are $\lambda_1 = 1$ and $\lambda_2 = 3$, with corresponding eigenvectors $\vb {x_1} = (1,0)$ and $\vb {x_2} = (1,2).$  This gives us the following values for $S, \Lambda, $ and $S^{-1}:$
$$S = \twobytwo 1 1 0 2, \Lambda = \twobytwo 1 0 0 3 , S^{-1} = \twobytwo 1 {-1/2} 0 {1/2}. $$
We use these to find $e^{At}:$
$$S e^{\Lambda t} S^{-1} = \twobytwo 1 1 0 2 \twobytwo {e^t} 0 0 {e^{3t}} \twobytwo 1 {-1/2} 0 {1/2}= \twobytwo {e^t} {.5e^{3t} - .5e^{t}} 0 {e^{3t}}= e^{At}.$$

Check: 
$$e^{A t} = \twobytwo {e^t} {.5e^{3t} - .5e^{t}} 0 {e^{3t}} \text{ equals } I \text{ when } t = 0. \hspace{4pt}\checkmark$$
$$\frac{de^{A t}}{dt} = \twobytwo {e^t} {1.5 e^{3t} - .5e^t} 0 {3e^{3t}}.$$
$$ \left. \frac{d e^{At}}{dt}\right|_{t=0} = \twobytwo 1 1 0 3 = A. \hspace{4pt}\checkmark $$
}

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