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If I wanted to calculate the matrix exponential of $\left ( \begin{array}{cc} -3 & 4 \\ -4 & -3 \end{array} \right )$, I could do this by calculating the eigenvectors and values and then use $e^{At}=P\ \mbox{diag}(e^{\lambda t}) P^{-1}$ where $P$ is the matrix with the eigenvectors of $A$ as its columns and I get $$e^{At}= e^{-3t}\left ( \begin{array}{cc} \cos(4t) & \sin(4t) \\ -\sin(4t) & \cos(4t) \end{array} \right )$$ but I was wondering how I do this using the definition, I'm sure it shouldn't be tough but I'm getting a bit lost; I have: $e^{At}=\sum\limits^{\infty}_{m=0} \frac{1}{m!}(At)^m$ but then looking at the powers of this matrix I just get rubbish i think? Thanks very much for any help.

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    $\begingroup$ Cayley-Hamilton theorem gives a quadratic polynomial $\phi(x)$ for which the matrix vanish. So the problem reduces to the calculation of the remainder of $x^n$ divided by $\phi(x)$. But this method also requires the knowledge about eigenvalues, so it's not so different from the method you know. There are several methods that circumvents this problems, which I know not so well. $\endgroup$ Mar 31, 2012 at 12:03
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    $\begingroup$ Your $e^{At}$ does not correspond to your $A$. $\endgroup$
    – Did
    Mar 31, 2012 at 12:54
  • $\begingroup$ @Didier sorry I edited it, should be -3 in the top left, however my tex was right just the - didnt show up. I have had to put 2 in now so that one shows up which is a bit strange, sorry about that should have checked my post but as I say it something to do with the tex. $\endgroup$
    – hmmmm
    Mar 31, 2012 at 13:10
  • $\begingroup$ Funny bug... A somewhat surer way to encode this is to use \begin{array}{cc}, see the code of the revised version of your post. // The new version of your question has an even simpler solution than the previous one, see the second part of my post. $\endgroup$
    – Did
    Mar 31, 2012 at 14:52
  • $\begingroup$ @Didier Thanks for the answer and again sorry for that bug, i'll keep the {cc} in mind next time thanks $\endgroup$
    – hmmmm
    Mar 31, 2012 at 15:01

3 Answers 3

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For a solution to the second version of the question, see below.

This applies to the first version of the question, where $\color{red}{A=\begin{pmatrix}3 & 4 \\ -4 & -3\end{pmatrix}}$. Since $\text{tr}(A)=0$ and $\det(A)=7$, Cayley-Hamilton theorem yields $A^2=-7I$. Hence, for every $n\geqslant0$, $A^{2n}=(-7)^n\cdot I$ and $A^{2n+1}=(-7)^n\cdot A$. This yields $$ \mathrm e^{tA}=\sum\limits_{n=0}^{+\infty}\frac{t^n}{n!}\cdot A^n=\sum\limits_{n=0}^{+\infty}(-1)^n7^n\frac{t^{2n}}{(2n)!}\cdot I+\sum\limits_{n=0}^{+\infty}(-1)^n7^n\frac{t^{2n+1}}{(2n+1)!}\cdot A, $$ that is, $$ \color{red}{\mathrm e^{tA}=\cos(\sqrt7t)I+\frac{\sin(\sqrt7t)}{\sqrt7}A}. $$


The second version of the question, where $\color{blue}{A=\begin{pmatrix}-3 & 4 \\ -4 & -3\end{pmatrix}}$, can be solved even more easily. Since $A+3I=4J$ with $J=\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}$, one has $\mathrm e^{tA}=\mathrm e^{-3t}\cdot\mathrm e^{4tJ}$. Note that $J^2=-I$ (and/or that $J$ is nothing but the matrix of the clockwise rotation by $90°$), hence, for every $n\geqslant0$, $J^{2n}=(-1)^nI$ and $J^{2n+1}=(-1)^nJ$. This yields $$ \mathrm e^{tJ}=\sum\limits_{n=0}^{+\infty}\frac{t^n}{n!}\cdot J^n=\sum\limits_{n=0}^{+\infty}(-1)^n\frac{t^{2n}}{(2n)!}\cdot I+\sum\limits_{n=0}^{+\infty}(-1)^n\frac{t^{2n+1}}{(2n+1)!}\cdot J, $$ that is, $$ \mathrm e^{tJ}=\cos(t)I+\sin(t)J. $$ Coming back to $A$ and using the formula for $\mathrm e^{4tJ}$, one gets $$ \color{blue}{\mathrm e^{tA}=\mathrm e^{-3t}\cdot\left(\cos(4t)I+\sin(4t)J\right)}. $$

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  • $\begingroup$ Thanks for the response, sorry about the question being wrong -I think that changes the answer somewhat. Don't know what is going on with the tex $\endgroup$
    – hmmmm
    Mar 31, 2012 at 13:11
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Post problem change; i.e $A=\begin{bmatrix}-3&4\\-4&-3\end{bmatrix}$. $A^2+6A+25I=0$ therefore $A^n=a_nA+b_nI$ for some $a_n$ and $b_n$. Note that $$ \begin{align} a_nA+b_n &=(a_{n-1}A+b_{n-1}I)A\\ &=a_{n-1}(-6A-25I)+b_{n-1}A\tag{1} \end{align} $$ Therefore, $a_n=-6a_{n-1}+b_{n-1}$ and $b_n=-25a_{n-1}$; that is, $$ A^n=a_nA-25a_{n-1}I\tag{2} $$ where $$ a_n=-6a_{n-1}-25a_{n-2}\tag{3} $$ Solving $(3)$ with $a_1=1$ and $a_0=0$ yields $$ a_n=\frac{(-3+4i)^n-(-3-4i)^n}{8i}\tag{4} $$ Now $$ \begin{align} e^{tA} &=\sum_{n=0}^\infty\frac{1}{n!}t^nA^n\\ &=\sum_{n=0}^\infty\frac{t^n}{n!}\left(\frac{(-3+4i)^n-(-3-4i)^n}{8i}A-25\frac{(-3+4i)^{n-1}-(-3-4i)^{n-1}}{8i}I\right)\\ &=\frac14e^{-3t}\sin(4t)A-\frac{1}{4}e^{-3t}(-3\sin(4t)-4\cos(4t))I\\ &=e^{-3t}\begin{bmatrix}\cos(4t)&\sin(4t)\\-\sin(4t)&\cos(4t)\end{bmatrix}\tag{5} \end{align} $$

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I know you explicitly wanted to use the definition, but another useful way to get $e^{tA}$ would be to use the Laplace transform. $$ e^{tA}=\mathcal{L}^{-1}\left((sI-A)^{-1}\right)= \mathcal{L}^{-1}\left(\begin{bmatrix}s+3 & -4 \\ 4 & s+3\end{bmatrix}^{-1}\right) =\mathcal{L}^{-1}\left(\frac{1}{(s+3)^2+16}\begin{bmatrix}s+3 & 4 \\ -4 & s+3\end{bmatrix}\right)= e^{-3t}\mathcal{L}^{-1}\left(\frac{1}{s^2+16}\begin{bmatrix}s & 4 \\ -4 & s\end{bmatrix}\right) =e^{-3t}\begin{bmatrix}\cos{4t} & \sin{4t} \\ -\sin{4t} & \cos{4t}\end{bmatrix}. $$

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