Find $e^{xA}$ by definition

Question

$A =\left [ \begin{matrix} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ \end{matrix} \right ] $

I have to find $e^{xA}$ by definition.

$A^1 =\left [ \begin{matrix} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ \end{matrix} \right ] $,$A^2 =\left [ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 4 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right ] $,$A^3 =\left [ \begin{matrix} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 6 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ \end{matrix} \right ] $...

Then, $A^n =\left [ \begin{matrix} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 2n & 1 & 0 \\ 1 & 0 & 0 & 0 \\ \end{matrix} \right ]$ for odd $n$

, $A^n =\left [ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 2n & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right ] $ for even $n$

For odd $n$:

$e^{xA}=\left [ \begin{matrix} 0 & 0 & 0 & e^x \\ 0 & e^x & 0 & 0 \\ 0 & 2xe^x & e^x & 0 \\ e^x & 0 & 0 & 0 \\ \end{matrix} \right ]$

For even $n$:

$e^{xA}=\left [ \begin{matrix} e^x & 0 & 0 & 0 \\ 0 & e^x & 0 & 0 \\ 0 & 2xe^x & e^x & 0 \\ 0 & 0 & 0 & e^x \\ \end{matrix} \right ]$

Is it correct ?

How can I find the general $e^{xA}$? Is it the sum of both ?

Thanks !

Answer 1

You've done well, but you stumbled a little at the last hurdle. Consider the following:

$$\begin{eqnarray}e^{xA} & = & \sum_{n = 0}^\infty \frac{x^n A^n}{n!} \\ & = & \sum_{n \text{ even}} \frac{x^n A^n}{n!} + \sum_{n \text{ odd}} \frac{x^n A^n}{n!} \\ & = & \sum_{k = 0}^\infty \frac{x^{2k} A^{2k}}{(2k)!} + \sum_{k = 0}^\infty \frac{x^{2k+1} A^{2k+1}}{(2k+1)!} \end{eqnarray}$$

What you did was nearer to $\sum_{k=0}^\infty \frac{x^k A^{2k}}{k!}$ for the even terms and similarly for the odd terms, but notice that this doesn't line up at all.

I won't give the final answer, but as mentioned in the comments the sum of the even terms of the expansion of $e^x$ gives you $\cosh x$ and the sum of the odd terms gives you $\sinh x$, which should be enough for you to get a final answer.

Answer 2

If $n$ is odd, then$$(xA)^n=\begin{bmatrix}0&0&0&x^n\\0&x^n&0&0\\0&2nx^n&x^n&0\\x^n&0&0&0\end{bmatrix}$$and, if $n$ is even, then$$(xA)^n=\begin{bmatrix}x^n&0&0&0\\0&x^n&0&0\\0&2nx^n&x^n&0\\0&0&0&x^n\end{bmatrix}.$$Therefore, if $e^{xA}=B=(b_{ij})_{1\leqslant i,j\leqslant4}$, then:

• $b_{11}=b_{44}=1+\frac1{2!}x^2+\frac1{4!}x^4+\cdots=\cosh(x)$;
• $b_{12}=b_{13}=b_{21}=b_{23}=b_{24}=b_{31}=b_{34}=b_{42}=b_{43}=0$;
• $b_{14}=b_{41}=x+\frac1{3!}x^3+\frac1{5!}x^5+\cdots=\sinh(x)$;
• $b_{22}=b_{33}=e^x$;
• $b_{32}=2x+\frac4{2!}x^2+\frac6{3!}x^3+\cdots=2x\left(1+x+\frac1{2!}x^2+\frac1{3!}x^3+\cdots\right)=2xe^x$.

In other words,$$e^{xA}=\begin{bmatrix}\cosh(x)&0&0&\sinh(x)\\0&e^x&0&0\\0&2xe^x&e^x&0\\\sinh(x)&0&0&\cosh(x)\end{bmatrix}.$$