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I am trying to find the eigen value of matrix $e^A$ such that the matrix A is given as

$$ \begin{bmatrix} a & 0 & 0\\ 0 & 0 & a\\ 0 & a & 0 \end{bmatrix} $$

I found $A^2 = \begin{bmatrix} a^2 & 0 & 0 \\ 0 & a^2 & 0 \\ 0 & 0 & a^2 \end{bmatrix}$, $A^3 = \begin{bmatrix} a^3 & 0 & 0 \\ 0 & 0 & a^3 \\ 0 & a^3 & 0 \end{bmatrix}$ and so forth...

So that I can find, $e^A$ by substituting above matrices in its power series expansion \begin{equation} e^A = 1 + A + \frac{1}{2!} A^2+\dots \end{equation}

which gives

\begin{equation} e^A = \begin{bmatrix} 1+a+\frac{1}{2}a^2+\frac{1}{6}a^3+\dots & 0 & 0 \\ 0 & 1+\frac{1}{2}a^2+\frac{1}{24}a^4+\dots & a+\frac{1}{6}a^3+\frac{1}{120}a^5\dots \\ 0 & a+\frac{1}{6}a^3+\frac{1}{120}a^5\dots & 1+\frac{1}{2}a^2+\frac{1}{24}a^4+\dots \end{bmatrix} \end{equation}

which can be written as

\begin{equation} e^A = \begin{bmatrix} e^a & 0 & 0 \\ 0 & \frac{e^a}{2}+\frac{e^{-a}}{2} & \frac{e^a}{2}-\frac{e^{-a}}{2} \\ 0 & \frac{e^a}{2}-\frac{e^{-a}}{2} & \frac{e^a}{2}+\frac{e^{-a}}{2} \end{bmatrix} \end{equation}

I am able to solve this, but I am wondering if there is a quicker way or a trick that I can use to find $e^A$ quickly that this method?

hmakholm left over Monica
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orionphy
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    One thing you can do as well is to note that your matrix is diagonalisable. The exponential of a diagonal matrix is very simple. – s.harp Mar 23 '19 at 13:07
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    Please avoid double dollar signs or displays on titles. Regards, – Pedro Mar 23 '19 at 13:15
  • Your matrix exponent $A$ is real symmetric, so it has a complete basis of eigenvalues, i.e. $A$ is diagonalizsble. If you are interested in the eigenvalues of $e^A$, you should work with the eigenvalues of $A$. – hardmath Mar 23 '19 at 13:38

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By diagonalization, $$\exp\begin{pmatrix} 0 & a \\ a & 0 \end{pmatrix} = \begin{pmatrix} \cosh(a) & \sinh(a) \\ \sinh(a) & \cosh(a)\end{pmatrix} $$ is just De Moivre/Euler's formula $e^{i\alpha}=\cos(\alpha)+i\sin(\alpha)$ written in a equivalent way.

Jack D'Aurizio
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