# Matrix exponential question

if a matrix A is diagonal

$$A=\begin{bmatrix} a_1 & 0 & \ldots & 0 \\ 0 & a_2 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & a_n \end{bmatrix}$$

then its exponential can be obtained by exponentiating each entry on the main diagonal:

$$e^A=\begin{bmatrix} e^{a_1} & 0 & \ldots & 0 \\ 0 & e^{a_2} & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & e^{a_n} \end{bmatrix}$$

My question is will this statement holds of any (Real) $n$ by $n$ matrix? Or if the entries are on the other diagonal? That is,

$$A=\begin{bmatrix} 0 & \ldots & 0 & a_1 \\ 0 & \ldots &a_2 &0 \\ \vdots & \ddots & \vdots & \vdots \\ a_n & \ldots & 0 &0 \end{bmatrix}$$

My question is will this statement holds of any (Real) $n$ by $n$ matrix?

If you mean for any real diagonal matrix, then yes, your understanding is right. If you mean any real $n\times n$ matrix, then no, in general you must use the definition $e^A = \sum\limits_{k=0}\frac{A^k}{k!}$. This always works because every finite dimensional square matrix is a bounded linear operator, thus has finite norm so that it follows that $\|e^A\|\leq \exp(\|A\|)$ on applying the triangle inequality to the partial sums of the defining series.

.. Or if the entries are on the other diagonal? That is,$$A=\begin{bmatrix} 0 & \ldots & 0 & a_1 \\ 0 & \ldots &a_2 &0 \\ \vdots & \ddots & \vdots & \vdots \\ a_n & \ldots & 0 &0 \end{bmatrix}$$

No, surprisingly there is no easy formula for this "forward slash" diagonal. To see this, apply the definition (the exponential power series) to the $2\times2$ example $A = \begin{pmatrix} 0 & -\theta\\\theta&0 \end{pmatrix}$ for $\theta\in\mathbb{R}$. In this case, the characterstic equation for $A$ is $A^2 + \theta^2\mathrm{id}=0$, so you can use this to simplify the exponential series to:

$$e^A = \cos\theta\mathrm{id} + \sin\theta \begin{pmatrix} 0 & -1\\1&0 \end{pmatrix} = \begin{pmatrix} \cos\theta & - \sin\theta\\\sin\theta& \cos\theta \end{pmatrix}$$

and we get the $2\times 2$ rotation matrix, which is not of the "forward slash diagonal" form.

1) Ies it holds for any square diagonal matrix with real (or complex) entries.

2) No. For non diagonal matrix $A$ you have to diagonalize it and, if $A=PDP^{-1}$ we have $e^A = Pe^DP^{-1}$ . If $A$ is not diagonalizable then we can compute the exponential using the Jordan canonical form, but , in general the solution can be complicated and numerical methods need to be used ( see here for a discussion)