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Wiki https://en.wikipedia.org/wiki/Matrix_exponential said:

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,

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

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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.

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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)

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