If $$M=\begin{pmatrix} 1 & 1 & 0\\ 0 & 1 & 1\\ 0 & 0 & 1\\ \end{pmatrix}$$ is a matrix and $e^M=\sum_\limits{n=0}^{\infty}\frac{M^n}{n!}$ then find $\frac{1}{e}\sum_\limits{i=1}^{3}\sum_\limits{j=1}^{3} b_{ij}$ where $b_{ij}$ are entries of $e^M$.
I know the definition of $e^M$ and how to find $e^M$ for diagonalisable matrix, nilpotent matrix but I've no idea how to solve this please give some hints to solve this problem. Thanks in advance.
On
HINT
You know how to diagonalize a matrix, that is, to find a diagonal matrix $D$ and an orthogonal matrix $P$ (which will be the matrix whose columns are the eigenvectors of $M$) such that $$ M = PDP^{-1} $$ ($D$ will have the eigenvalues of $M$ on the diagonal elements.)
Then for all $k$, $$ M^k = PD^kP^{-1} $$ since pairs of $P^{-1}P$ cancel out except at the left and right sides of the product.
Finally, notice how this lets you relate $e^M$ to $P e^D P^{-1}$ since you know how to exponentiate the diagonal matrix $D$.
We can write $M=I+N$ where
$$ I = \begin{pmatrix}1&0&0 \\ 0&1&0 \\0&0&1 \end{pmatrix} \qquad N = \begin{pmatrix}0&1&0 \\ 0&0&1 \\0&0&0 \end{pmatrix} $$
Since $I$ is the identity, $N$ and $I$ commute; hence $e^{I+N} = e^I e^N$ and $N$ is nilpotent with index $3$. So, $N^3=0$
$$ e^I e^N = eI\cdot(I+N+\frac{1}{2}N^2) = e\begin{pmatrix}1&1&\frac{1}{2} \\ 0&1&1 \\0&0&1 \end{pmatrix}$$
Thus the sum of the entries of the matrix $M$ is $\displaystyle 5.5e$.