$\tiny M=\begin{bmatrix}1 & 1 &0\\0 & 1 & 1\\0 & 0 & 1\end{bmatrix}$ and $e^M$=$[a_{ij}]$. Then $\sum_{i=1}^3 \sum_{j=1}^{3}a_{ij}=?$

Question

let $M=\begin{bmatrix}1 & 1 &0\\0 & 1 & 1\\0 & 0 & 1\end{bmatrix}$ and $e^M$=$[a_{ij}]$.Then $\sum_{i=1}^3 \sum_{j=1}^{3}a_{ij} =?$ I tried my way and got the answer as $5e$ but that's not correct because the correct answer is $5.5e$.Can you suggest any better procedure to solve this problem$?$Thank you.

Answer 1

Note that your matrix $M$ is in Jordan-Normal-Form, hence 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 both factors are easy to calculate since $N$ is nilpotent. We get:

$$ e^I e^N = eI\cdot(I+N+\frac{1}{2}N^2) = e\begin{pmatrix}1&1&\frac{1}{2} \\ 0&1&1 \\0&0&1 \end{pmatrix}$$

Thus the sum of the entries is $\displaystyle\frac{11e}{2}$.

Answer 2

$ \lambda=1$ is the only eigenvalue of $A$, hence by Cayley-Hamilton: $(A-I)^3=0$. Then:

$\frac{1}{e}e^A=e^{A-I}=A+\frac{1}{2}(A-I)^2$

Its your turn to compute $(A-I)^2$