It is asked to evaluate the matrix exponential of
$$A=\begin{bmatrix} 0 & 1 & 0\\ 1 & 0 & 2 \\ 0 & 1 & 0 \end{bmatrix}$$
It is not hard to do this, since this matrix have 3 distinct eigenvalues and I can write it as $A=P \Lambda P^{-1}$ for $\Lambda$ diagonal. I evaluated some powers of it and it is not nilpotent. But it seems to have a "good" form, I mean, it has a couple of zeros so I was wondering if there is a better way of calculating the matrix exponential of $A$ without using its diagonal form.
Thanks in advance!
The matrix is degenerate (two rows are the same). This helps a little (even with eigenvalue decomposition).
In some cases, it's beneficial to construct a special basis, for instance identity, upper diagonal, lower diagonal, upper diagonal squared, lower diagonal squared. In this case, this would probably require an entire basis of $9$ matrices and you'd gain nothing. But the following works:
$$A=[0,1,0]^T [1,1,2] + [1,-1,1]^T [0,1,0]$$ This is a (guessed) SVD decomposition. Note only 2 terms, as the matrix has rank $2$. Write it symbolically $$A=a^T b + c^T a,\quad a=[0,1,0], b=[1,1,2], c=[1,-1,1]$$
Note $a^2=1$, $b^2=6$, $c^2=3$, $ab=1$, $ac=-1$, $bc=2$.
When you process all this, you find $A^3=3A$. Then you're finished. You could have done it without my decomposition, using only patience.
Bottom line:
$$B=A^2=\begin{bmatrix}1&0&2\\0&3&0\\1&0&2\end{bmatrix}$$
Consecutive powers:
$$(1,A,B,3A,3B,9A,9B,\ldots)$$