Find $e^{At}$ where $A=\left(\begin{smallmatrix} -7&8&2\\-4&5&1\\-23&21&7 \end{smallmatrix}\right)$ Let $$A=\begin{pmatrix}
-7&8&2\\-4&5&1\\-23&21&7  
\end{pmatrix}$$
Find $e^{At}$.
I've found that $p(\lambda)=(\lambda-1)(\lambda-2)^2$
$\underline{\lambda=1}$
$$\begin{pmatrix}
-8&8&2\\-4&4&1\\-23&21&6  
\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}\to v_1=\begin{pmatrix}3\\1\\8\end{pmatrix}$$
$\underline{\lambda=2}$
$$\begin{pmatrix}
-9&8&2\\-4&3&1\\-23&21&5  
\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}\to v_2=\begin{pmatrix}2\\1\\5\end{pmatrix}$$
$\begin{pmatrix}
-9&8&2\\-4&3&1\\-23&21&5  
\end{pmatrix}=\begin{pmatrix}2\\1\\5\end{pmatrix}\to v_3=$some vector
setting $C=\begin{pmatrix}|&|&|\\v_1&v_2&v_3\\|&|&|\end{pmatrix}$
Then the solution is $$e^{At}=C\begin{pmatrix}e^{t}&0&0\\0&e^{2t}&te^{2t}\\
0&0&e^{2t}\end{pmatrix}C^{-1}$$
My question is if $\begin{pmatrix}e^{t}&0&0\\0&e^{2t}&te^{2t}\\
0&0&e^{2t}\end{pmatrix}$ has the right form?
 A: You cannot set $v_3$ to be arbitrary, you must pick $v_3$ to be such that conjugating gives you a matrix of the form
$$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix}.$$
But if you do, you will have found the Jordan Normal Form of the matrix.  If $A$ is a sum of Jordan blocks, then $e^{At}$ will also be a sum of blocks, and to figure out what those blocks are, it suffices to understand what $e^{At}$ is when $A=J_n(\lambda)$. is a single Jordan block.  There is a nice trick to this: When can write $J_n(\lambda)=\lambda I + N$ where $N$ is the matrix of $1$s just above the diagonal.  Since $\lambda I$ and $N$ commute, we can actually apply the bionomial theorem to compute powers of $J_n(\lambda)$ (which does not in general work for calculating powers of sums of matrices).
Before we can do this, we need a small calculation: $N^k$ will be the matrix which is all $0$ except for $1$s along the diagonal $k$ above the main diagonal (or $0$ everywhere if $k\geq n$).
I will not work out the general case here, but let us see what it gives in your particular case, where $\lambda=2$ and $N^2=0$.
$$(2I+N)^n=\sum_{i+j=n}\binom{n}{i}2^iN^j=2^nI+n2^{n-1}N.$$
Then
$$\begin{align}e^{t(2I+N)}&=\sum_{n\geq 0} \frac{t^n}{n!}(2I+N)^n \\
             &=\sum_{n\geq 0} \frac{t^n}{n!}(2^nI+n2^{n-1}N)\\
             &=\sum_{n\geq 0} \frac{(2t)^n}{n!}I+\sum_{n\geq 0} \frac{n2^{n-1}t^n}{n!}N\\ 
             &=e^{2t}I+te^{2t}N.          
\end{align}$$
The upshot of this calculation is that yes, your answer is correct (if you make the proper choice of $v_3$). It is a worthwhile exercise to do this calculation for a $3\times 3$ block.
