$A=\begin{bmatrix}3&2\\0&3\end{bmatrix}$
Since it is upper triangular, I right away know that its eigenvalues are 3 and 3. Therefore, if $A$ is diagonalizable then $\exists$ matrix $V$ such that
$A=VDV^{-1}=V\begin{bmatrix}3&0\\0&3\end{bmatrix}V^{-1}$
After emailing my professor, he replied: Can you say anything about the right hand side of the last equation knowing D but not what V is?
And I have no idea what that means or implies.
Is there something obvious I am not seeing? Help! Thanks!
$$V \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} V^{-1} = 3V \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} V^{-1} = 3VV^{-1} = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \neq \begin{bmatrix} 3 & 2 \\ 0 & 3 \end{bmatrix}$$