Find an invertible matrix V and a diagonal matrix D so that A=VDV^-1
$\displaystyle \left(\begin{array}{rrr}5&8\\2&-3\end{array}\right) $
I got the eigenvalue as u=1 and the eigenvector as $\displaystyle \left(\begin{array}{rrr}1\\-2\end{array}\right)$..but I'm not sure what to do after that. Since the question also asks "or explain why this cannot be done" that is what I am thinking... honestly I'm not sure.
Either you have written the matrix incorrectly or you have the wrong eigenvalues. The "characteristic equation" is
$\displaystyle \left|\begin{array}{cc}5-\lambda & 8 \\ 2 & -3-\lambda\end{array}\right|= 0$
$\displaystyle (5-\lambda)(-3-\lambda)- 16= \lambda^2- \lambda- 15-16= \lambda^2- \lambda- 31= 0$
$\displaystyle \lambda= 1$ does NOT satisfy that. In fact, the eigenvalues are $\displaystyle \lambda= 1\pm 4\sqrt{2}$. Since there are two distinct eigenvalues, there are two independent eigenvectors and the matrix is "diagonalizable" with the $\displaystyle D= \begin{bmatrix}1+ 4\sqrt{2} & 0 \\ 0 & 1- 4\sqrt{2}\end{bmatrix}$ and V equal to the matrix having the eigenvectors as columns.
On the other hand, if either the "2" or "8" in the matrix is negative, then the characteristic equation is $\displaystyle \lambda^2- 2\lambda+ 1= (\lambda-1)^2= 0$ and has $\displaystyle \lambda= 1$ is a "double" eigenvalue. Now you need to find the eigenvectors. If there are two independent eigenvectors corresponding to eigenvalue 1, then A is "diagonalizable": $\displaystyle D= VAV^{-1}$ with D being the diagonal matrix $\displaystyle \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$, with the eignvalues along the diagonal, and V is the matrix having two independent eigenvectors as columns. If there are not two independent eigenvalues (all eigenvalues can be written as a multiple of one eigenvector) then the matrix is not diagonalizable.
(But, in that case, it could be written in "Jordan Normal Form" $\displaystyle \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$.)