Determine if matrices A and B are similar Determine if matrices $A$ and $B$ are similar.

$A=\begin{pmatrix}1 &amp; 1 &amp; 0 &amp; 0 \\0 &amp; 1 &amp; 0 &amp; 0 \\0 &amp; 0 &amp; 3 &amp; 0 \\0 &amp; 0 &amp; 0 &amp; 3\end{pmatrix}\: $

and

$B=\begin{pmatrix}1 &amp; 1 &amp; 0 &amp; 0 \\0 &amp; 1 &amp; 0 &amp; 0 \\0 &amp; 0 &amp; 3 &amp; 1 \\0 &amp; 0 &amp; 0 &amp; 3\end{pmatrix}$

I found the characteristic polynomials to be the same for both matrices

$(1-λ)(3-λ)(3-λ)(1-λ)$

so I went to plug in the values to find the eigenvectors for matrix $A$ and I think I may have screwed up because 

$\begin{pmatrix}0 &amp; 1 &amp; 0 &amp; 0 \\0 &amp; 0 &amp; 0 &amp; 0 \\0 &amp; 0 &amp; 2 &amp; 0 \\0 &amp; 0 &amp; 0 &amp; 2\end{pmatrix}$

I got an eigenvector for $λ=1$ to be $\begin{pmatrix}0 \\0 \\0 \\0\end{pmatrix}$ which I'm not even sure if I did it correctly or not because I don't think this is possible.

for $λ=3$ I got $\:\begin{pmatrix}1 \\\frac{1}{2} \\0 \\0\end{pmatrix}$

If I did do this correctly then is the geometric multiplicity for both = 1?
Both matrices are in Jordan normal form. This normal form is unique for all similar matrices. Henece $A$  and $B$ are not similar.

Besides, these Jordan normal forms tell us the eigenspace $E_1$ (for the eigenvalue $1$ has dimension $1$ for  both matrices, while the eigenspace $E_3$ has dimension $2$ for $A$ and $1$ for $B$.
Hint: you can use Jordan normal form.