Determine if matrices A and B are similar Determine if matrices $A$ and $B$ are similar.
$A=\begin{pmatrix}1 & 1 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & 3 & 0 \\0 & 0 & 0 & 3\end{pmatrix}\: $
and
$B=\begin{pmatrix}1 & 1 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & 3 & 1 \\0 & 0 & 0 & 3\end{pmatrix}$
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 
$\begin{pmatrix}0 & 1 & 0 & 0 \\0 & 0 & 0 & 0 \\0 & 0 & 2 & 0 \\0 & 0 & 0 & 2\end{pmatrix}$
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?
 A: 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 $B$.
A: If you look at the nullspace of $(A-3I)$ and of $(B-3I)$, you'll find that they have different dimensions.  Hence the set of eigenvectors of the eigenvalue 3 (called an eigenspace) is a subspace of dimension 2 for the first matrix and dimension 1 for the second matrix.  Hence these two matrices are not similar.
I'll prove now that if two matrices are similar, then their eigenspaces (for each eigenvalue) must have the same dimension. Recall that in general, if $A$ and $B$ are similar matrices, then they represent the same linear transformation but with respect to (possibly) different bases.  In other words, there is some change of basis matrix $P$ such that $B=P^{-1}AP$. Suppose $x$ is an eigenvector of $A$ corresponding to eigenvalue $\lambda$. Then $Ax=\lambda  x$.  Hence $(PBP^{-1})x = \lambda x$, whence $B (P^{-1}x) = \lambda (P^{-1} x)$.  Thus, $P^{-1}x$ is an eigenvector of $B$ associated with the same eigenvalue $\lambda$. Similarly we can show that every eigenvector of $B$ gives rise to an eigenvector of $A$ associated with the same eigenvalue.  If the eigenspace of $A$ corresponding to eigenvalue $\lambda$ has dimension $r$ ($r=2$ in your example), then the set of vectors $\{P^{-1}x: x \mbox{ is an eigenvector of $A$ } \}$ must also have dimension $r$. This is because   $P^{-1}$ is an invertible linear transformation, and hence it preserves the dimension of subspaces. In your example, it can be verified that the set of eigenvectors of $B$ corresponding to eigenvalue 3 is a subspace of dimension only 1, not 2.  Hence, $A$ and $B$ are not similar. 
The Jordan canonical form theorem is useful but I think a bit overkill for this problem.  If you know the theorem, you can just read off (without doing computations by hand) that the dimensions of the eigenspaces (corresponding to the eigenvalue 3) in your two matrices are different.  In particular,  the $2 \times 2$ submatrix in the lower-right corner of $B$ is known to not be diagonalizable (and the theorem says that that is about as close to diagonal as we can make any complex matrix, using similarity transformations), whereas the corresponding submatrix in $A$ is in diagonal form.  
A: Hint: you can use Jordan normal form.
