The matrices are defined with $A = \left(\begin{array}{rrrr} -1&2&-3&-1\\% 8&-7&12&4\\% 6&-6&10&3\\% 2&-2&3&2\\% \end{array}\right)$ and $B= \left(\begin{array}{rrrr} -1&-4&7&-6\\% 0&11&-25&25\\% -4&8&-25&28\\% -4&4&-16&19\\% \end{array}\right) \in \mathbb{Q}^{4 \times 4}$
Now I have to proof or disproof that $A$ and $B$ are similar. The characteristic polynoms and the eigenvalues are the same. However the eigenvectors are $a_1=\left(\begin{matrix}1\\1\\0\\0\end{matrix}\right), a_2=\left(\begin{matrix}-3\\0\\2\\0\end{matrix}\right), a_3=\left(\begin{matrix}-1\\0\\0\\2\end{matrix}\right)$ and $b_1=\left(\begin{matrix}-3\\5\\2\\0\end{matrix}\right), b_2=\left(\begin{matrix}4\\-5\\0\\2\end{matrix}\right)$
My thoughts were the following:
There exists two invertable matrices $S,T\in \mathbb{Q}$ with $S^{-1}AS=D, T^{-1} BT=D$. Where $D$ is diagonalmatrix with the eigenvalues on the diagonal. The matrices $S,T$ contain the eigenvectors of $A,B$. Therefore we get $S^{-1}AS=T^{-1}BT \Rightarrow [ST^{-1}]^{-1}A(ST^{-1})=B$ and we can conclude that the matrices are similar.
The problem
My problem is, that if the matrices $S,T$ contains the eigenvectors, how can they have an invertable, if they aren't square?
If two matrices are similar, then the size of a maximal set of linearly independent eigenvectors will be the same for both of them. It follows from this that, if your computations are correct, then your matrices are not similar.