No two 2x2 matrices in Jordan form are similar? Let $S$ be the set of 2x2 matrices in Jordan Normal form $\begin{pmatrix}x&a\\ 0&y\end{pmatrix}$ with $a=0$ or $1$ and $x \leq y$. How do I show that no two matrices in $S$ are similar? 
Thank you!
PS – If anyone knows how to edit so the matrix is displayed properly that would be much appreciated!
 A: I'll show that if two such matrices $J_1,J_2$ are both Jordan Normal Forms, and they are similar, then they are equal.
Similar matrices have the same characteristic polynomial, and the characteristic polynomial of your matrix is $(X-x)(X-y)$ so the hypothesis implies $\{x_1,y_1\}=\{x_2,y_2\}$. Now if one has $x_1\neq y_1$, and therefore $x_2\neq y_2$, one must have $a_1=0$ and $a_2=0$ for these matrices to be Jordan Normal Forms in the first place; then given the ordering imposed on the diagonal entries we see that they are in fact equal. The remaining case is when $x_1=y_1$, and therefore $x_2=y_2$. If either $a_1=0$ or $a_2=0$, then one of the matrices is a multiple of the identity, and therefore only similar itself, so necessarily $J_1=J_2$. But in the remaining case $a_1=1$ and $a_2=1$ and we have $J_1=J_2$ anyway.
A: If two such matrices have different diagonal entries, then one has an eigenvalue that the other does not have, hence they are not conjugate. If the two matrices have the same diagonal entries but different upper right corner, then one has a basis of eigenvectors and the other does not, so they cannot be conjugate.
A: Two similar matrices have the same characteristic polynomial, so they have same determinant, same trace, same eigenvalues. Testing one of this we see if two matrix are similar. If the matrices are in Jordan form any of this test is easy.
