# 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!

• Note that this is not a Jordan Normal Form for $a=1$ unless $x=y$. – Marc van Leeuwen Feb 20 '15 at 15:14

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.
• @skim You need to be very careful about what you mean by "this"; the most straightforward generalisation does not hold beyond the $2\times2$ case. Also, induction is in any case probably not the way to go. – Marc van Leeuwen Dec 9 '17 at 20:51