# Classifying complex $2\times 2$ matrices up to similarity

I would like to prove the following proposition, which is given as an exercise in Hoffman and Kunze:

If $A$ is a $2\times 2$ matrix with coefficients in $\mathbb{C}$, then $A$ is similar either to a matrix of the form $\begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}$ or to a matrix of the form $\begin{pmatrix} a & 0 \\ 1 & a \end{pmatrix}$.

A hint directs the reader to prove that if $N$ is an nilpotent matrix (also in $M_{2}(\mathbb{C})$), then either $N=0$ or $N$ is similar to $\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$. I have proven this claim (by supposing that $N \neq 0$ and showing that the transformation induced by multiplication by $N$ has matrix representation $\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$ under a particular basis).

However, I'm not really sure how to use this knowledge to prove the proposition in question. Any steps in the right direction would be appreciated. Also, I'd love to see any other proofs of the proposition.

• You wrote idempotent and you meant nilpotent. – Mariano Suárez-Álvarez Nov 3 '13 at 1:59
• Indeed, thanks for pointing it out, Mariano! Post edited with corrections. – Bachmaninoff Nov 3 '13 at 2:21
• Link to the hint. – leo Sep 27 '14 at 1:02

If a matrix has two distinct eigenvalues, it is similar to $\begin{pmatrix} a&0 \\ 0&b\end{pmatrix}$ with $a\neq b$. Now, suppose that $A$ has the eigenvalue $a$ of algebraic multiplicity $2$. We note that $A - aI$ must be nilpotent. From there, note that for any invertible $S$, $$S(A - aI)S^{-1} = SAS^{-1}-SaIS^{-1} = SAS^{-1}-aI$$
• It definitely helps, thank you for responding. Is there any need to deal with the case of A having no eigenvectors? I'm guessing not because the base field is $\mathbb{C}$? (sorry if this is a dumb question, I'm still getting used to this stuff) – Bachmaninoff Nov 4 '13 at 3:49
• Your guess is spot on. You can define the eigenvalues of a matrix $A$ as being the roots of the polynomial given by $\det(tI - A)$. Because every polynomial is reducible over $\mathbb{C}$, an $n\times n$ matrix will always have $n$ eigenvalues, up to "algebraic" multiplicity. – Omnomnomnom Nov 4 '13 at 4:07
• Ah ok! That makes sense. I guess I'm struggling to see why $A-aI$ is nilpotent (probably just a little thing I'm not seeing). But I think I can complete the proof from there. – Bachmaninoff Nov 4 '13 at 4:38
• A matrix $N$ is nilpotent iff all its eigenvalues are zero. – Omnomnomnom Nov 4 '13 at 4:53