Is $A-cI$ nilpotent matrix of index 2?

Question

I've studied Classifying complex $2\times2$ matrices up to similarity.

There it's said that when $c$ is the only eigenvalue of $A$, then $A-cI$ is nilpotent. But to solving that problem we have to use the fact that if a matrix is nilpotent of index 2, it is similar to $\begin{pmatrix}0&0\\1&0\end{pmatrix}$.I've proved that $A-cI$ has zero eigenvalue. But how to show that it's nilpotent of index 2?

Answer 1

Let $A$ be a $2\times2$ matrix having only $0$ as eigenvalue. Then its trace and determinant are $0$, so $$ A=\begin{bmatrix} u & v \\ w& -u \end{bmatrix},\qquad u^2+vw=0 $$ Now compute $$ A^2=\begin{bmatrix} u^2+vw & uv-uv \\ uw-uw & vw+u^2 \end{bmatrix} $$

Answer 2

The characteristic polynomial of $A$ is $(\lambda-c)^2$. By Cayley-Hamilton:

$$(A-cI)^2=0.$$

Answer 3

$A$ is either similar to the Jordon canonical form $J=\begin{pmatrix}c&0\\0&c\end{pmatrix}$ or Jordon block $J_1=\begin{pmatrix}c&1\\0&c\end{pmatrix}$

First case, $(A-cI)$ is similar to the zero matrix , which have nilpotent of index $1$ , not possible.

Second case , $(A-cI)$ is similar to $\begin{pmatrix}0&1\\0&0\end{pmatrix}$, which have nilpotent of index $2$.