# Show that a nonzero $2 \times 2$ matrix $A$ such that $A^2 = 0$ is similar to $\begin{pmatrix}0&1\\0&0\end{pmatrix}$

Let $A$ be a $2 \times 2$ non-zero matrix such that $${A}^{2}=0.$$ How do I find an invertible matrix P such that $${P}^{-1}AP=\begin{bmatrix}0&1\\0&0\end{bmatrix} ?$$

Anyone? Please provide a clear explanation, I really appreciate it!

Thank you.

• @Travis look at ques thx! – UnusualSkill May 5 '15 at 17:20
• I cannot talk for him although I think Travis was looking at the question just as I am now: you must state $\;A\neq 0\;$ otherwise the claim is false. – Timbuc May 5 '15 at 17:22

## 1 Answer

Hint Since $A \neq 0$ we can pick an element ${\bf e}_2$ such that $A {\bf e}_2 \neq 0$.

Additional hint Since the only eigenvalue of $A$ is $0$, $A {\bf e}_2$ cannot be a multiple of ${\bf e}_2$.

• why the only eigenvalue of A is 0? – UnusualSkill May 5 '15 at 17:33
• @UnusualSkill If $\lambda$ is an eigenvalue of $A$, say with eigenvector $\bf x$, then $$A^2 {\bf x} = A(A{\bf x}) = A (\lambda{\bf x}) = \lambda A{\bf x} = \lambda(\lambda {\bf x}) = \lambda^2 {\bf x},$$ and in particular $\lambda^2$ is an eigenvalue of $A^2$. Since $A^2 = 0$, the only eigenvalue of $A^2$ is $0$, and hence all eigenvalues $\lambda$ of $A$ are $0$. – Travis May 5 '15 at 17:35
• I still cant catch the hint. Can you provide me with more steps please?thx! – UnusualSkill May 5 '15 at 17:41
• Sure: What is the matrix for the linear transformation in the basis $\{{\bf e}_1, {\bf e}_2\}$, where ${\bf e}_1 := A{\bf e}_2$? – Travis May 5 '15 at 17:43