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Let $A$ be a $2\times 2$ complex matrix such that $A^2=0$. Prove that either $A=0$ or $A$ is similar over $\mathbb{C}$ to $$\left(\begin{array}{cc} 0 & 0 \\1 & 0 \end{array}\right) $$

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Note that A is not invertible as its determinant is zero. So either A has rank 0 or 1. In the former case it is itself 0 and in the latter case it is similar to your matrix above. – Shahab Sep 16 '12 at 13:13
I started with finding out the characterstic values and characteristic vectors. – neha Sep 22 '12 at 12:19

If $A^2=0$, then $det(A^2) = [det(A)]^2 = 0$, implying $det(A) = 0$. Since $A$ is a $2 \times 2$ matrix, what can you conclude about $A$?

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If $A$ is non-zero,there is $X$ such that $AX$ is non-zero.Now $\{X,AX\}$ is linearly independent,forms a basis of $C^2$.Then the matrix representation of $A$ with respect to the basis is the given matrix. [QED]

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... where linear independene is the point where the condition $A^2=0$ is made use of. – Hagen von Eitzen Dec 24 '12 at 11:02

If $Av=\lambda v$ with $v\ne0$ then $$0=0v=A^2v=A\lambda v=\lambda^2v$$ so $\lambda=0$. That is, the only eigenvalue $A$ has is zero. Either there are two linearly independent eigenvectors $v$ and $w$, in which case $Ax=0$ for all $x$, and $A=0$; or, there's only one eigenvector $v$, in which case you can show there's a vector $w$ with $Aw=v$. Then if $P$ is a matrix whose columns are $v$ and $w$ you should get $AP=PD$ where $D$ is (the transpose of) your second possibility.

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If $A$ is similar to a projection, i.e., $P = S A S^{-1}$ and $P^2 = P$, then $S A^2 S^{-1} = P^2 = P = S A S^{-1}$, so $A^2 = A$, i.e., $A$ is a projection. But, for $A \ne 0$, this is a contradiction with $A^2 = 0$. Hence, your title is wrong.

As for the body of your question: if you know what Jordan normal form is, the answer is straightforward.

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The title to which you refer is not the original, but a recent edit. – Gerry Myerson Sep 27 '14 at 6:14

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