1
$\begingroup$

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 $$\left(\begin{array}{cc} 0 & 0 \\1 & 0 \end{array}\right) $$

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Shahab Sep 16 '12 at 13:13
  • $\begingroup$ I started with finding out the characterstic values and characteristic vectors. $\endgroup$ – neha Sep 22 '12 at 12:19
7
$\begingroup$

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 $A$?

$\endgroup$
3
$\begingroup$

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]

$\endgroup$
  • 1
    $\begingroup$ ... where linear independene is the point where the condition $A^2=0$ is made use of. $\endgroup$ – Hagen von Eitzen Dec 24 '12 at 11:02
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.