Is it true that if $A$ is nilpotent, then $A^{*}A$ is also nilpotent? ($A$ here is an $n$ by $n$ complex matrix)
If $A$ is nilpotent, then $A^{*}A$ is nilpotent as well?
1
$\begingroup$
linear-algebra
-
0@Kannappan Sampath: Yes, if they commute, then we have: $(A^*A)^{k}=A^{k}(A^*)^{k}=O$, which proves that $(A^*A)$ is nilpotent. Thanks. – 2012-04-20
1 Answers
2
Not in general: take $A:=\pmatrix{0&1\\ 0&0}$, then $A$ is nilpotent since $A^2=0$, but $A^*A=\pmatrix{0&0\\1&0}\cdot\pmatrix{0&1\\ 0&0} =\pmatrix{0&0\\ 0&1}$ which is not nilpotent.
In fact if $A^*A$ is nilpotent, since it's a Hermitian matrix it's diagonalizable, so we can write $A^*A=P^*DP$ where $D$ is diagonal. Then fact that $A^*A$ is nilpotent gives us that $D=0$ hence $A^*A=0$.
-
1Isn't it enough to have one of them nilpotent? @DavideGiraudo (That is relax that $B$ is nilpotent..., we may prove the same result with less assumptions.) – 2012-04-20