Matrix norm relation between $A^A$ and $AA^$

Question

Let $A^*$ denote the complex conjugate transpose of a matrix $A$. In the Euclidean norm, if

$$||A^*A+AA^*||=||A^*A||$$

does this imply that

$$||A^*A+AA^*||=||A^*A-AA^*||$$

What do you mean by "Euclidean norm"? Do you mean the operator norm induced by the standard euclidean norm or do you mean the Frobenius norm? — 5xum, Feb 01 '15 at 14:31

Daniel · Accepted Answer · 2015-02-01T18:13:31.973

1

Hints:

If $B$ is Hermitian then $\|B\|=\sup_{\|v\|=1}|tr(Bv\overline{v}^t)|$. Also, $tr(A^*Av\overline{v}^t)\geq0$ and if $tr(A^*Av\overline{v}^t)=0$ then $Av=0$.
Use item 1 to prove that $\|AA^*-A^*A\|\leq\|AA^*+A^*A\|$
Use item 1 and $\|AA^*+A^*A\|=\|AA^*\|$ to prove that if $AA^*v=\|AA^*\|v$ then $A^*Av=0$.

Finally, $(AA^*-A^*A)v=\|AA^*\|v$. Thus, $\|AA^*-A^*A\| \geq \|AA ^*\|=\|AA^*+A^*A\|$.

So $\|AA^*-A^*A\|=\|AA^*+A^*A\|$.

edited Feb 01 '15 at 18:13

answered Feb 01 '15 at 16:07

Daniel

I can prove the four statements you've set up, but I'm afraid I don't follow from "Finally, ..." There seems to be a parenthesis missing? – PeterA Feb 01 '15 at 18:00
Do you mean $(AA^-A^A)(v-w)=|AA^*|(v-w)$? – Daniel Feb 01 '15 at 18:03
Notice that $|AA^|$ is an eigenvalue of $AA^-A^A$. Thus, $|AA^-A^A|\geq|AA^|$. – Daniel Feb 01 '15 at 18:04
Yes, that's the one. – PeterA Feb 01 '15 at 18:04
You are right. I made a mistake. – Daniel Feb 01 '15 at 18:09
It is easy to fix – Daniel Feb 01 '15 at 18:12
I believe it is fixed now. – Daniel Feb 01 '15 at 18:13

1 Answers1