Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$A\in \mathbb{C}^{n\times n}$. Set: $$ r(A) = \max_{||u||_2=1}|u^* Au| $$

Prove the following statements:

(1) $||A||_2 \leq 2 r(A)$

(2)if $A^*A=AA^*$ then $r(A)=||A||_2$

share|improve this question
    
The way this is phrased makes it sound like these are exercises from a book, or a course. If so, then ideally this should be acknowledged (at least in my personal opinion). –  user16299 Mar 1 '12 at 4:08

2 Answers 2

Write $A=\frac{A+A^*}2+\frac{A-A^*}2$. Let $H_1:=\frac{A+A^*}2$, then $H_1$ is hermitian and is therefore diagonalizable. Thanks to that we can see that $||H_1||=r(H_1)$. Since $H_2:=\frac{A-A^*}2$ is anti-hermitian it's in particular a normal matrix, so $H_2$ is unitary diagonalizable and $||H_2||=r(H_2)$. Since $r(H_1)\leq r(A)$ and $r(H_2)\leq r(A)$ we have $$||A||_2\leq ||H_1||_2+||H_2||_2=r(H_1)+r(H_2)\leq 2r(A).$$

share|improve this answer
    
could you please help me again? The problem is now reedited. –  BerSerK Feb 26 '12 at 3:16
    
"Since $H_2$ is anti-hermitian it's in particular an unitary matrix", are you saying every anti-hermitian matrix is a unitary matrix? I don't believe that is true. –  Calle Feb 26 '12 at 3:28
    
These are certainly not all unitary (only those with eigenvalues $\pm i$) but that is also not required. $H_2$ is diagonalizable with respect to an orthonormal basis and that suffices. (To see it note that $iH_2$ is hermitian or, more general, $H_2$ is normal.) –  WimC Feb 26 '12 at 7:47
    
I probably made a confusion, thinking that if $A$ is anti-hermitian, $A^*A=-AA=A(-A)=AA^*$, but I still don't know why I wrote unitary instead of normal. –  Davide Giraudo Feb 26 '12 at 9:55

For part (2) note that $A$ is normal in this case and hence is diagonalizable with respect to an orthonormal basis. (See here for more about normal operators.) As noted above, this implies that $||A||_2 = r(A)$: If $||v||_2=1$ and $v = a_1e_1 + \dotsc + a_ne_n$ in this basis then

$$ |v^{\ast}Av| = |\langle Av, v \rangle| = \left|\lambda_1 |a_1|^2 + \dotsc + \lambda_n |a_n|^2 \right| \leq \max_k(|\lambda_k|) \cdot (|a_1|^2 + \dotsc + |a_n|^2) = \max_k(|\lambda_k|) $$

and this maximum is attained for some $e_k$ and so $r(A) = \max_k(|\lambda_k|)$. A similar computation shows that this indeed equals $||A||_2$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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