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.

Let $A$ be normal $n\times n$ matrix. $c$ is eigen value of A. Than must exist:

  1. if $A+A^*$ invertible then $c=bi$ (b real).

  2. if $A$ invertible then $A+A^*$ invertible.

  3. if $c$ is the only eigenvalue of $A$ and $c=bi$ (b is real) than $A+A^*$ is the zero matrix.

  4. if $A+A^*$ invertible than for every $b$ that is real $c\not=bi$.

  5. all answers 1-4 are false.

The correct answers are 3+4 but I can't understand why

share|improve this question
add comment

2 Answers

up vote 3 down vote accepted
  1. Take $A=I_n$.
  2. Take $A=diag(i,-i)$.
  3. Because $A$ is normal and $c$ is the only eigenvalue, we have $A=U^*DU$, for some unitary matrix $U$ and $D=diag(c,\cdots,c)$. Therefore $A+A^*=(U^*DU)+(U^*DU)^*=U^*(D+D^*)U=U^*diag(ib-ib, \cdots , ib-ib)U=U^*0_{\mathbb{C}^n}U=0_{\mathbb{C}}$
  4. Prove the contrapositive: suppose $c=ib$. It follows from $A$ being normal, that $A=U^*DU$, for some unitary matrix $U$ and $D=diag(c,\lambda_2\cdots,\lambda_n)$, where $\lambda _i$ are all the eigenvalues of $A$. So $A+A^*=U^*DU + (U^*DU)^*=U^*DU+U^*D^*(U^*)^*=U^*DU+U^*D^*U=U^*(D+D^*)U$=$U^*diag(c+\bar{c},\lambda _2+\bar{\lambda _2},\cdots,\lambda_n+\bar{\lambda_n})U=U^*diag(0,\lambda _2+\bar{\lambda _2},\cdots,\lambda_n+\bar{\lambda_n})U$, therefore $0$ is an eigenvalue of $A+A^*$ and $A+A^*$ isn't invertible.
share|improve this answer
1  
are you sure? let $A$ be the 1 by 1 matrix $i$. Is $c=i=0$? –  Bombyx mori Jan 13 '13 at 13:54
    
You're right. OP, my answer for 3. is correct only if $A$ is a real matrix. Because $c=0$ necessarily, only if $A$ is real. –  Git Gud Jan 13 '13 at 17:29
    
I wasn't claiming you're the OP. Sorry for my confusing message. –  Git Gud Jan 13 '13 at 17:32
1  
The solution is now correct. –  Git Gud Jan 13 '13 at 17:39
    
yes. Now I agree. –  Bombyx mori Jan 13 '13 at 17:43
add comment

It would help if you can point out what you do not understand.

1) is not correct if you consider diagonal matrix with entries $(1,1)$.

2) is not correct if you consider diagonal matrix with entries $(1,-1)$.

3) is correct since there exist a basis such that $A,A^{*}$ are spontaneously diagonalizable. So we have $A=D,A^{*}=\overline{D^{T}}=\overline{D}$. Under the condition give $A+A^{*}=0$.

4) If $A+A^{*}$ is invertible then $A$ cannot have any imaginary eigenvalues. So $c\not=bi$.

5) cannot be correct since we know 3 and 4 are correct.

share|improve this answer
add comment

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.