Linear algeba - eigenvalues and zero matrix

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

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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.
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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
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

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.

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