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Let $A$ be a $n \times n$ complex unitary matrix. I want to show that the eigenvalues $\lambda$ of the matrix $A+A^{\star}$ are real numbers that satisfy the relation $-2 \leq \lambda \leq 2$.

I have looked up the definitions and I read that a unitary matrix is a square matrix for which $AA^{+}=I$.

(The transpose matrix of $A^{\star}$ is symbolized with $A^{+}$.)

($A^{\star}$: complex conjugate)

In order to show that the eigenvalues $\lambda$ of the matrix $A+A^{\star}$ are real numbers and satisfy that $-2 \leq \lambda \leq 2$, do we maybe have to find the minimal polynomial of the matrix $A+A^{\star}$ ? If so, how? Is there a relation? Or do we have to do it somehow else?

Jose Brox
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Evinda
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  • https://math.stackexchange.com/questions/1717713/show-that-the-eigenvalues-of-a-unitary-matrix-have-modulus-1 – RozaTh Sep 03 '18 at 15:03
  • @RozaTh You should read the question first. – amsmath Sep 03 '18 at 15:05
  • Evinda Your definition of "unitary" is incorrect. A matrix is unitary if $AA^* = I$. Here, $A^*$ is the conjugate transpose of $A$, i.e., the transpose with each entry conjugated. – amsmath Sep 03 '18 at 15:06
  • @amsmath A ok. And how can we find a relation for the eigenvalues? – Evinda Sep 03 '18 at 15:12
  • Is $A^\star$ just the complex conjugate or the hermitian conjugate (i.e. transpose of complex conjugate)? Your definitions of $A^\star$ and $A^+$ are confusing. – Christoph Sep 03 '18 at 15:17
  • @Christoph Read my comment. The stuff in the question is definitely wrong. – amsmath Sep 03 '18 at 15:20
  • @amsmath If $A^+$ is the hermitian conjugate, the definitions in the question are at least not contradictory. I guess the statement about $A+\overline A$ doesn't hold, but I didn't think about a counter example. – Christoph Sep 03 '18 at 15:22
  • @Christoph Uhh aahhh. Sorry. My bad. Usually, $A^*$ is the conjugate transpose and not just the conjugate. So, my answer does not refer to the problem as stated... – amsmath Sep 03 '18 at 15:26

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In the following I let $A^*$ denote the complex conjugate transpose of $A$ and by $\overline A$ just the complex conjugate.

The original statement about $A+\overline A$ in the question is false. Just take $A = (\begin{smallmatrix}0 & 1\\-1 & 0\end{smallmatrix})$ which is unitary. Then $A+\overline A = 2A$ which has eigenvalues $\pm 2i$.

Let us prove that the statement holds for $A+A^*$. You may want to prove first as an exercise that $A$ unitary implies that $\|Ax\| = \|x\|$ for each vector $x$.

Now, let $\lambda$ be an eigenvalue of $A+A^*$ with eigenvector $x$. We may assume WLOG that $\|x\|=1$. Then (since for all vectors $y,z$ we have $(A^*y,z) = (y,Az) = \overline{(Az,y)}$) \begin{align*} \lambda &= (\lambda x,x) = (Ax+A^*x,x) = (Ax,x) + (A^*x,x)\\ &= (Ax,x)+(x,Ax) = 2\operatorname{Re}(Ax,x). \end{align*} In particular, $\lambda$ is real. Now, this implies $$ |\lambda|\le 2|(Ax,x)|\le 2\|Ax\|\|x\| = 2\|Ax\| = 2\|x\| = 2. $$

amsmath
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