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?
Best Answer
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 $$ |\lambda|\le 2|(Ax,x)|\le 2\|Ax\|\|x\| = 2\|Ax\| = 2\|x\| = 2. $$