Eigenvalues are real numbers and satisfy inequality

[evinda]

Hello! (Wave)

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? (Thinking)

 

[I like Serena]

Hey evinda!

Suppose we pick $A=\begin{pmatrix}0&-1\\ 1&0\end{pmatrix}$.
Then aren't the eigenvalues of $A+A^*$ imaginary? (Worried)

Can it be that $A+A^+$ was intended?

To find an upper magnitude of 2, did you consider that $\|A\mathbf x\| = \|\mathbf x\|$, which is a property of a unitary matrix? (Wondering)