Suppose $A$ is a complex Hermitian matrix which then can be expressed as
$A = B + iC$
for $B$ and $C$ are real matrix.
I have proved that $A$ is Hermitian iff the following matrix $M$ is symmetric:
$$
M =
\begin{pmatrix}
B & -C\\
C & B
\end{pmatrix}
$$
let $\lambda $ be an eigenvalue of $A$ then is $\lambda$ also an eigenvalue of $M$?
My first approach was the following:
since I proved that $M$ is symmetric and wants to show that the eigenvalue of $A$ is also an eigenvalue of $M$, I got some ideas from here and here.
Since the $det(M) = |det(B+iC)|^2 = |det(A)|^2$ and we also know that the determinant is a product of eigenvalues,
$det(A) = \lambda \Delta$, (where $\Delta$ is the product of all the eigenvalues of $A$ except $\lambda$)
$\lambda$ has to involve in the $det(M)$. But since $det(M) = |det(A)|^2$, I think it will have $\lambda ^2$ instead of $\lambda$.
I am kind of stuck here and my mind is tunneled right now. Is this whole process wrong or am I missing something?
Best Answer
The map $\phi:M_n(\mathbb{C})\rightarrow M_{2n}(\mathbb{R})$, mapping $X+iY$ to $$ \begin{bmatrix} X & -Y\\ Y & X \end{bmatrix} $$ has the property that for a complex vector $z=x+iy\in\mathbb{C}^n$, we have $$ Mz = \tilde{x}+i\tilde{y} $$ where $$ \phi(M)\begin{bmatrix} x\\ y \end{bmatrix}=\begin{bmatrix} \tilde{x}\\ \tilde{y} \end{bmatrix}. $$ So if $\lambda$ is a real eigenvalue of $M$, then it is an eigenvalue of $\phi(M)$, too.
Edit: To understand what happens to eigenvalues under this map, it is enough to evaluate it on diagonal matrices. If $M=diag(\lambda_1,\dots,\lambda_n)$, then there is a $2\times 2$-block in $\phi(M)$ for each $\lambda_i$: If $\lambda_i$ is real, then it is $\lambda_i I_2$. If $\lambda_i=x+iy$ is complex, then the block is $$ \begin{bmatrix} x & -y\\ y & x \end{bmatrix} $$ and this $2\times 2$-matrix has characteristic polynomial $t^2-2tx+x^2+y^2$. Since $y\neq 0$, it has no real roots for any $x,y$.