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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?

glS
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TUC
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  • Perhaps $det(M) = det(A) \times det(A) = \lambda^2 \Delta^2= \lambda \Delta \lambda \Delta$ hence $\lambda$ is an eigenvalue of M. Does this make sense? I am uncertain whether the $\lambda$ exists as an eigenvalue of $M$, not as $\lambda^2$ – TUC Apr 25 '21 at 15:23

1 Answers1

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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 $x,y$.

Levent
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  • Thank you for the answer! I'm getting kind of confused with $M = diag(\lambda_1, ..., \lambda_n)$. Is this M a new matrix or is it the same as the one above, i.e. $\begin{pmatrix} B & -C\ C & B \end{pmatrix}$

    sorry about typo

     – TUC Apr 25 '21 at 15:44
  • @TUC I just restricted the map $\phi$ to diagonal matrices as it gives us all the information about the eigenvalues we need. So $M=B+iC$ is again a complex matrix, just a diagonal one. – Levent Apr 25 '21 at 15:45
  • I think I just confused myself with the changed variables. In your answer, M is just $M = X + iY$ and $\phi(M) = \begin{bmatrix} X & -Y\ Y & X \end{bmatrix}$ ? – TUC Apr 25 '21 at 15:52
  • Yes. Sorry for the confusion :) – Levent Apr 25 '21 at 16:03
  • I'm sorry but I have one last question: In the line $Mz = \overline{x} + i\overline{y}$, are $\overline{x}$ and $\overline{y}$ some new values and the overline is just to denote that they are different from $x$ and $y$? I learned $\overline{x}$ is the conjugate of $x$ in complex numbers. I guess this has nothing to do with it but I'd like to understand this thoroughly. Thank you – TUC Apr 25 '21 at 16:15
  • @TUC the notation is $\tilde$, not $\bar$, i.e., it is not complex conjugate just some other vectors. I could have used $a,b$ or $\alpha,\beta$ instead of $\tilde{x},\tilde{y}$. – Levent Apr 25 '21 at 16:22
  • Wow. again, thank you for all the answers / replies. Have a great day! – TUC Apr 25 '21 at 16:24
  • If you are satisfied with the answer, you may consider to accept it so it does not look like an unanswered question. – Levent Apr 25 '21 at 16:26