The relation between the eigenvalue of a Hermitian matrix and the block matrix that composed by it real and imaginary part

Question

Recently I am reading a paper. In their "Proof of Lemma 1" on page 24, they have:

$$\lambda_+(\mathbf{Q})=2\lambda_+(\tilde{\mathbf{Q}})$$

where $\mathbf{Q}$ is a Hermtian matrix, $\tilde{\mathbf{Q}}=\frac{1}{2}\left[\begin{array}{cc} \operatorname{Re}\{\mathbf{Q}\} & -\operatorname{Im}\{\mathbf{Q}\} \\ \operatorname{Im}\{\mathbf{Q}\} & \operatorname{Re}\{\mathbf{Q}\} \end{array}\right]$ is a real symmetric matrix, $\lambda_+ = \max\{\lambda_{\max}(\mathbf{-Q},0)\}$, and $\lambda_{\max}$ denotes the maximum eigenvalue.

I haven't figured out why it's that.

Answer 1

First of all, it's well known (from the min-max theorem for example) that for a Hermitian matrix $H$ the maximum eigenvalue is given by $\sup\{\mathbf{z}^* H \mathbf{z}: \mathbf{z}^*\mathbf{z}=1\}$ and, similarly, for a real symmetric matrix $A$, the maximum eigenvalue is given by $\sup\{\mathbf{x}^T A \mathbf{x}: \|\mathbf{x}\|_2=1\}$. Now for any $\mathbf{z}=\mathbf{u}+i\mathbf{v}$ you can see by writing $Q = \Re Q + i \Im Q$ and noticing that $\Re Q = \Re Q^T$ and $\Im Q = -\Im Q^T$ that $$\mathbf{z}^* Q \mathbf{z} = 2\cdot [\mathbf{u}^T, \mathbf{v}^T] \;\tilde{Q} \begin{bmatrix}\mathbf{u}\\\mathbf{v}\end{bmatrix}$$

Finally, notice that $\mathbf{z}^*\mathbf{z}=1$ if and only if $\left\|\begin{bmatrix}\mathbf{u}\\\mathbf{v}\end{bmatrix}\right\|_2 = 1$ and so by taking supremum in the above equality we get that the maximum eigenvalues of $Q$ and $2\tilde{Q}$ are the same. I'll leave it to you to fill out the details.

Answer 2

You can represent the matrix entries of a complex matrix as $2\times 2$ blocks with real components, where $a+b i$ then becomes the block: $$ \begin{pmatrix} a & -b \\ b & a \end{pmatrix} $$ By doing so you get a real $2n \times 2n$ matrix. And then you change the ordering of the basis vectors $e_m$ by: $$ e'_m = e_{2m-1} \ \ \ \mbox{for} \ \ m=1..n $$ $$ e'_{n+m} = e_{2m} \ \ \ \mbox{for} \ \ m=1..n $$ to get the block form of the matrix $\bar{Q}$.

Before this change of basis it should be obvious that the representation of $Q$ in its eigenvector basis has a block-diagonal matrix equivalent for the $2n \times 2n$ matrix and, subsequently, the $2\times 2$ blocks can be diagonalized to the same eigenvalues that $Q$ has plus their conjugates. After the change of basis the eigenvalues should still be the same. (And the factor $1/2$ is just a choice of the authors, of course).

Answer 3

Short calculations with block matrices show that the block matrix $U=\frac{1}{\sqrt{2}} \begin{pmatrix} iI & I \\ -iI & I \end{pmatrix}$ is unitary, and that $U^*\begin{pmatrix}A-Bi & 0\\0 & A+Bi\end{pmatrix} U=\begin{pmatrix}A & -B\\B & A\end{pmatrix}$ holds for for arbitrary $A, B$, where $^*$ denotes the conjugate transpose operation. It follows that the eigenvalues of any block matrix of the form $\begin{pmatrix}A & -B\\B & A\end{pmatrix}$ are the eigenvalues of the matrix $A + Bi$ together with the eigenvalues of the matrix $A - Bi$ (not just as sets of complex numbers, but even taking multiplicities into account).

When $A, B$ are the real/imaginary parts of the self adjoint matrix $Q$, what you are calling $2\widetilde{Q}$ is thus unitarily equivalent to a block sum of $Q := A+Bi$ and $A - Bi$, the entrywise complex conjugate of $Q$ (which I will denote by $\overline{Q}$). But the matrix $\overline{Q}$ is also self-adjoint (because $Q$ is), and has the same eigenvalues (with the same multiplicities) as $Q$. This is because complex conjugation maps each eigenspace for $Q$ onto an eigenspace for $\overline{Q}$ corresponding to the same eigenvalue (the eigenvalues of self adjoint matrices are real, and $Q \xi = \lambda \xi$ with $\lambda$ real iff $\overline{Q} \overline{\xi} = \lambda \overline{\xi}$, where $\overline{\xi}$ is the entrywise conjugate of the vector $\xi$).

Conclusion: the matrices $2 \widetilde{Q}$ and $Q$ have the same eigenvalues (and while it does not affect your problem, the multiplicity of each eigenvalue of $\widetilde{Q}$ is twice that of $Q$).

The desired result follows, e.g. because then one has $\lambda_{\operatorname{max}}(-Q) = \lambda_{\operatorname{max}} (-2\widetilde{Q}) = 2 \lambda_{\operatorname{max}} (-\widetilde{Q})$.

This approach amounts to the same idea as Jos Bergervoet's answer. I post separately because this is too large to be a comment on his answer. Also, the matrix algebra framing may make it slightly easier to see how that there is one part of this setup (i.e., the "block diagonalization" expressed in the first paragraph) that works for any matrices $A$ and $B$ whatsoever, and one part of this setup (that the eigenvalues of $A - Bi$ and $A + Bi$ are the same, second paragraph) where the assumed self-adjointness is "really" being used.