# Eigenvalues of $A$ and $A + A^T$

This question has popped up at me several times in my research in differential equations and other areas:

Let $A$ be a real $N \times N$ matrix. How are the eigenvalues of $A$ and $A + A^T$ related? Of course, if $A$ is symmetric, the answer is easy: they are the same up to factor or $2$, since then $A + A^T = 2A$. But if $A \ne A^T$?

I'm particularly interested in the question of the real parts of the eigenvalues. How are the real parts of the eigenvalues of $A$ related to the (necessarily) real eigenvalues of $A + A^T$?

Answers for complex matrices appreciated as well.

Any references, citings, or explanations at any level of detail will be appreciated.

• If $A$ is diagonalizable, then $\rho(A + A^T) \leq \rho(A) + \rho(A^T) = 2 \rho(A)$, where $\rho$ is the radius (equal to the matrix norm for $A$ diagonalizable). Jul 27 '13 at 21:58
• @Eric Auld: yes, of course; the challenging part comes when $A$ is not real-diagonalizable, so the real Jordan form has blocks like $\begin{bmatrix} a & b \\ -b & a \end{bmatrix}$ etc. Jul 27 '13 at 22:04
• I don't think there's much you can say about the eigenvalues of $A+A^T$ solely in terms of those of $A$. For example, consider $A = \left(\begin{array}{cc}\lambda&t\\0&\mu\end{array}\right)$ where $\lambda$ and $\mu$ are fixed, and $t$ is a real parameter. So the eigenvalues of $A$ are $\lambda$ and $\mu$. Using the quadratic formula you can check that the eigenvalues of $A+A^T$ are $\lambda+\mu \pm \sqrt{(\lambda-\mu)^2+t^2}$ which depend on $t$ and not just $\lambda$ and $\mu$. Jul 27 '13 at 22:27
• @RobertLewis In response to your comment , I was wondering in case when $A$ is not real diagonalizable, is it always possible to transform $A$ into a structure $\begin{bmatrix}a & b\\{-b} & a\end{bmatrix}$? In extension to that, should an even order matrix will always be transformed into a matrix with each diagonal block having structure $\begin{bmatrix}a & b\\{-b} & a\end{bmatrix}$. Jun 11 '18 at 16:13
• @jbgujgu: No, but a $2 \times 2$ real matrix will transform either to $\begin{bmatrix} a & b \\ -b & a \end{bmatrix}$ or $\begin{bmatrix} a & b \\ 0 & a \end{bmatrix}$. This is Jordan canonical form stuff. Cheers! Jun 11 '18 at 16:34

Here is Tao's comment from, now deleted, post 2:

"Note that if A is strictly upper triangular, then its eigenvalues are all zero, whereas $A+A^T$ is an arbitrary symmetric matrix with zero diagonal, which constrains the trace of the matrix but otherwise imposes almost no conditions on the spectrum whatsoever (the only other constraint I can see is that the matrix cannot be rank one). So, apart from the trace $tr(A+A^T)=2tr(A)$, there appears to be essentially no relationship."

• The closed post has been deleted. Dec 22 '14 at 19:02
• I'm acepting this answer since I have learned a lot from the references an it has been around long enough to be "ripe", as it were. But I do not think the last word has been said. I have been thinking of posting a seperate, reformulated version, but I am currently struggling with a backlog of about sixty answers-in-preparation, and they bug me. Anyway, thanks for the helpful remarks. Holiday Cheers to one and all! Dec 22 '14 at 23:15
• @Studiosus: and a separate word of gratitude to you, good sir! Cheers! Dec 22 '14 at 23:17

EDIT: Let $\lambda\in spectrum(A)$; then there is $\mu \in spectrum(A+A^T)$ s.t. $|\lambda-\mu|\leq ||A||_2$ (spectral norm).

Proof: Since $A+A^T$ is real symmetric, according to the Bauer–Fike Theorem, there is $\mu\in spectrum(A+A^T)$ s.t. $|\lambda-\mu|\leq ||-A^T||_2=||A||_2$. cf. http://en.wikipedia.org/wiki/Bauer%E2%80%93Fike_theorem

• Wow! Very nice; useful and interesting! Thanks so much! Yul Tov! As you can see from my pic on my user page, I am Santa's brother! I'll put in a good word, so watch your stocking! Salud! Dec 23 '14 at 0:23

A special case is when $A$ is a unitary matrix: eigenvalues of $A$ lie on unit circle in complex plane. If $e^{i\theta}$ is an eigenvalue of $A$, then $2 \cos(\theta)$ is an eigenvalue of $A + A^T$. Similarly, $2i\sin(\theta)$ is an eigenvalue of $A - A^T$.