From the spectrum theorem, we know real symmetric matrices have real eigenvalues.

But can real non-symmetric matrix have real eigenvalues?

What are the necessary and sufficient conditions for a real matrix to have real eigenvalues?

  • 3
    $\begingroup$ Here's a short paper claiming sufficiency conditions: ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=1448531 $\endgroup$ – Moss Murderer Jan 17 '18 at 22:21
  • 3
    $\begingroup$ Just conjugate any real diagonal matrix by any invertible but non-orthogonal matrix to produce an example. Here's minimal R code: a <- matrix(c(1,1,0,1),2); b <- a %*% diag(c(2,-1)) %*% solve(a); eigen(b) You will find that the non-symmetric matrix b has real eigenvalues $2,-1$ as specified. $\endgroup$ – whuber Jan 17 '18 at 22:59
  • $\begingroup$ Is 0 real? ((0,0),(0,1)) $\endgroup$ – Bananach Aug 7 '20 at 11:50
  • $\begingroup$ Any real upper triangular matrix is non-symmetric and has real eigenvalues (the eigenvalues are the diagonal entries) $\endgroup$ – Jake Levi Dec 15 '20 at 15:29

Yes, I asked myself the same question, and the condition:

  • real eigenvalues
  • real non symmetric matrix

can be more strict:

$\lambda_i \in\mathbb{N}, i=1, \ldots, n $ and non symmetric matrix $M \in \mathbb N^{n\times n}$


$\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{A}_{R} \subset \mathbb{R}$

Here is one such matrix.

$M = \left[\begin{array}{ll}2 & 3 \\ 1 & 4\end{array}\right]$

with the $\lambda_1= 1$ and $\lambda_2= 5$, $\{\lambda_1, \lambda_2\}$ $\in \mathbb N$.

Let's show there are infinitely many matrices that are non symmetric with exactly the same eigenvalues (similar).

If we decompose matrix $M$ via Schur method we get:

$\begin{array}{c} Q\to \left[ \begin{array}{cc} -0.948683 & -0.316228 \\ 0.316228 & -0.948683 \end{array} \right] \\ T\to \left[ \begin{array}{cc} 1 & 2 \\ 0 & 5 \end{array} \right] \\ M\to Q\cdot T\cdot Q^{\text{-1}} \end{array}$

So now we are having the matrix $T$ that is non symmetric, and we can alter the upper triangular value $2$, to anything other than $0$ (because of the non symmetric condition) and still the eigenvalues will be $1$ and $5$.


This is an old question, but a more complete answer than the ones given may still be of interest. Necessary and sufficient conditions are given below.

The basic story is that in a quantum mechanical (QM) system the energy levels are given by the eigenvalues of a matrix $\mathbf{H}$ called the Hamiltonian. In the usual formulation of QM, $\mathbf{H}$ is required to be Hermitian to ensure real energy eigenvalues but one can ask what wider class of matrices would be eligible. It turns out that if the physical system satisfies parity (spatial reflection) and time-reversal symmetry, $\mathbf{H}$ may have a real spectrum. This is formulated in mathematical language below.

Theorem 1 (necessary condition): Suppose a matrix $\mathbf{H}\in\mathbb{C}^{N \times N}$ has real or complex conjugate eigenvalues. Then there exist two matrices/operators $\mathbf{P}$ and $\hat{\mathbf{T}}$ s.t.:

  • $\mathbf{P}^2=\mathbf{I}$ and $\mathbf{P}$ is also a Hermitian and unitary matrix
  • $\hat{\mathbf{T}}^2=\mathbf{I}$ and $\hat{\mathbf{T}}=\hat{K}\mathbf{U}$ where $\hat{K}$ is the complex conjugation operator and $\mathbf{U}$ is a unitary matrix
  • $[\mathbf{P},\hat{\mathbf{T}}]=\mathbf{0}$
  • $[\mathbf{P}\hat{\mathbf{T}},\mathbf{H}]=\mathbf{0}$.

Theorem 2 (sufficient condition): Suppose that there exists in addition to the above operators/matrices a further matrix which has $\mathbf{C}^2=\mathbf{I}$ and $[\mathbf{C},\mathbf{H}]=\mathbf{0}$ but is itself not the unit matrix. Then $\mathbf{H}$ has real spectrum only if $\mathbf{C}$ commutes with $\mathbf{P}\hat{\mathbf{T}}$.

The proofs to the above two theorems are given in Bender, Carl M., and Philip D. Mannheim. "PT symmetry and necessary and sufficient conditions for the reality of energy eigenvalues." Physics Letters A 374.15-16 (2010): 1616-1620

Note 1: There is a more general version of Theorem 1 given in Bender, Carl M., M. V. Berry, and Aikaterini Mandilara. "Generalized PT symmetry and real spectra." Journal of Physics A: Mathematical and General 35.31 (2002): L467.

Definition: $\hat{\mathbf{A}}$ is antiunitary if $(\hat{\mathbf{A}}\mathbf{u})^\dagger(\hat{\mathbf{A}}\mathbf{v})=(\mathbf{u}^\dagger\mathbf{v})^*$. (Any antiunitary $\hat{\mathbf{A}}$ can be expressed as the complex conjugation operator times a unitary matrix.)

Theorem 1.1: A matrix $\mathbf{H}$ has real or complex conjugate eigenvalues if it commutes with an antiunitary $\hat{\mathbf{A}}$ that also satisfies $\hat{\mathbf{A}}^{2k}=\mathbf{I}$ for $k$ odd integer. ($\mathbf{P}\hat{\mathbf{T}}$ corresponds to $k=1$.)

Note 2: A nice sufficient condition is indeed given by the paper referenced by @MossMurderer, albeit it's less general (only works for real matrices): Bose, N. K. "On real eigenvalues of real nonsymmetric matrices." Proceedings of the IEEE 56.8 (1968): 1380-1380.

Note 3: A non-trivial non-Hermitian operator which has real and positive eigenvalues is e.g. $\hat{H}=\frac{\mathrm{d}^2}{\mathrm{d}x^2}+m^2x^2-(ix)^N$ with a real parameters $m$ and $N$. More examples are given in many works by Carl Bender et. al., but one needs a bit (or sometimes a lot) of QM to understand them. Check out e.g. Bender, Carl M., and Stefan Boettcher. "Real spectra in non-Hermitian Hamiltonians having P T symmetry." Physical Review Letters 80.24 (1998): 5243. .


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.