Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I realize that when a matrix is symmetric, then it must have all real eigenvalues. However, I am doing research on matrices for my own pleasure and I cannot find a mathematical proof or explanation when a matrix will have all real eigenvalues except for when it is symmetric. I am dealing with matrices such as A below and I want to know what is it about A and its characteristic polynomial that gives it real eigenvalues (0, 0, -2)? Similarly, what is it about matrix B that gives it only one real eigenvalue (0) and the other two complex? enter image description here

enter image description here

share|cite|improve this question
up vote 2 down vote accepted

There are so-called $\mathcal{PT}$-symmetric matrices which may have purely real eigenvalues. A square matrix $M$ is called $\mathcal{PT}$-symmetric iff it satisfies the property: $$ [\mathcal{PT},M] = 0 \Leftrightarrow \mathcal{P}M = M^*\mathcal{P} $$ where $\mathcal{P}$ is and $\mathcal{T}\equiv*$ is the complex conjugation operator. Further $[\mathcal{P},\mathcal{T}]=0$ and $\mathcal{P}^2=1$, $\mathcal{T}^2=1$ $\Rightarrow (\mathcal{PT})^2=1$. A $\mathcal{PT}$-symmetric matrix is said to have 'unbroken' $\mathcal{PT}$ symmetry iff any eigenvector of $M$ is also an eigenvector of $\mathcal{PT}$.

Claim: If $M$ has unbroken $\mathcal{PT}$ symmetry, this implies that $M$ has real eigenvalues.

Proof: First note that the eigenvalues of $\mathcal{PT}$ are non-zero since the combination is an involution: $\mathcal{PT} u = \mu u \Rightarrow \mathcal{PT}^2 u = u = \mu^*\mu u \Rightarrow |\mu|=1$.

Now let $Mv = \lambda v \Rightarrow \mathcal{PT}Mv = \mathcal{PT} \lambda v$. Thus since the combination $\mathcal{PT}$ is an anti-linear operator and $M$ commutes with $\mathcal{PT}$: $\mathcal{PT} M v = \lambda^*\mathcal{PT}v\Rightarrow \lambda \mu = \lambda^*\mu$. Since we've shown that $\mu\neq0$, this implies $\lambda^* = \lambda$. QED

More on $\mathcal{PT}$-symmetry and related concepts in this article: The English in there isn't perfect but the content looks good.

share|cite|improve this answer

Here is one example of sufficient conditions.

share|cite|improve this answer
Posting just a link as an answer is rather fragile; links tend to become stale some day. At least you could try to lift from the paper the statement of the conditions you refer to. – Marc van Leeuwen Apr 9 '14 at 6:16

Do you know of companion matrices? See the Wikipedia link here:

They are made-to-order matrices which will have the polynomial you want as its characteristic polynomial. They are far from symmetric matrices. Now start with a polynomial having your favorite real numbers as its roots, and construct the Companion matrix for that polynomial.

share|cite|improve this answer
I don't think the question is about how to construct matrices with real eigenvalues, but on how to recognise them. Construction is simple: just take any real triangular matrix and conjugate it by any real invertible matrix (moreover all examples can be obtained in this way). – Marc van Leeuwen Apr 9 '14 at 6:19
Ok, I see the distinction in the question which my answer does not address. – P Vanchinathan Apr 9 '14 at 6:35

Another approach is to construct a triangular matrix with pre-determined diagonal entries; they will be the eigenvalues, and the matrix is not symmetric.

share|cite|improve this answer

A necessary and sufficient condition for a matrix$~A$ to have only real eigenvalues (that is, not have any non-real complex eigenvalues) is the existence of a polynomial $P$ that splits into linear factors over the real numbers and such that $P[A]=0$. If such a polynomial exists at all, one can take the characteristic polynomial for$~A$ (but not necessarily the minimal polynomial) as$~P$. This gives a trivially valid, but fairly hard to check condition. Without using eigenvectors, it is actually not so obvious why the characteristic polynomial of a symmetric matrix should always allow such a factorisation. But I don't think one can do much better to completely characterise the case of real-only eigenvalues.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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