# Besides being symmetric, when will a matrix have ONLY real eigenvalues?

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?

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 $$\mathcal{T}\equiv*$$ is the complex conjugation operator and $$\mathcal{P}$$ is a matrix satisfying $$[\mathcal{P},\mathcal{T}]=0$$ (implying that $$\mathcal{P}$$ is a real matrix) and $$\mathcal{P}^2=1$$, $$\mathcal{T}^2=1$$ $$\Rightarrow (\mathcal{PT})^2=1$$. Since $$\mathcal{P}$$ is an involution, its eigenvalues are $$\pm1$$ and one may find a basis such that $$\mathcal{P}=\mathrm{diag}(1,1,1,1,...,-1,-1,-1)$$. 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: http://arxiv.org/abs/1212.1861 The English in there isn't perfect but the content looks good.

• What is PT? Thanks Commented Jun 29, 2019 at 18:06
• It comes from Physics, there it stands for Parity (position $x \to -x$) and Time inversion operator (time $t \to -t$). Their properties a $\mathcal{P}^2=1$ and $\mathcal{T}^2=1$ are derived in the context of PT-symmetric quantum mechanics. Commented Jul 18, 2019 at 20:44
• Thanks, but I don't understand what is the definition of the operators. Can you update the answer? Currently, it just says "where P is and T". I think this is a mistake. Commented Jul 19, 2019 at 13:56

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

Do you know of companion matrices? See the Wikipedia link here: http://en.wikipedia.org/wiki/Companion_matrix

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.

• 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). Commented Apr 9, 2014 at 6:19
• Ok, I see the distinction in the question which my answer does not address. Commented Apr 9, 2014 at 6:35

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.

• 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. Commented Apr 9, 2014 at 6:16

I do not understand what is the OP's question. Yet, I think that the interesting question is : if I randomly choose a large square matrix, then should I wait for a long time before getting a matrix whose all eigenvalues ​​are real? The answer is: yes, a very long time.

Let $$N_n$$ be the number of real roots of a random polynomial (its coefficients are iid and follow the same law) of degree $$n$$. In this paper (one of the authors is Van Vu who wrote papers with Terrific Tao...),

https://arxiv.org/pdf/1402.4628.pdf

we find this beautiful result

$$\textbf{Theorem}$$. For any random variable $$\xi$$ with mean $$0$$ and variance $$1$$ and bounded $$(2 +\epsilon)$$-moment, one has

$$E(N_n)=2/π\log n+O(1)$$ where $$O(1)$$ depends on $$\xi$$.

$$\textbf{Remark}$$. Generally, the standard deviation of $$N_n$$ is in $$O(\sqrt{\log(n)})$$.

Thus, the probability when $$n$$ is large that $$N_n=n$$ is $$\approx 0$$. For example, if $$n=10$$ and the coefficients are uniformly random in $$[[-100,100]]$$, then $$10^5$$ tests give $$0$$ such matrix!!