It is well-known that the solutions of $$|M-\lambda I|=0$$ are purely real if $M$ is real symmetric.

Conjecture This is still true if we replace the identity matrix by a diagonal matrix $D$ whose diagonal elements are either $0$ or $1$:

$$\left. \begin{matrix}|M-\lambda D|=0 \\ M\text{ is real symmetric}\\M\text{ and }D\text{ do not have a common nullspace}\end{matrix}\right\}\implies \lambda \in \mathbb{R}\text{ (if a solution exists)}$$

Partial proof

Without loss of generality we can assume $$D=\text{diag}(1,1,\cdots,1,1,0,0,\cdots,0,0)$$

If $M$ is symmetric, $$|M-\lambda D|=\left|\begin{matrix}M_1-\lambda I & M_2 \\ M_2^T & M_3\end{matrix}\right|$$ with $M_1,M_3$ symmetric. If $M_3$ is invertible, $$\left|\begin{matrix}M_1-\lambda I & M_2 \\ M_2^T & M_3\end{matrix}\right|=|M_3||M_1-M_2 M_3^{-1}M_2^T -\lambda I|$$ $|M_3|\neq 0$ by hypothesis, so $$|M-\lambda D|=0\implies |M_1-M_2 M_3^{-1}M_2^T -\lambda I|=0$$ thus $\lambda$ is an eigenvalue of the symmetric matrix $M_1-M_2 M_3^{-1}M_2^T$, and is therefore real.


What if $M_3$ is not invertible? The condition that $M_3$ is invertible does not appear to be necessary, for example $$\left| \begin{array}{ccc} e-\lambda & a & b \\ a & d-\lambda & c \\ b & c & 0 \\ \end{array} \right|=0$$ has a single real solution. Another example, where $M_3$ is $2\times 2$ and nonzero: $$\left| \begin{array}{cccc} 1-\lambda & 2 & 5 & 4 \\ 2 & 1-\lambda & 3 & 6 \\ 5 & 3 & 1 & 1 \\ 4 & 6 & 1 & 1 \\ \end{array} \right|=0$$ $$\lambda=-\frac{151}{5}$$

  • $\begingroup$ Can you find an example where $M_3$ is not invertible and non-zero? In the zero case the problem becomes an eigenvalue problem for $M_1$ with the constant term in the characterstic polynomial adjusted by an amount determined by $M_2$, which is a special case. $\endgroup$
    – Eddy
    Sep 30 '17 at 10:00
  • $\begingroup$ This is an example with $M_3$ not invertible and not zero: $$\left| \begin{array}{cccc} 1-\lambda & 2 & 5 & 4 \\ 2 & 1-\lambda & 3 & 6 \\ 5 & 3 & 1 & 1 \\ 4 & 6 & 1 & 1 \\ \end{array} \right|=0$$ $$\lambda=-\frac{151}{5}$$ $\endgroup$
    – Wouter
    Sep 30 '17 at 10:48
  • $\begingroup$ If you put it as an edit to your question then it will be more human readable. $\endgroup$
    – Eddy
    Sep 30 '17 at 10:49
  • $\begingroup$ Is it not enough to consider the other case in which the block matrix is not invertible? $\endgroup$
    – user480945
    Oct 2 '17 at 10:50

The result can be proven using the same arguments as in the proof that the eigenvalues of symmetric matrices are real.

Let $x\ne0$ solve $Mx=\lambda Dx$ for some possibly complex $\lambda$.

If $Dx=0$, then it holds $Mx=0$, and $\lambda$ can be arbitrary. This corresponds to an infinite generalized eigenvalue of the matrix pencil $(M,D)$. Under your assumption, that $M$ and $D$ do not have common null space, then this case cannot happen.

Now let $Dx\ne0$. Since $D$ is diagonal with non-negative entries, it follows $x^HDx\ne0$. Then $$ \lambda x^HDx = x^HMx = (Mx)^Hx= (\lambda Dx)^Hx=\bar\lambda x^HDx. $$ This implies $\lambda=\bar\lambda$, and $\lambda$ is a real number.

Note, that this argument also works if $D$ is assumed to be symmetric positive semidefinite, as then $Dx\ne0$ implies $x^HDx\ne0$.


This is a very simple density argument unless I missed something. Note that the set of matrices with only real eigenvalues is closed in $M_n({\mathbb R})$ with the usual topology.

Consider the matrices $M_t=M+t(I-D)$ for $t\in{\mathbb R}$. There is an $\varepsilon >0$ such that $M_3+tI_m$ is invertible for $t\in (0,\varepsilon)$ (here $m$ is the size of $M_3$). For those $t$, all you say in your OP applies to $M_t$, so $M_t$ has only real eigenvalues. Now $M=\lim_{t\to 0} M_t$, so $M$ has only real eigenvalues also.


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