Conditions under which a $2\times2$ block matrix has complex eigenvalues

Question

Consider a matrix $A$ in $\mathbb{R}^2$:

\begin{equation} A=\begin{bmatrix} 0 & -c \\ 1 & -b \end{bmatrix} \end{equation}

Then it can be shown that the matrix has complex eigenvalues if $b^2-4c < 0$. Can a similar relation be derived also for the elements of a similarly shaped 2x2 block matrix, ie:

\begin{equation} M=\begin{bmatrix} 0_n & -C \\ I_n & -B \end{bmatrix} \end{equation}

where $B, C$ are matrices in $\mathbb{R}^n$, such that $\lambda_M$ are complex?

Answer 1

First we note that the elements $(2,1)$ and $(2,2)$ of matrix $M$ commute. Then we consider:

\begin{equation} (\lambda I_{2n}-M)v = 0 \end{equation}

which is the eigenvalue problem of $M$. The equation admits a solution $v \neq 0$ when $|\lambda I_{2n}-M| = 0$, so we resort to computing the determinant of the characteristic matrix:

\begin{equation} \begin{bmatrix} \lambda I_n & -C \\ I_n & \lambda I_n + B \\ \end{bmatrix} \end{equation}

We assume that $(1,1)$ is invertible, i.e., that $0$ is not an eigenvalue of $M$. Then since $I_n$ commutes with any matrix, we can write:

\begin{equation} |\lambda I_{2n} - M| = |\lambda ( \lambda I_n + B) + C | = |\lambda^2 I_n + \lambda B + C| = 0 \end{equation}

This is a Quadratic Eigenvalue Problem and has been intensively studied. Particularly, in (Tisseur, 2001), properties of the solutions $\lambda$ have been given in terms of the coefficient matrices $M, C, K$ of $\lambda^2 M + \lambda C + K = 0$. When the three matrices are Hermitian, then the eigenvalues are some solutions of:

\begin{equation} \lambda = \frac{-c(x) \pm \sqrt{c(x)^2 - 4k(x)m(x)}}{2m(x)} \end{equation}

with $m(x) = x^*Mx,c(x) = x^*Cx, k(x) = x^*Kx$. Then in this case it can be seen that a sufficient, but not necessary condition for $\lambda$ to be complex is that $c(x)^2 < 4k(x)m(x)$ $\forall x$, which in our case reduces to

\begin{equation}b(x)^2 < 4c(x) \text{ } \forall x\end{equation}

with $b(x) = x^* B x, c(x) = x^* C x$ (pay attention to the different meaning of $c(x)$ in the problem of Tisseur and the answer's problem).

If $B, C$ are Hermitian (which, being them real, implies symmetry), then $\lambda_{C,n} \leq c(x) \leq \lambda_{C,1}$ and $\min(|\lambda_{B}|)^2 \leq b(x)^2 \leq \max(|\lambda_{B}|)^2$, so a (very) broad sufficient condition is that:

\begin{equation} \min(\lambda_C) > \bigg(\frac{\max(|\lambda_{B}|)}{2}\bigg)^2 \end{equation}

If they are not Hermitian, then the same condition can be stated in terms of their symmetric parts since $b(x) = x^*Bx = \frac12 x^*(B+B^T)x$ and similarly for $c(x)$.

Answer 2

It depends on matrices $B$ and $C$. When you solve for $\lambda$ to find the eigenvalues, you generally have to use the quadratic formula, which involves square roots. For real numbers $b$ and $c$, that's easy, but what does it mean to square root matrices? If a matrix $D$ is diagonal, then its square root is simply $$\sqrt{D}=\left(\begin{bmatrix} d_{11} & 0 & \ldots & 0 \\ 0 & d_{22} & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & d_{nn} \end{bmatrix}\right)^{\frac{1}{2}} = \begin{bmatrix} \sqrt{d_{11}} & 0 & \ldots & 0 \\ 0 & \sqrt{d_{22}} & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & \sqrt{d_{nn}} \end{bmatrix}$$ If the matrix is not diagonal, then you will need to find all matrices $K$ such that $K^2$ is equal to that matrix. If there exists a $K$ with complex entries, then, in the context of this problem, at least one of the eigenvalues is complex. For now, let's assume the result of the matrix calculation $D=B^2-4C$ is diagonal, where $D$ represents the "discriminant" of the quadratic formula, even though it is a matrix. You will know that $\sqrt{D}$ has complex entries if and only if any of the entries of $D$ are negative (or complex-valued, but that won't happen since $B$ and $C$ have real entries themselves).

But, what if $D$ isn't diagonal? One helpful reference may be this Wikipedia entry, and probably your best option is to square root by diagonalization. From there, it may be tricky to determine any conditions on $B$ and $C$ for there to be complex eigenvalues.

So, to know whether there are complex eigenvalues based on the matrices $B$ and $C$, you must determine whether $D=B^2-4C$ is diagonal (note that this could only mean that $C$ should be diagonal). If $D$ is diagonal, then determine whether any entry of $D$ is negative. If it is not diagonal, then you will need to do a bit more work.

Answer 3

I don't think this is a complete answer. Using the formula for the characteristic polynomial of such a $2$ by $2$ block matrix, found for example here: https://en.wikipedia.org/wiki/Determinant#Block_matrices, one gets that the characteristic polynomial of this block matrix is: $$p(s) = \det(s^2I + sB + C).$$ So the problem is now reduced to finding necessary and sufficient conditions for $p(s)$ to only have real roots.

Since we are assuming that the matrices $B$ and $C$ are real, it thus follows that the characteristic polynomial has real coefficients and hence, as is well known, the eigenvalues are either real, or occur as part of a complex conjugate pair. Knowing this, it thus follows that the eigenvalues of $M$ are real iff either all eigenvalues of $M$, or all those of $-M$, have nonnegative imaginary part, which is equivalent to either all eigenvalues of $iM$, or all eigenvalues of $-iM$, have nonpositive real parts. This can probably be tested by looking at the dynamical properties of the system of ODEs given by $$ \frac{dx}{dt} = \pm iM x(t),$$ where $x(t)$ is a smooth function of the real variable $t$ with values in $\mathbb{C}^n$. I am not sure if this answers the OP's question.