# eigenvalues of $C=\begin{bmatrix}−I &-I\\L&0\end{bmatrix}$

Consider the following matrix $$C=\begin{bmatrix}−I &-I\\L&0\end{bmatrix}$$ where for $L$ we have: $$L\mathbf{1}=0$$ $$\mathbf{1}^TL=0$$ $$rank(L)=dim(L)-1$$ $$L+L^T\geq 0$$ zero is a simple eigenvalue of $L$ and $L+L^T$.

We cans show that C has a zero eigenvalue and another one equal to -1, what does we can say about the rest of eigenvalues?

For example, we know that when$L=L^T$,C has only one simple zero and the rest of the eigenvalues have negative real parts.

Numerical examples show that most of the time C has only one simple zero and the rest of the eigenvalues have negative real part, but sometimes there is a pair of eigenvalues with zero real part also appearing. Under what conditions does this pair appear?

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We can show $\lambda$ is an eigenvalue of $C$ if and only if $-\lambda(\lambda + 1)$ is an eigenvalue of $L$.

We know $L$ has a one-dimensional nullspace, and setting $-\lambda(\lambda+1) = 0$ shows that $\lambda = 0,-1$ are each eigenvalues of $C$.

Of course that is one way to obtain an eigenvalue of $C$ having zero real part, namely zero. Is there another possibility? Let $r$ be an eigenvalue of $L$, perhaps complex. If $bi$ were a purely imaginary eigenvalue of $C$, we must have:

$$-bi(bi+1) = r$$

$$r = b^2 - bi$$

If $L$ is a real matrix, its complex eigenvalues occur in conjugate pairs. So $\overline{r} = b^2 + bi$ gives rise to another purely imaginary eigenvalue $-bi$ of $C$.

That the characteristic polynomial of $C$ has the form $det(\lambda(\lambda+1)I + L)$ follows from a brief computation:

$$\lambda I - C = \begin{pmatrix} (\lambda+1)I & I \\ -L & \lambda I \end{pmatrix}$$

Hence our claim, that $\lambda$ is an eigenvalue of $C$ if and only if $\lambda(\lambda+1)$ is an eigenvalue of $-L$.

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