No repeated eigenvalues or the real part of any eigenvalue is not zero

Question

I have an $n$ x $n$ matrix

$M=\begin{bmatrix}-1 & -1\\ \frac{1}{2} & 0 & -\frac{1}{2}\\ & \ddots & \ddots & \ddots\\ & & \frac{1}{2} & 0 & -\frac{1}{2}\\ & & & \frac{1}{2} & 0 & -\frac{1}{2}\\ & & & & 1 & -1 \end{bmatrix}$

I need to show for $n\geq3$, either that the real part of all eigenvalues are negative or that there are no repeated eigenvalues with real part zero.

Numerical tests for different values of $n$ show that the first statement is true, and hence the second as well, but how to prove it?

I am able to show that the real part of all eigenvalues of $M$ are non-positive as for

$H=\begin{bmatrix}\frac{1}{2}\\ & 1\\ & & \ddots\\ & & & 1\\ & & & & 1\\ & & & & & \frac{1}{2} \end{bmatrix}$,

the matrix, $HM+M^{*}H$, is negative semidefinite. $^*$ denotes conjugate transpose of the matrix. The proof of this could be found in any control theory book. I have copied the same below.

Another observation: $det(M)=(-1)^{n}(\frac{1}{2})^{n-3}$

Proof (if $HM+M^{*}H$ is negative semidefinite, real part of all eigenvalues of $M$ are non-positive):

Suppose $\lambda_{i}$ is an eigenvalue of $M$ and let $\mathbf{v}_{i}$ be the corresponding eigenvector, then

$\mathbf{v}_{i}^{*}(HM+M^{*}H)\mathbf{v}_{i}=\mathbf{v}_{i}^{*}H(M\mathbf{v}_{i})+(M\mathbf{v}_{i})^{*}H\mathbf{v}_{i}=\mathbf{v}_{i}^{*}H(\lambda_{i}\mathbf{v}_{i})+(\lambda_{i}\mathbf{v}_{i})^{*}H\mathbf{v}_{i}=(\lambda_{i}+\lambda_{i}^{*})\mathbf{v}_{i}^{*}H\mathbf{v}_{i}$.

Since $HM+M^{*}H$ is negative semidefinite,

$\mathbf{v}_{i}^{*}(HM+M^{*}H)\mathbf{v}_{i}\leq0\Rightarrow(\lambda_{i}+\lambda_{i}^{*})\mathbf{v}_{i}^{*}H\mathbf{v}_{i}\leq0\Rightarrow Re(\lambda_{i})\leq0$.

The last inequality holds as $H$ is positive definite.

Answer 1

Hint: assuming normalized eigenvectors, $\mathbf{v}_{i}^{*}(HM+M^{*}H)\mathbf{v}_{i}= - \beta$ and $\mathbf{v}_{i}^{*}H\mathbf{v}_{i}= Re(\lambda_i)(2-\beta)$ where $\beta = |v_{i,1}|^2+|v_{i,n}|^2$ (sum of the squared modulus of the first and last elements of the eigenvector). So $Re(\lambda)=\beta/(\beta-2)$. This is non-positive, and is zero iff $v_{i,1}=0$ and $v_{i,n}=0$. Show that an eigenvector of $M$ can't have $v_{i,1}=0$ for $n\ge3$