Eigenvectors and eigenvalues of a symmetric matrix and its entry-wise absolute value

Question

The modulus of matrices is meant componentwise in the following. Let $H$ be a sqaure matrix that satisfies the following assumptions:

1. $H$ is real-valued, symmetric, and positive-definite..
2. $H$ is irreducible.
3. The largest eigenvalue $\lambda$ of $H$ is simple, say with eigenvector $\mathbf{v}$.
4. All element of $\mathbf{v}$ can be taken to be non-negative, i.e. $v_i \geq 0 $. (Even if $\mathbf{v}$ doesn't satisfy this, we can use replace $H$ with $SHS$ where $S_{ij} = \delta_{ij} \frac{v_i}{|v_i|}$ to satisfy this condition.)
5. The vector $\mathbf{v}$ is also an eigenvector of $|H|$; denote the associated eigenvalue by $\lambda^+$.

Question: Does it follow that $\lambda = \lambda^+$?

In case that this might be helpful, let me provide the properties I found so far:

• $\lambda \leq \lambda^+$

• Assumptions 4 and 5 and the Perron-Frobenius theorem imply that each entry $v_i$ of $\mathbf{v}$ is strictly positive.

• Also by the Perron Frobenius theorem, $\lambda^+$ is the largest eigenvalue of $|H|$.

• Given the assumptions of the problem setting, $\mathbf{v}$ must also be an eigenvector for the largest eigenvalue of $H - |H|$, which might be helpful in constructing counterexamples.

• Under the given assumption, the property $\lambda = \lambda^+$ that I'm interested in is equivalent to $H = |H|$. This can be seen as follows:

  If $\lambda = \lambda^+$ then $\sum_{i,j} v_i H_{ij} v_j = \sum_{i,j} v_i |H|_{i,j} v_j$ and since $v_i \geq 0$ and $|H|_{i,j} = |H_{i,j}|$, every term in summation must be equal to each other, which leads to $H_{ij} = |H|_{ij}$.

Thank you very much in advance.

Answer 1

Here is a counterexample:

We start with the matrices $$ A = \begin{pmatrix} 2 & 0 & 0 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 1 & 2 & 0 \\ 1 & 1 & 0 & 1 \end{pmatrix} \quad \text{and} \quad B = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix} . $$ The matrix $A$ is irreducible, so its leading eigenvalue $3$ is simple. The corresponding eigenvector is $v = (1,1,1,1)$. As the spectral radius of $A$ is $3$, the matrix $A+4I$ is positive definite; its leading eigenvalue $7$ is simple with eigenvector $v$ (by $I$ I denote the identity matrix). Moreover, $v$ is also an eigenvector of $B$ for the eigenvalue $1$.

Now consider the matrix $H := A - \varepsilon B$ for $\varepsilon > 0$.

The vector $v$ is an eigenvector of $H$ for the eigenvalue $7-\varepsilon$. Since the eigenvalues of matrices behave continuously under perturbation, we know that if $\varepsilon > 0$ is sufficiently small, then $H$ is positive definite and $7- \varepsilon$ is the largest eigenvalue of $H$ and is simple.

Since there is no entry at which both $A$ and $B$ are non-zero, we have $\lvert H \rvert = A + \varepsilon B$. So the largest eigenvalue of $\lvert H \rvert$ is $7+\varepsilon$ with eigenvector $v$.