2
$\begingroup$

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.

Improve this question
$\endgroup$
12
  • $\begingroup$ I'm having difficulties to understand the question. In the third paragraph, why is $v$ unique (i.e., why is the eigenvalue $\lambda$ of $H$ simple?) and why are the $v_i$ strictly positive? Is this an additional assumption? Regarding paragraphs 5 and 6: why is the question asked in paragraph 6 equivalent to the one asked in parsgraph 4? $\endgroup$ Commented Oct 12, 2023 at 0:20
  • $\begingroup$ Apart from this, please allow me to point out that the notation $H^+$ for the componentwise modulus is misleading. In vector lattice theory (which is, in a sense, married to Perron-Frobenius theory) a superscript + is always used to denote the positive part, not the modulus. I'd thus recommend to replace the notation $H^+$ with $|H|$. $\endgroup$ Commented Oct 12, 2023 at 0:25
  • $\begingroup$ @JochenGlueck sorry for confusing sentences. Uniquness and positiveness of $\mathbf{v}$ is assumption. and for your advice, isn't it also misleading since $|H|$ usually mean determinant? $\endgroup$ Commented Oct 12, 2023 at 8:15
  • $\begingroup$ @ChristianRemling thank you for the comment. However, I think $\mathbf{v} = (1, 1)$ is not a ev for largest eigenvalue of $H$ and not suitable conterexample for my problem setting. $\endgroup$ Commented Oct 12, 2023 at 8:25
  • 1
    $\begingroup$ At keisukemurota: As you can see from the discussion with @ChristianRemling and me, we got a bit confused. Hence, I edited your question to streamline it and to make the logical structer a bit clearer. Could you please check whether the edited version still reflects your intentions? $\endgroup$ Commented Oct 12, 2023 at 15:16

1 Answer 1

Reset to default
1
$\begingroup$

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 $v$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ That's a perfect counterexample I was looking for! I really appreciate it! Btw, do you think we can ensure $\lambda = \lambda^+$ if we impose some additional condition on $H$? $\endgroup$ Commented Oct 12, 2023 at 16:37
  • $\begingroup$ @keisukemurota: Unfortunately, I'm currently not aware of a good condition which ensures $\lambda = \lambda^+$. $\endgroup$ Commented Oct 12, 2023 at 16:59

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.