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Let $\mathcal{N}_{n}$ be the set of symmetric nonnegative irreducible matrices. For a matrix $A \in \mathcal{N}_{n}$ let $v^{A}$ be its Perron vector, normalized so that $||v^{A}||_{2}=1$.

Define the function $f:\mathcal{N}_{n} \rightarrow \mathbb{R}$ by $$ f(A)=||v^{A}||_{1}. $$

According to my observations, $f$ is apparently convex.

Is this true/known?

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What's the Perron vector? – Deane Yang Feb 24 '14 at 1:12
@DeaneYang It's the positive eigenvector corresponding to the spectral radius. – Felix Goldberg Feb 24 '14 at 1:15
Thanks for that. – Deane Yang Feb 24 '14 at 2:04
up vote 4 down vote accepted

Maybe I am misinterpreting something, because according to my experiments, this function is neither convex nor concave.

The following is a counterexample (EDIT: I changed the example to use symmetric matrices):

\begin{equation*} A=\begin{pmatrix}8 &4\\ 4 & 6\end{pmatrix},\quad B=\begin{pmatrix} 4 & 4\\ 4 & 6\end{pmatrix},\quad C=\frac{A+B}{2}=\begin{pmatrix}6 & 4\\ 4 &6 \end{pmatrix}. \end{equation*} For this choice, we have

\begin{equation*} v(A) = (.7882, .6154),\quad v(B)=(.6154,.7882),\quad v(C)=(.7071, .7071). \end{equation*} But $\|v(C)\|_1 > 0.5\|v(A)\|_1 + 0.5\|v(B)\|_1$ (notice all there vectors have unit 2-norm as required).

A similar counterexample to potential concavity is also easy to find.

EDIT 2: Here is a counterexample to concavity.

\begin{equation*} A = \begin{pmatrix}16&2\\ 2&16\end{pmatrix},\quad B = \begin{pmatrix}14&8\\8 &2 \end{pmatrix},\quad C = (A+B)/2 \end{equation*} Then, we have \begin{equation*} v(A) = \begin{pmatrix}\tfrac{1}{\sqrt{2}}\\\tfrac{1}{\sqrt{2}}\end{pmatrix},\quad v(B) =\begin{pmatrix}\tfrac{2}{\sqrt{5}}\\\tfrac{1}{\sqrt{5}}\end{pmatrix},\quad v(C)= \begin{pmatrix}\tfrac{3+\sqrt{34}}{\sqrt{68+6 \sqrt{34}}}\\\tfrac{5}{\sqrt{68+6 \sqrt{34}}} \end{pmatrix}. \end{equation*} Doing the numerics with this shows that $\|v(C)\|_1-0.5(\|v(A)\|_1+\|v(B)\|_1) = -0.0150285...$.

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But $A$ is not symmetric (the question requires symmetric matrices) – Suresh Venkat Feb 24 '14 at 2:55
@Suresh: thanks; changed example to use symmetric $A$; the claim still fails. Maybe Felix had something else in mind... – Suvrit Feb 24 '14 at 3:30
Actually, I've been getting $\|v(C)\|_1 > 0.5\|v(A)\|_1 + 0.5\|v(B)\|_1$ all the time (which is of course concave, not convex, silly me). Are you sure you can produce examples of the opposite kind? I'm unable to replicate them... – Felix Goldberg Feb 24 '14 at 6:55
Thanks - this put paid to my guess! :) – Felix Goldberg Feb 25 '14 at 1:00
@Felix: it was an attractive conjecture. A related geometric convexity does hold though for Perron vectors (but under a rather special situation): see here:… – Suvrit Feb 25 '14 at 1:23

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