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The following is proved in these lecture notes.

Let $A$ be an $n\times n$ real matrix with all entries positive. Then $A$ has a unique positive eigenvector (up to positve scaling), and the eigenvalue of $A$ corresponding to this eigenvector is greater than every other real eigenvalue in absolute value.

By a positive vector we mean a vector all of whose entries are positive.

(I am alluding the statement made alongside "Perron-Frobenius Theorem" in the link provided. What I have stated is not quite what is stated in the PDF).

The proof given goes as follows.

Let $X$ be the set of all the points $(x_1, \ldots, x_n)\in\mathbf R^n$ such that each $x_i\geq 0$, and $\sum x_i^2=1$. Define $T:X\to X$ as $Tv=Av/\|Av\|_2$. Then $T$ is a contraction (I haven't yet verified this but this seems intuitively reasonable). Therefore, by Banach Fixed Point theorem, there is a unique vector $v\in X$ which is fixed by $T$. Thus $v$ is an eigenvector of $A$ and say $Av=\lambda v$.

Now let $\mu$ be any other real eigenvalue of $A$, and let $w=(w_1, \ldots, w_n)$ be a corresponding eigenvector. At least one entry of $w$ is negative. Write $|w|$ to denote the vector whose $i$-th entry is $|w_i|$.

Here is the statement I am unable to understand

Then $A(|w|)$ has smaller length than $\lambda L$, where $L$ is the length of $w$.

Can somebody explain why is this true?

It seems what is really used here is that $\|Au\|\leq \lambda \|u\|_2$ for each $u\in X$, but I don't see why is this true.

A related discussion is here.

user1551
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caffeinemachine
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1 Answers1

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This is not true. Consider the following nonnegative matrix first: $$ A=\pmatrix{2&1\\ 0&1}. $$ Its two eigenvalues are $\lambda:=\rho(A)=2$ and $\mu=1$, and $w=(1,-1)$ an eigenvector corresponding to the unit eigenvalue $\mu$. Hence $A|w|=(3,1)^T$ and $\dfrac{\|A|w|\|}{\|w\|}=\sqrt{5}>\lambda$.

Now, as both $\lambda$ and $\mu$ are simple eigenvalues, if we perturb $A$ slightly by adding a sufficiently small positive multiple of the all-one matrix to it, the eigenvalues and eigenvectors of the new $A$ will be perturbed slightly from the old ones as well. Therefore, a counterexample with a positive $A$ is obtained.

While their claim that $\|A|w|\|\le\lambda\|w\|$ is false and their proof of the inequality $|\mu|\le\lambda$ is invalid, the latter inequality itself is true. The usual proof goes something like this. If we consider left eigenvectors instead of right eigenvectors, the same contraction map argument implies that there exists a positive left eigenvector $u$ of $A$ corresponding to some eigenvalue $\lambda'$. We must have $\lambda=\lambda'$ because $\lambda'u^Tv=(u^TA)v=u^T(Av)=\lambda u^Tv$ and $u^Tv>0$. It follows that $$ |\mu|\,u^T|w|=u^T|Aw|\le u^TA|w|=(u^TA)|w|=\lambda u^T|w|. $$ As $u$ is positive and $w$ is nonzero, $u^T|w|$ is positive. Hence $|\mu|\le\lambda$.

user1551
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