Significance of the length of the Perron eigenvector

Question

Let $A$ be a positive square matrix. Perron-Frobenius theory says that there exist $\lambda,v$ with $Av=\lambda v$ and $\lambda$ equals the spectral radius of $A$, $\lambda$ is simple, and $v$ is positive.

Now consider also the left Perron eigenvector $u^T A=\lambda u^T$. Another result of Perron-Frobenius theory is that

$$\lim_{m\to \infty} \frac{A^m}{\lambda^m} = \frac{v u^T}{u^T v}.$$

Suppose $\|v\|=1$. The above result says that the "correct" normalization for u is $u^T v=1$ rather than the more usual $u^T u=\|u\|^2=1$. This motivates the question: what is the significance of the ratio

$$\frac{u^T v}{u^T u} ?$$

Are there matrices $A$ for which this ratio is arbitrarily large? Arbitrarily small? Does this ratio determine any properties of $A$? Note that if $A$ is symmetric, then $u=v$ and this ratio is always equal to $1$, but that's not the case in general for arbitrary $A$. Could it be the case that this ratio is measuring how far $A$ is from being symmetric?

Note too that this normalization is necessary so that the limit $\frac{v u^T}{u^T v}$ is a projection matrix (i.e. that its only non-zero eigenvalue is one). In this context, I understand why the normalization is necessary, but I'm interested in the amount of normalization necessary with respect to the length of $u$.

Any pointers appreciated. Thanks!

EDIT In the comments, it is argued that the real quantity of interest in this setup is

$$\frac{\left( u^T v \right)^2}{\left(u^T u \right) \left( v^T v \right)}.$$

This quantity is also of interest to me, and an acceptable replacement for my original question.

Answer 1

That quantity $s = \frac{|u^Tv|}{\|u\|\|v\|}$ is the inverse of the eigenvalue condition number. The smaller it is, the more sensitive to perturbation the Perron value is.

More precisely, any perturbed matrix $A+E$ with $\|E\| \leq \varepsilon$ has a Perron value $\tilde{\lambda}$ that satisfies $|\tilde{\lambda}-\lambda| \leq \frac{\varepsilon}{s} + \mathcal{O}(\varepsilon^2)$. See e.g. Section 7.2.2 of Golub and Van Loan's Matrix Computations 4th ed.

In addition, note that if $A$ is normal then $s=1$ (its maximum possible value) and the Perron value is perfectly conditioned; while if $\lambda$ is a defective eigenvalue (e.g. $A = \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$) then $s=0$. So rather than a "distance from symmetric" I'd say that $1-s$ is a "distance from normal" or $s$ is a "distance from defective".

Answer 2

For example, if $$ \eqalign{A &= \pmatrix{1 & t\cr 1 & 1\cr},\ v = \pmatrix{\sqrt{t}\cr 1},\ u =\pmatrix{1\cr \sqrt{t}},\cr \frac{(u^T v)^2}{(u^T u)(v^T v)} &= \frac{4t}{(1+t)^2} \to 0 \ \text{as}\ t \to \infty} $$ By Cauchy-Schwarz we always have $$ 0 < \frac{(u^T v)^2}{(u^T u)(v^T v)} \le 1$$ with equality on the right iff $u$ is a scalar multiple of $v$.

Note also that if $A$ is doubly stochastic, $u = v = (1,\ldots,1)^T$. Not all doubly stochastic matrices are symmetric.

Answer 3

I'm not sure how helpful, but you can say the following: in general, $u$ (or the 1-dimensional eigenspace spanned by it) is the orthogonal complement to the span of all the (right-) eigenvectors except $v$, call this $W$. The quantity I mentioned is $\cos^2$ of the angle between $v$ and $u$, which is $1 - \cos^2$ of the angle between $v$ and $W$.

So if you know things about just the right eigenvectors, you can say things about this quantity. For example, it's close to 0 if $v$ is close to $W$, and close to 1 if $v$ is close to being perpendicular to $W$.