# Significance of the length of the Perron eigenvector

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.

• No significance. It's just a question of scaling. $u^Tv=1$ ensures that $u$ and $v$ are dual eigenvectors. With this scaling, there is still one degree of freedom: you can multiply $u$ by a constant $t$ and divide $v$ by the same constant (notice that this rescaling scales your ratio by a factor of $1/t^2$). – Anthony Quas Feb 18 at 6:12
• Perhaps the true quantity to understand is $(u^T v)^2/((u^T u)(v^T v))$ then? – Kevin Casto Feb 18 at 11:17
• @AnthonyQuas the ratio is still uniquely defined as stated (with $v$ having unit length). In any case, Kevin Casto proposes an alternative. – Leo Feb 18 at 12:28
• @KevinCasto, yes, studying that alternative would also be of interest! – Leo Feb 18 at 12:28

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.

• Interestingly, the ratio equals $1$ exactly when the matrix is symmetric. – Leo Feb 18 at 20:18
• well $u=v$ if matrix is symmetric, so it is not a surprise ? – Piyush Grover Feb 18 at 20:24
• I didn't say it was a surprise, I said it was interesting in light of my conjecture (see original post) that this ratio measures how far $A$ is from being symmetric. – Leo Feb 18 at 20:36
• @RobertIsrael - I just saw your latest edit - perhaps this ratio being equal to $1$ is an indication of normality rather than symmetricity. – Leo Feb 18 at 20:38
• Doubly stochastic matrices need not be normal either. – Robert Israel Feb 18 at 21:02

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 $$W$$.

• This is very helpful! I take it the other way around though: knowing only $v$ and $u$ can say something about $v$ and $W$. Now I'm wondering whether $v$ being close to $W$ affects the convergence rate of power methods for computing eigenvalues... – Leo Feb 20 at 12:33