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.

  • 3
    $\begingroup$ 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 $1/t^2$). $\endgroup$ – Anthony Quas Feb 18 at 6:12
  • 1
    $\begingroup$ Perhaps the true quantity to understand is $(u^T v)^2/((u^T u)(v^T v))$ then? $\endgroup$ – Kevin Casto Feb 18 at 11:17
  • $\begingroup$ @AnthonyQuas the ratio is still uniquely defined as stated (with $v$ having unit length). In any case, Kevin Casto proposes an alternative. $\endgroup$ – Leo Feb 18 at 12:28
  • $\begingroup$ @KevinCasto, yes, studying that alternative would also be of interest! $\endgroup$ – 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.

| cite | improve this answer | |
  • $\begingroup$ Interestingly, the ratio equals $1$ exactly when the matrix is symmetric. $\endgroup$ – Leo Feb 18 at 20:18
  • $\begingroup$ well $u=v$ if matrix is symmetric, so it is not a surprise ? $\endgroup$ – Piyush Grover Feb 18 at 20:24
  • $\begingroup$ 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. $\endgroup$ – Leo Feb 18 at 20:36
  • $\begingroup$ @RobertIsrael - I just saw your latest edit - perhaps this ratio being equal to $1$ is an indication of normality rather than symmetricity. $\endgroup$ – Leo Feb 18 at 20:38
  • $\begingroup$ Doubly stochastic matrices need not be normal either. $\endgroup$ – 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 $W$.

| cite | improve this answer | |
  • $\begingroup$ 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... $\endgroup$ – Leo Feb 20 at 12:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.