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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.

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    $\begingroup$ 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
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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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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 $W$.

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  • $\begingroup$ 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

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