Powers of a Positive Matrix in the Limit

Question

I'm trying to prove a standard result: for a positive $n \times n$ matrix $A$, the powers of $A$ scaled by its leading eigenvalue $\lambda$ converge to a matrix whose columns are just scalar multiples of $A$'s leading eigenvector $\mathbf{v}$. More precisely, $$ \lim_{k \rightarrow \infty} \left(\frac{A}{\lambda}\right)^k = \mathbf{v}\mathbf{u},$$ where $\mathbf{v}$ and $\mathbf{u}$ are the leading right and left eigenvectors of $A$, respectively (scaled so that $\mathbf{u}\mathbf{v} = 1$).

These notes give a nice compact proof, pictured below. (It covers the more general case where $A$ is primitive, but I'm happy to assume it's positive for my purposes.)

I'm stuck on the highlighted step. Having established the existence of a matrix $M$ that (a) fixes $\mathbf{u}$ and $\mathbf{v}$, and (b) annihilates all other generalized eigenvalues of $A$, how do we know $M$ is unique? Why couldn't there be other matrices satisfying (a) and (b)?

I don't know much about generalized eigenvectors. I gather they're linearly independent, hence form a basis for $\mathbb{R}^n$. So each column of $M$ must be a unique linear combination of $A$'s generalized right eigenvectors. Is there some path I'm not seeing from there to the conclusion that only one matrix can satisfy both (a) and (b)?

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Answer 1

According to PF theorem, the characteristic polynomial of $A$ is in the form $\chi_A(x)=(x-\lambda)f(x)$ where $\lambda>0$ and the roots of $f$ have modulus $<\lambda$. Moreover, there is a (unique up to a factor) vector $v>0$ s.t. $Av=\lambda v$. Since $A^T$ is primitive, there is a (unique up to a factor) vector $u>0$ s.t. $u^TA=\lambda u^T$. Since $u^Tv>0$, we can choose the above factors s.t. $u^Tv=1$.

There is a basis in the form $v,\cdots$ s.t., for this change of basis of matrix $P\in M_n(\mathbb{R})$, $A=Pdiag(\lambda,B_{n-1})P^{-1}$ where $\chi_B(x)=f(x)$; then $B$ has a spectral radius $\rho(B)<\lambda$, that is, $\rho(\lambda^{-1}B)=\mu<1$.

Thus $(\lambda^{-1}A)^k=Pdiag(1,(\lambda^{-1}B)^k)P^{-1}$ tends, when $k\rightarrow\infty$, to the rank $1$ projector $M=Pdiag(1,0_{n-1})P^{-1}$; moreover, for every $\epsilon>0$, $||(\lambda^{-1}A)^k-M||=O((\dfrac{\mu+\epsilon}{\lambda})^k)$.

A projector is uniquely defined by $im(M)$ and $\ker(M)=(im(M^T))^{\perp}$, that is, by $im(M), im(M^T)$.

Notice that $Mv=\lim_k (\lambda^{-1}A)^kv=v,u^TM=\lim_k u^T(\lambda^{-1}A)^k=u^T$. Then the rank $1$ projector $M$ is uniquely defined by $im(M)=span(v),im(M^T)=span(u)$.

Now $R=vu^T$ is also a projector with image $span(v)$ and $im(R^T)=span(u)$. Then $M=vu^T$.

Answer 2

You may dispense with left eigenvectors when proving the result, but calling for the one of them is a comfortable path to the explicit expression for $$M=\lim_{k\to\infty}J^k$$ where $\,J=\lambda^{-1}\!A\,\in M_n\big(\mathbb R^{>0}\big)\,$ for notational convenience.
All the good properties from $A$, due to Perron & Frobenius, carry over to $J$: Its leading eigenvalue is $1$, the eigenspace is spanned by one positive vector $v$, and all its other eigenvalues live in the interior of the complex unit disc.
This is the reason why $M\,$ has an $n\!-\!1$-dimensional kernel, as worked out in M. Boyle's notes, cf the question-embedded screen-shot, and only the fixed points of $J$ are not annihilated in the limit, i. e., the $1$-dimensional eigenspace to the eigenvalue $1$ which equals $\operatorname{span}(v)$. Hence $M\,$ has rank $1$, and it is uniquely determined because $M$ is known on all of $\,\mathbb R^n$.

Notice that left eigenvectors of a matrix are in 1-to-1 correspondence to the right eigenvectors of its transpose. The transpose $J^T$ satisfies the assumptions for Perron & Frobenius as well as $J$, whence there exists a positive left eigenvector $u$ of $J$ to the eigenvalue $1$. Analogous to $v$, it survives in the limit.

Since $M\,$ has rank $1$, let's make the ansatz $M=v\,\langle ?|\cdot\rangle\,$. Then $M^T\!=\, ?\,\langle v|\cdot\rangle\,$, and $$Mv =v\,\langle ?|v\rangle = v\quad\text{and}\quad M^T\!u=\, ?\,\langle v|u\rangle = u$$ yield $M=v\,\langle u|\cdot\rangle \equiv vu\,$ with $\,uv\equiv\langle u|v\rangle =1$.

Note that $M$ is idempotent. It is selfadjoint (hence an orthogonal projector) if and only if $v$ and $u$ are scalar multiples of each other (thus linear dependent).

Furthermore, one has $v=\lim\limits_{k\to\infty}\,(J^kw)\,$ where $w>0\,$ is an arbitrary positive vector.

PS & IMHO
$(1)\:$ If you seek to master this result and its proof, you shouldn't consider projectors, kernels, images as "additional machinery" but integrate them into your picture.
$(2)\:$ In this same spirit I'd like to add that loup blanc's answer follows M. Boyle's path more than it's different to it.