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)?
Best Answer
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 $M=vu^T$.