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Let $A=(a_{ij})$ be a non-negative square matrix of dimension $N$ and $n$ a natural number so that $a_{ij}^{(n)}>0$ for all $i,j \in \{1,...,N \}$. Here $a_{ij}^{(n)}$ denotes an entry of $A^n$.

Then the Perron-Frobenius theorem states, that $A$ has a positive eigenvalue $r > 0$ with geometric and algebraic multiplicity equal to $1$ and whose absolut value is strictly bigger than the absolute value of any other eigenvalue of $A$. Further more, there are two vectors $v$ and $w$ with strictly positive entries such that $v^{T}A = rv^{T}$ and $Aw=rw$. So $A$ has strictly positive left and right eigenvectors belonging to $r$.

Now I want to show that

$$ \lim_{n \rightarrow \infty} A^n/r^n = wv^{T}/v^{T}w. $$

First I want to looks at the case where $A$ is diagonalizable. I chose a matrix $U$ consisting of a basis of eigenvectors and replaced $A$ by $UDU^{-1}$, where $D$ is the diagonal matrix with all the eigenvalues on $A$ on the diagonal. W.l.o.g. I assumed that the first column of $U$ is the vector $w$. If I then use the fact that the absolute value of $r$ is bigger than the absolute value of any other eigenvalue, I get the following result:

$$ \lim_{n \rightarrow \infty} A^n/r^n = wv^{T} .$$

What am I doing wrong? Why is the $v^{T}w$ missing? Thanks for any advice!

Joker123
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2 Answers2

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It is not easy to spot mistakes in your derivation of the limit because it is not very explicit.

If $D$ is the diagonalised form of $A$ with the largest eigenvalue $r$ in the upper left corner then $D^n/r^n$ converges to a matrix with zeroes everywhere except 1 in the upper left corner.

Furthermore, if $A$ is diagonal(isable) then $v$ and $w$ are multiples of each other, so dividing by $v^Tw$ is just normalisation. Perhaps in your derivation of the limit you assumed that $v$ and $w$ were unit vectors? (in which case $wv^T$ would be a matrix with zeroes everywhere and $\pm1$ in the upper left corner)

Justpassingby
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  • hey. thanks for the answer. I also obtained the matrix which only has $0$ in the upper left corner. Lets call this matrix $B$. I however did NOT assume that $v$ and $w$ are unit vectors, but I still obtain the result without $w*v^{T}$. Did you obtain a different result?

    And why are $v$ and $w$ multiples of each other when $A$ is diagonalisable? Thanks!

     – Joker123 Dec 27 '15 at 14:06
  • Then you need to show your work or we are stuck. Are you sure that your U was not assumed to be a rotation? – Justpassingby Dec 27 '15 at 14:08
  • $v$ and $w$ are eigenvectors for the same eigenvalue that has multiplicity 1. – Justpassingby Dec 27 '15 at 14:10
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Let $B=A/r.\,$ Applying this matrix as many times as we like leaves the eigenvector $w$ unchanged, i.e. $$\eqalign{ B^n w &= w \cr }$$ but drives any other eigenvector to zero $$\eqalign{ \lim_{n\rightarrow\,\infty } B^n x &= 0 \cr }$$ Similarly for the left eigenvectors, only $v^T$ is spared $$\eqalign{ v^TB^n &= v^T \cr \lim_{n\rightarrow\,\infty } y^TB^n &= 0^T \cr }$$ So the limit of $B^n$ must be a rank-1 matrix (only 1 vector is not in the nullspace), which satisfies the 2 non-zero equations. The only possibility is $\,\,wv^T/v^Tw$

john316
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