# Limit of powers of $3\times3$ matrix

Consider the matrix

$$A = \begin{bmatrix} \frac{1}{2} &\frac{1}{2} & 0\\ 0& \frac{3}{4} & \frac{1}{4}\\ 0& \frac{1}{4} & \frac{3}{4} \end{bmatrix}$$

What is $$\lim_{n→\infty}A^n$$ ?

A)$$\begin{bmatrix} 0 & 0 & 0\\ 0& 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$$ B)$$\begin{bmatrix} \frac{1}{4} &\frac{1}{2} & \frac{1}{2}\\ \frac{1}{4}& \frac{1}{2} & \frac{1}{2}\\ \frac{1}{4}& \frac{1}{2} & \frac{1}{2}\end{bmatrix}$$ C)$$\begin{bmatrix} \frac{1}{2} &\frac{1}{4} & \frac{1}{4}\\ \frac{1}{2}& \frac{1}{4} & \frac{1}{4}\\ \frac{1}{2}& \frac{1}{4} & \frac{1}{4}\end{bmatrix}$$ D)$$\begin{bmatrix} 0 &\frac{1}{2} & \frac{1}{2}\\ 0 & \frac{1}{2} & \frac{1}{2}\\ 0 & \frac{1}{2} & \frac{1}{2}\end{bmatrix}$$ E) The limit exists, but it is none of the above

The given answer is D). How does one arrive at this result?

• Did you try doing for small values of $n$. Take $n=2, 3,4$ and post your observations (if any) as well. – Vizag May 24 at 11:12

By this question, we know that

$$\begin{equation} A^n= \begin{pmatrix} 2^{-n} & n\cdot 2^{-n-1} - 2^{-n-1} + \frac12 & {1-\frac{n+1}{2^n}\over2}\\ 0 & {2^{-n}+1\over2} & {1-2^{-n}\over2} \\ 0 & {1-2^{-n}\over2} & {2^{-n}+1\over2} \end{pmatrix}. \end{equation}$$

It is thus clear that $$\lim_{n\to\infty} A^n = \begin{pmatrix} 0 &\frac{1}{2} & \frac{1}{2}\\ 0 & \frac{1}{2} & \frac{1}{2}\\ 0 & \frac{1}{2} & \frac{1}{2}\end{pmatrix}$$.

• $2^{-n}$ tends to 0, when $n->\infty$.....right?? – Srestha May 25 at 19:27
• @Srestha that is correct – Maximilian Janisch May 25 at 19:40

If you are in $$1$$, you have same probability to stay there or to pass to $$2$$, but no way to get back from there. Thus you are finally drifting to $$2$$.

States $$2$$ and $$3$$ are symmetrical: at long they will tend to be equally populated, independently of the starting conditions.

Therefore also starting from $$1$$ you will at long be split between $$2$$ and $$3$$.

It’s often worth examining a matrix for obvious eigenvectors and eigenvalues, especially in artificial exercises, before plunging into computing and solving the characteristic equation. From the first column of $$A$$, we see that $$(1,0,0)^T$$ is an eigenvector with eigenvalue $$\frac12$$. The rows of $$A$$ all sum to $$1$$, so $$(1,1,1)$$ is an eigenvector with eigenvalue $$1$$. The remaining eigenvalue $$\frac12$$ can be found by examining the trace.
$$A$$ is therefore similar to a matrix of the form $$J=D+N$$, where $$D=\operatorname{diag}\left(1,\frac12,\frac12\right)$$ and $$N$$ is nilpotent of order no greater than 2. (If $$A$$ is diagonalizable, then $$N=0$$.) $$D$$ and $$N$$ commute, so expanding via the Binomial Theorem, $$(D+N)^n=D^n+nND^{n-1}$$. In the limit, $$D^n=\operatorname{diag}(1,0,0)$$ and the first column of $$N$$ is zero, so the second term vanishes. Thus, if $$A=PJP^{-1}$$, then $$\lim_{n\to\infty}A^n=P\operatorname{diag}(1,0,0)P^{-1}$$, but the right-hand side is just the projector onto the eigenspace of $$1$$. Informally, repeatedly multiplying a vector by $$A$$ leaves that vector’s component in the direction of $$(1,1,1)^T$$ fixed, while the remainder of the vector eventually dwindles away to nothing.
Since $$1$$ is a simple eigenvalue, there’s a shortcut for computing this projector that doesn’t require computing the change-of-basis matrix $$P$$: if $$\mathbf u^T$$ is a left eigenvector of $$1$$ and $$\mathbf v$$ a right eigenvector, then the projector onto the right eigenspace of $$1$$ is $${\mathbf v\mathbf u^T\over\mathbf u^T\mathbf v}.$$ (This formula is related to the fact that left and right eigenvectors with different eigenvalues are orthogonal.) We already have a right eigenvector, and a left eigenvector is easily found by inspection: the last two columns both sum to $$1$$, so $$(0,1,1)$$ is a left eigenvector of $$1$$. This gives us $$\lim_{n\to\infty}A^n = \frac12\begin{bmatrix}1\\1\\1\end{bmatrix}\begin{bmatrix}0&1&1\end{bmatrix} = \begin{bmatrix}0&\frac12&\frac12\\0&\frac12&\frac12\\0&\frac12&\frac12\end{bmatrix}.$$