Limit of matrix powers

Question

Suppose to consider a matrix $P$ whose entries $p_{ij} \in [0,1)$.

Can we always say that

$$\lim_{k \rightarrow \infty} P^k = 0$$ ?

Thanks for your help!

Answer 1

I don't think that is true. Consider a $2 \times 2$ matrix $A$ with all the entries $\frac{1}{2}$, i.e. $$A=\begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix}$$ Then every power of that matrix is the original matrix itself, i.e. $$A^n=A \quad \forall n \in \mathbb{N}^+$$

Answer 2

If $P$ is a squared matrix and $\rho(P)$ its spectral radius, then $$\lim_{k\to \infty} P^k = 0 \qquad \iff \qquad \rho(P)<1$$ Now, note that for a nonnegative matrix $P$, by the Collatz-Wielandt principle, we have $$ \max_{\substack{x\geq 0\\ x\neq 0}}\min_{x_i\neq 0}\frac{(Px)_i}{x_i}\leq \rho(P)\leq \min_{x> 0}\max_{i}\frac{(Px)_i}{x_i},$$ where the inequalities $x\geq 0$ and $x>0$ have to be understood component-wise.

Combining these facts is useful for at least two reasons:

1) It allows you to relatively easily check if $\rho(P)<1$.
Example: If $P=\dfrac{1}{6}\begin{pmatrix}2 & 3 \\ 2 & 2 \end{pmatrix}$ and $x=\begin{pmatrix}1\\ 1 \end{pmatrix}$, then $$ \rho(P) \leq \max_{i=1,2}\frac{(Px)_i}{x_i} = \frac{2+3}{6}<1 \qquad \implies \quad \lim_{k\to \infty} P^k =0.$$

2) It allows you to easily build counter-examples for your claim.
Example: If $P=\dfrac{1}{4}\begin{pmatrix}2 & 3 \\ 3 & 3 \end{pmatrix}$ and $x=\begin{pmatrix}1\\ 1 \end{pmatrix}$, then $$ \rho(P) \geq \min_{i=1,2}\frac{(Px)_i}{x_i} = \frac{2+3}{4}>1 \qquad \implies \quad \lim_{k\to \infty} P^k \neq 0.$$

Answer 3

If the spectral radius is less than unity the sequence will converge.

Look at the largest singular value.

This post shows what happens as a matrix is raised to a sequence of powers. What is the mechanism of Eigenvector?

Consider the matrix $$ \mathbf{A} = \left[ \begin{array}{rr} 1 & -1 \\ 0 & 1 \\ \end{array} \right]. $$ which has singular values $$ \sigma = \left( \sqrt{\frac{1}{2} \left(\sqrt{5}+3\right)},\sqrt{\frac{1}{2} \left(3-\sqrt{5}\right)}\right) \approx (1.61803, 0.618034) $$ The map of a matrix onto a unit circle (left) creates an ellipse (right).

This next image overlays the column vectors from the SV decomposition matrix $\mathbf{V}$, scaled by the singular values. The matrix grows in directions where the singular value exceeds unity.