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Power sequence
The spectral radius is closely related to the behavior of the convergence of the power sequence of a matrix; namely as shown by the following theorem.
Theorem. Let $A ∈ ℂ^{n×n}$ with spectral radius $ρ(A)$. Then $ρ(A) < 1$ if and only if\[\lim_{k \to \infty} A^k = 0.\]
On the other hand, if $ρ(A) > 1$, $\lim_{k \to \infty} \|A^k\| = \infty$. The statement holds for any choice of matrix norm on $ℂ^{n×n}$.
Proof
Assume that \(A^k\) goes to zero as \(k\) goes to infinity. We will show that $ρ(A) < 1$. Let $(\mathbf v,λ)$ be an eigenvector-eigenvalue pair for $A$. Since $A^k 𝐯=λ^k 𝐯$, we have
\begin{align*}
0 &= \left(\lim_{k \to \infty} A^k \right) \mathbf{v} \\
&= \lim_{k \to \infty} \left(A^k\mathbf{v} \right ) \\
&= \lim_{k \to \infty} \lambda^k\mathbf{v} \\
&= \mathbf{v} \lim_{k \to \infty} \lambda^k
\end{align*}Since $𝐯 ≠ 0$ by hypothesis, we must have
\[\lim_{k \to \infty}\lambda^k = 0,\]which implies $|λ| < 1$. Since this must be true for any eigenvalue λ, we can conclude that $ρ(A) < 1$.
Now, assume the radius of $A$ is less than $1$. From the Jordan normal form theorem, we know that for all $A∈ ℂ^{n×n}$, there exist $V,J ∈ ℂ^{n×n}$ with $V$ non-singular and $J$ block diagonal such that:
\[A = VJV^{-1}\]with
\[J=\begin{bmatrix}
J_{m_1}(\lambda_1) & 0 & 0 & \cdots & 0 \\
0 & J_{m_2}(\lambda_2) & 0 & \cdots & 0 \\
\vdots & \cdots & \ddots & \cdots & \vdots \\
0 & \cdots & 0 & J_{m_{s-1}}(\lambda_{s-1}) & 0 \\
0 & \cdots & \cdots & 0 & J_{m_s}(\lambda_s)
\end{bmatrix}\]where
\[J_{m_i}(\lambda_i)=\begin{bmatrix}
\lambda_i & 1 & 0 & \cdots & 0 \\
0 & \lambda_i & 1 & \cdots & 0 \\
\vdots & \vdots & \ddots & \ddots & \vdots \\
0 & 0 & \cdots & \lambda_i & 1 \\
0 & 0 & \cdots & 0 & \lambda_i
\end{bmatrix}\in \mathbf{C}^{m_i \times m_i}, 1\leq i\leq s.\]It is easy to see that
\[A^k=VJ^kV^{-1}\]and, since $J$ is block-diagonal,
\[J^k=\begin{bmatrix}
J_{m_1}^k(\lambda_1) & 0 & 0 & \cdots & 0 \\
0 & J_{m_2}^k(\lambda_2) & 0 & \cdots & 0 \\
\vdots & \cdots & \ddots & \cdots & \vdots \\
0 & \cdots & 0 & J_{m_{s-1}}^k(\lambda_{s-1}) & 0 \\
0 & \cdots & \cdots & 0 & J_{m_s}^k(\lambda_s)
\end{bmatrix}\]Now, a standard result on the $k$-power of an \(m_i \times m_i\) Jordan block states that, for \(k \geq m_i-1\):
\[J_{m_i}^k(\lambda_i)=\begin{bmatrix}
\lambda_i^k & {k \choose 1}\lambda_i^{k-1} & {k \choose 2}\lambda_i^{k-2} & \cdots & {k \choose m_i-1}\lambda_i^{k-m_i+1} \\
0 & \lambda_i^k & {k \choose 1}\lambda_i^{k-1} & \cdots & {k \choose m_i-2}\lambda_i^{k-m_i+2} \\
\vdots & \vdots & \ddots & \ddots & \vdots \\
0 & 0 & \cdots & \lambda_i^k & {k \choose 1}\lambda_i^{k-1} \\
0 & 0 & \cdots & 0 & \lambda_i^k
\end{bmatrix}\]Thus, if \(\rho(A) < 1\) then for all $i$, \(|\lambda_i| < 1\). Hence for all $i$ we have:
\[\lim_{k \to \infty}J_{m_i}^k=0\]which implies
\[\lim_{k \to \infty} J^k = 0.\]Therefore,
\[\lim_{k \to \infty}A^k=\lim_{k \to \infty}VJ^kV^{-1}=V \left (\lim_{k \to \infty}J^k \right )V^{-1}=0\]On the other side, if \(\rho(A)>1\), there is at least one element in $J$ that does not remain bounded as $k$ increases, thereby proving the second part of the statement. |
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