Editing Spectral radius (section)

==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 {{math|''A'' ∈ '''C'''<sup>''n''×''n''</sup>}} with spectral radius {{math|''ρ''(''A'')}}. Then {{math|''ρ''(''A'') < 1}} if and only if
:<math>\lim_{k \to \infty} A^k = 0.</math>
On the other hand, if {{math|''ρ''(''A'') > 1}}, <math>\lim_{k \to \infty} \|A^k\| = \infty</math>. The statement holds for any choice of matrix norm on {{math|'''C'''<sup>''n''×''n''</sup>}}.

'''Proof'''

Assume that <math>A^k</math> goes to zero as <math>k</math> goes to infinity. We will show that {{math|''ρ''(''A'') < 1}}. Let {{math|('''v''', ''λ'')}} be an [[eigenvector]]-[[eigenvalue]] pair for ''A''. Since {{math|''A<sup>k</sup>'''''v''' {{=}} ''λ<sup>k</sup>'''''v'''}}, we have

:<math>\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}</math>

Since {{math|'''v''' ≠ 0}} by hypothesis, we must have

:<math>\lim_{k \to \infty}\lambda^k = 0,</math>

which implies <math>|\lambda| < 1</math>. Since this must be true for any eigenvalue <math>\lambda</math>, we can conclude that {{math|''ρ''(''A'') < 1}}.

Now, assume the radius of {{mvar|A}} is less than {{math|1}}. From the [[Jordan normal form]] theorem, we know that for all {{math|''A'' ∈ '''C'''<sup>''n''×''n''</sup>}}, there exist {{math|''V'', ''J'' ∈ '''C'''<sup>''n''×''n''</sup>}} with {{mvar|V}} non-singular and {{mvar|J}} block diagonal such that:

:<math>A = VJV^{-1}</math>

with

:<math>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}</math>

where

:<math>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.</math>

It is easy to see that

:<math>A^k=VJ^kV^{-1}</math>

and, since {{mvar|J}} is block-diagonal,

:<math>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}</math>

Now, a standard result on the {{mvar|k}}-power of an <math>m_i \times m_i</math> Jordan block states that, for <math>k \geq m_i-1</math>:

:<math>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}</math>

Thus, if <math>\rho(A) < 1</math> then for all {{mvar|i}} <math>|\lambda_i| < 1</math>. Hence for all {{mvar|i}} we have:

:<math>\lim_{k \to \infty}J_{m_i}^k=0</math>

which implies

:<math>\lim_{k \to \infty} J^k = 0.</math>

Therefore,

:<math>\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</math>

On the other side, if <math>\rho(A)>1</math>, there is at least one element in {{mvar|J}} that does not remain bounded as {{mvar|k}} increases, thereby proving the second part of the statement.