Editing Spectral radius (section)

==Gelfand's formula==

Gelfand's formula, named after [[Israel Gelfand]], gives the spectral radius as a limit of matrix norms.

=== Theorem ===

For any [[matrix norm]] {{math|{{!!}}⋅{{!!}},}} we have<ref>The formula holds for any [[Banach algebra]]; see Lemma IX.1.8 in {{harvnb|Dunford|Schwartz|1963}} and {{harvnb|Lax|2002|pp=195–197}}</ref>
:<math>\rho(A)=\lim_{k \to \infty} \left \|A^k \right \|^{\frac{1}{k}}</math>.

Moreover, in the case of a [[matrix norm|consistent]] matrix norm <math>\lim_{k \to \infty} \left \|A^k \right \|^{\frac{1}{k}}</math> approaches <math>\rho(A)</math> from above (indeed, in that case <math>\rho(A) \leq \left \|A^k \right \|^{\frac{1}{k}}</math> for all <math>k</math>).

===Proof===
For any {{math|''ε'' > 0}}, let us define the two following matrices:

:<math>A_{\pm}= \frac{1}{\rho(A) \pm\varepsilon}A.</math>

Thus,

:<math>\rho \left (A_{\pm} \right ) = \frac{\rho(A)}{\rho(A) \pm \varepsilon}, \qquad \rho (A_+) < 1 < \rho (A_-).</math>

We start by applying the previous theorem on limits of power sequences to {{math|''A''<sub>+</sub>}}:

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

This shows the existence of {{math|''N''<sub>+</sub> ∈ '''N'''}} such that, for all {{math|''k'' ≥ ''N''<sub>+</sub>}},
:<math>\left\|A_+^k \right \| < 1.</math>
Therefore,
:<math>\left \|A^k \right \|^{\frac{1}{k}} < \rho(A)+\varepsilon.</math>

Similarly, the theorem on power sequences implies that <math>\|A_-^k\|</math> is not bounded and that there exists {{math|''N''<sub>−</sub> ∈ '''N'''}} such that,  for all {{math|''k'' ≥ ''N''<sub>−</sub>}},
:<math>\left\|A_-^k \right \| > 1.</math>
Therefore,
:<math>\left\|A^k \right\|^{\frac{1}{k}} > \rho(A)-\varepsilon.</math>

Let {{math|''N'' {{=}} max{''N''<sub>+</sub>, ''N''<sub>−</sub>}}}. Then,
:<math>\forall \varepsilon>0\quad \exists N\in\mathbf{N} \quad \forall k\geq N \quad \rho(A)-\varepsilon < \left \|A^k \right \|^{\frac{1}{k}} < \rho(A)+\varepsilon,</math>
that is,
:<math>\lim_{k \to \infty} \left \|A^k \right \|^{\frac{1}{k}} = \rho(A).</math>
This concludes the proof.

===Corollary===
Gelfand's formula yields a bound on the spectral radius of a product of commuting matrices: if <math>A_1, \ldots, A_n</math> are matrices that all commute, then

:<math>\rho(A_1 \cdots A_n) \leq \rho(A_1) \cdots \rho(A_n).</math>

===Numerical example===
[[File:Gelfand's formula for a 3x3 matrix.svg|thumb|The convergence of all 3 matrix norms to the spectral radius.]]
Consider the matrix

:<math>A=\begin{bmatrix}
9 & -1 & 2\\
-2 & 8 & 4\\
1 & 1 & 8
\end{bmatrix}</math>

whose eigenvalues are {{math|5, 10, 10}}; by definition, {{math|''ρ''(''A'') {{=}} 10}}. In the following table, the values of <math>\|A^k\|^{\frac{1}{k}}</math> for the four most used norms are listed versus several increasing values of k (note that, due to the particular form of this matrix,<math>\|.\|_1=\|.\|_\infty</math>):