Editing Spectral radius (section)

===Matrices===
Let {{math|''λ''<sub>1</sub>, ..., ''λ<sub>n</sub>''}} be the eigenvalues of a matrix {{math|''A'' ∈ '''C'''<sup>''n''×''n''</sup>}}. The spectral radius of {{math|''A''}} is defined as

:<math>\rho(A) = \max \left \{ |\lambda_1|, \dotsc, |\lambda_n| \right \}.</math>

The spectral radius can be thought of as an infimum of all norms of a matrix. Indeed, on the one hand, <math> \rho(A) \leqslant \|A\| </math> for every [[matrix norm#Matrix norms induced by vector norms|natural matrix norm]] <math>\|\cdot\|</math>; and on the other hand, Gelfand's formula states that <math> \rho(A) = \lim_{k\to\infty} \|A^k\|^{1/k} </math>. Both of these results are shown below.

However, the spectral radius does not necessarily satisfy <math> \|A\mathbf{v}\| \leqslant \rho(A) \|\mathbf{v}\| </math> for arbitrary vectors <math> \mathbf{v} \in \mathbb{C}^n </math>. To see why, let <math>r > 1</math> be arbitrary and consider the matrix
:<math> C_r = \begin{pmatrix} 0 & r^{-1} \\ r & 0 \end{pmatrix} </math>. 
The [[characteristic polynomial]] of <math> C_r </math> is <math> \lambda^2 - 1 </math>, so its eigenvalues are <math>\{-1, 1\}</math> and thus <math>\rho(C_r) = 1</math>. However, <math>C_r \mathbf{e}_1 = r \mathbf{e}_2</math>. As a result,
:<math> \| C_r \mathbf{e}_1 \| = r > 1 = \rho(C_r) \|\mathbf{e}_1\|. </math> 
As an illustration of Gelfand's formula, note that <math>\|C_r^k\|^{1/k} \to 1</math> as <math>k \to \infty</math>, since <math>C_r^k = I</math> if <math>k</math> is even and <math>C_r^k = C_r</math> if <math>k</math> is odd.

A special case in which <math> \|A\mathbf{v}\| \leqslant \rho(A) \|\mathbf{v}\| </math> for all <math> \mathbf{v} \in \mathbb{C}^n </math> is when <math>A</math> is a [[Hermitian matrix]] and <math> \|\cdot\| </math> is the [[Euclidean norm]]. This is because any Hermitian Matrix is [[diagonalizable matrix|diagonalizable]] by a [[unitary matrix]], and unitary matrices preserve vector length. As a result,
: <math> \|A\mathbf{v}\| = \|U^*DU\mathbf{v}\| = \|DU\mathbf{v}\| \leqslant \rho(A) \|U\mathbf{v}\| = \rho(A) \|\mathbf{v}\| .</math>