Editing Matrix norm (section)

===Square matrices===
Suppose <math>\|\cdot\|_{\alpha, \alpha}</math> is an operator norm on the space of square matrices <math>K^{n \times n}</math>
induced by vector norms <math>\|\cdot\|_{\alpha}</math> and <math>\|\cdot\|_\alpha</math>.
Then, the operator norm is a sub-multiplicative matrix norm: 
<math display="block">\|AB\|_{\alpha, \alpha} \leq \|A\|_{\alpha, \alpha} \|B\|_{\alpha, \alpha}.</math>

Moreover, any such norm satisfies the inequality
{{NumBlk||<math display="block">(\|A^r\|_{\alpha, \alpha})^{1/r} \ge \rho(A) </math>  | {{EquationRef|1}}}}
for all positive integers ''r'', where {{math|''ρ''(''A'')}} is the [[spectral radius]] of {{mvar|A}}. For [[Symmetric matrix|symmetric]] or [[Hermitian matrix|hermitian]] {{mvar|A}}, we have equality in ({{EquationNote|1}}) for the 2-norm, since in this case the 2-norm ''is'' precisely the spectral radius of {{mvar|A}}. For an arbitrary matrix, we may not have equality for any norm; a counterexample would be
<math display="block">A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix},</math>
which has vanishing spectral radius. In any case, for any matrix norm, we have the [[Spectral radius#Gelfand's formula|spectral radius formula]]:
<math display="block">\lim_{r\to\infty}\|A^r\|^{1/r}=\rho(A). </math>