# Why does a matrix inside double bars mean variance?

This question arose from the Wikipedia page (https://en.wikipedia.org/wiki/Common_spatial_pattern). It says in the formula $${{w}}={\arg \max }_{{\mathbf {w}}}{\frac {\left\|{\mathbf {wX}}_{1}\right\|^{2}}{\left\|{\mathbf {wX}}_{2}\right\|^{2}}}$$ that we need to maximize the ratio between variances. So both the enumerator and denominator are variances. But X1 and X2 are simply matrices and, when multiplied by row vectors w become vectors. So why does this article say that the denominator and enumerator are variances? The way I see it, there should be something like E[(X-E[X])] as is explained in this Wikipedia article (https://en.wikipedia.org/wiki/Covariance_matrix#Conflicting_nomenclatures_and_notations) about variance $${ {E} \left[(\mathbf {X} -\mathrm {E} [\mathbf {X} ])(\mathbf {X} -\mathrm {E} [\mathbf {X} ])^{\rm {T}}\right]}$$.

So my problem is I don't see where this first formula above 'hides' variance. Is it some simplified representation? Please help.

## 1 Answer

I think I figured it out. If we ignore the w factors, bothe fenominator and the enumerator are variances. Wikipedia failed to mention that the signals are centered, i.e. have zero means. The (1/1-N) factors that are part of the variance formula cancel out since both signals have the same number of samples.