# Double vertical bars [duplicate]

I've been reading the Wikipedia article on CSP (https://en.wikipedia.org/wiki/Common_spatial_pattern) and one of the first formulas there has double vertical bars in it. $${{w}}={\arg \max }_{{\mathbf {w}}}{\frac {\left\|{\mathbf {wX}}_{1}\right\|^{2}}{\left\|{\mathbf {wX}}_{2}\right\|^{2}}}$$ The text right before the formula says both the enumerator and denominator are variances of two multivariate signals. What do double vertical bars here mean? I know that double bars should mean the norm of a vector. In this case we have matrices (X1 and X2), so how can that be the norm of a matrix? Please help.

## marked as duplicate by John, Daniel W. Farlow, Claude Leibovici, Parcly Taxel, WatsonSep 8 '16 at 8:48

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• I would guess that ${\bf w}$ is a row vector, so $\|{\bf w X_i}\|$ are norms of row vectors. – Robert Israel Sep 7 '16 at 18:55
• $\|~\|$ is used to denote an arbitrary norm of a normed space. Usually, we do not bother specifying which norm it is because it should be clear from context, however if it is ambiguous we will often write a subscript to specify which norm specifically we are using. For example the $\ell_p$ norm $\|f\|_p$ as compared to the euclidean norm $\|v\|_2$, etc... Regardless, $w$ here is a vector, and a vector times a matrix is again a vector, so you can use your usual notion of vector norms (specifically euclidean norms) – JMoravitz Sep 7 '16 at 18:57

## 1 Answer

The norm of a matrix $A$ can be defined similarly to the norm of a vector. In your case, this $\Vert A\Vert$ means $\left(\sum_{i,j} a_{i,j}^2\right)^{1/2}$, known as the Frobenius norm. However, since $\vec w$ is likely a row-vector, the product $\vec wX$ is itself a row-vector. Thus the norm $\Vert\cdot\Vert$ is just the ordinary norm of a vector.

• thank you! That finally makes sense. – MegaNightdude Sep 7 '16 at 19:02