Definition of the $M$ norm

Question

Let $x \in \mathbb{R}^n$, then $||x||_2^2 = x^Tx$ when using the L2-norm.

However, what does it mean to have a M-norm:

$||x||_M^2$ for a given matrix $M \in R^{n \times n}$, eg $M = I$ (identity matrix)?

How should I see $||x||_M$ in general?

Answer 1

I assume the full definition is $$\| x \|^2_M = x^t M x $$ for $x\in \mathbb R^n$, and $M \in \mathrm M_(\mathbb R^n)$ with some additional conditions. Suppose for simplicity that $n=2$. Then $Mx$ is the vector in $\mathbb R^2$ to which $x$ is sent by the linear transformation represented by $M$ in the canonical basis. Multiplying it on the left by $x^t$ means projecting this new vector onto the one-dimensional subspace spanned by $x$. For example, if $$M = \begin{bmatrix} 3 & 1 \\ 1&3\end{bmatrix} $$ then $$\|x\|^2_M = \begin{bmatrix} x_1&x_2\end{bmatrix} \begin{bmatrix} 3 & 1 \\ 1&3\end{bmatrix} \begin{bmatrix} x_1\\x_2\end{bmatrix} = \begin{bmatrix} x_1&x_2\end{bmatrix} \begin{bmatrix} 3x_1 +x_2 \\ x_1+3x_2\end{bmatrix} = 3x_1^2 + 2x_1x_2 + 3x_2^2.$$ Trying $(x_1,x_2)=(0,1)=e_2$ gives a norm squared of $3$, which is exactly the length of the projection of $Me_2=(1,3)$ onto the vertical axis.

Answer 2

From Trefethen the norm you are referring is defined as the following.

$$ \| x \|_{M} = \left( \sum_{i=1}^{m} |m_{i}x_{i} |^{2} \right)^{\frac{1}{2}}$$

It presupposes the $M$ is a diagonal matrix with $m_{i}$ being the weighting for the norm. Illustration follows. Standard two norm looks like this.

Whereas the weighted 2 norm stretches it