Linear algebra norm notation I was reading a paper where the authors used the following notation:
$$||b - \mathbf{A}x||^2_D = (b - \mathbf{A}x)^t \mathbf{D} (b - \mathbf{A}x)$$
where $\mathbf{D}$ is a diagonal matrix
I was curious about the subscript $D$ when taking the norm-$2$. Does this notation represent something special or is it just the author's way of expressing the quantity in the right hand side? Have you guys encountered this notation before?
Thanks!
 A: I hope I understood the question correctly. I think the superscript $2$ is just supposed to mean "squared", it has nothing to do with the $2$-norm, and is just the author's way of expressing the right hand side. The reason the notation is natural is the following: given a diagonal matrix $D$ with positive entries, we can define an inner product by
$$\langle x,y\rangle_D = x^TDy$$
Now every inner product $\langle \cdot, \cdot \rangle$ induces a norm by
$$\|x\| = \sqrt{\langle x, x \rangle }$$
or, in other words,
$$\|x\|^2 = \langle x, x \rangle$$
So what the author means is that the norm $\|x\|_D$ is defined by
$$\|x\|_D = \sqrt{\langle x,y\rangle_D} = \sqrt{x^TDx}$$
or, to avoid the square root notation,
$$\|x\|_D^2 = x^TDx$$
I guess the only correlation to the $2$-norm is that both are induced by an inner product.
A: if x and y are complex, then there is a name for this type of inner product
 $<x,y>:= y^T D x$, 
which is the Hermitian product, where D is any Hermitian positive-definite matrix.
And as wikipedia explains:  For the real case this corresponds to the dot product of the results of directionally different scaling of the two vectors, with positive scale factors and orthogonal directions of scaling. 
So in your case, matrix D gives different scaling of the elements in x and y.
