# 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!

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Yes, this is fairly standard notation with 'variable metric' methods of optimization. The expression above is the definition, except that $D$ is positive definite (otherwise it doesn't define a norm). – copper.hat May 25 '12 at 5:37

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.

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Ok. I understand that the superscript 2 means "squared." I can also see why this expressions is natural. My only question was about the subscript $D$. I think your answer makes sense i.e the author defined the $||x||_D$ as $x^T D x$. I was just wondering if this operation had a particular name. Thanks! – Damian May 25 '12 at 4:35
@Damian OK. I'm not aware of any particular name for this. Maybe someone else knows though. – user12014 May 25 '12 at 4:46
One might extend this notation to any positive definite symmetric matrix in place of $\mathbf D$, which (for a finite dimensional real vector space) will give you an arbitrary inner product, and norm defined by such an inner product. Although it is not mentioned, I suppose $\mathbf D$ has only positive diagonal entries (so is positive definite), otherwhise it does not define an inner product or norm. – Marc van Leeuwen May 25 '12 at 4:47
Yes, you are right. The paper does mention that the elements of $\mathbf{D}$ are positive. So you are saying that $\mathbf{D}$ does not have to be diagonal, right? – Damian May 25 '12 at 4:49