ChapterZero

Update: it turns out this isn’t quite right. Almost, but not quite. I need to fix this.
It turns out that the quantity I mentioned in the last post:

is indeed a norm on the set of symmetric matrices (here, is the largest (absolute) entry in ). Furthermore, it is the trace dual of the norm I looked at earlier. This interesting fact is about all I’ve been able to show for my time in looking at this problem, so I provide my proof here.

First, to see that the quantity is well-defined, note that you can diagonalize any symmetric matrix, and write it as a sum of positive and negative definite matrices, for each of which such a factorization exists by Cholesky decomposition. Construct and appropriately from these factorizations, and we see that the set being minimized over is nonempty for any symmetric

The only other nontrivial property is to see that the triangle inequality holds: let and be minimal factorizations, and note that

is a permissible factorization for This implies that

Now we upper bound the trace-dual norm of . By definition,

Since such a has a decomposition where and — although not all such decompositions are minimal for the matrices they describe—, we have

Note that . Since this last quantity is actually . The and norms are dual, so

Recall that when is symmetric, and we have

But note that if is the vector at which is achieved, then is its own minimal decomposition, so

Therefore, and are trace dual norms.

Possibly relevant posts:

• Equivalence of the norm and the norm (1/27/2010)
• A failed attempt to use an SDP to find a nonorthogonal factorization (1/17/2010)
• An observation on the norming functionals of the norm ball (2/24/2010)