Suppose I have two quadratic expressions
$$(a – b)^2$$ and $$(a-c)^2$$
where $a, b, c$ are real numbers, then I think the following holds:
$$(a – b)^2 < (a – c)^2$$
iff $|a – b| < |a – c|$.
Now suppose I have two expressions in matrix form:
$$((A – B)v)^T((A-B)v)$$
$$((A – C)v)^T((A-C)v)$$
where $A, B, C$ are $m\times n$ real matrices, and $v$ is a $n\times 1$ vector. In this case,
when does $$((A – B)v)^T((A-B)v) < ((A – C)v)^T((A-C)v)$$
hold? I was thinking that the above will hold if
$$\|A – B\| < \|A – C\|$$
where $||\cdot||$ is the Frobenius norm. However, I am not sure if this is correct. Is it correct to approach this by measuring the "similarity" between $A$ and $B$ using a matrix norm? If so, is the Frobenius norm the correct choice?
Best Answer
The Frobenius norm of a matrix is the sum of all squared elements from the matrix. To obtain the Frobenius norm you choose $v$ to be a vector of ones.
But I think you want to allow $v$ be any real valued vector. If $A-B$ and $A-C$ are positive definite then you can create following norm $||M||_v=v^TM^TMv=(Mv)^TMv$ where $M$ is some $m\times n$ matrix and your idea holds.
EDIT: I am not allowed to comment yet, so I edit. Anyways, insert A-B and A-C where M is in the suggested definition of norm. Instead of M use $||A-B||_v<||A-C||_v$. Remember A-B and A-C need to be positive definite too.
https://en.wikipedia.org/wiki/Definite_symmetric_matrix
If they are not positive definite or if you cannot know if they are positive definite, then we need another norm.