Q:

Is there a geometric meaning of the Frobenius norm?

I have a positive definite matrix $A$. I am going to choose its Frobenius norm $\|A\|_F^2$ as a cost function and then minimize $\|A\|_F^2$. But I think I need to find a reason to convince people it is reasonable to choose $\|A\|_F^2$ as a cost function. So I'm wondering if there are some geometric meanings of the Frobenius norm. Thanks.
Edit: here $A$ is a 3 by 3 matrix. In the problem I'm working on, people usually choose $\det A$ as a cost function since $\det A$ has an obvious geometric interpretation: the volume of the parallelepiped determined by $A$. Now I want to choose $\|A\|_F^2$ as a cost function because of the good properties of $\|A\|_F^2$. That's why I am interested in the geometric meaning of $\|A\|_F^2$.

A:

To give intuition, lets look at a particularly nice class of matrices.  Suppose the matrix $A$ is symmetric.  Then we can find a basis of $n$ orthonormal eigenvectors $v_1,\cdots v_n$ with eigenvalues $\lambda_1,\dots,\lambda_n$.  Each eigenvalue geometrically represents the "stretching factor" in the direction of its associated eigenvector.  
The Frobenius norm of $A$ in this case is $$\|A\|_F =\sqrt{\text{Tr}(AA^t)}=\sqrt{\text{Tr}(A^2)}=\sqrt{\lambda_1^2+\cdots+\lambda_n^2}.$$  Working in the basis $v_1,\cdots v_n$, consider the box which is the image of the unit cube under $A$.  The Frobenius norm is the diagonal of that box, and the determinant is the volume.
The usual norm defined as $\sup_{\|x\|=1}\|Ax\|$ corresponds to the longest side of the box.

