Is $\lVert Ax \rVert^2 - \lVert Bx \rVert^2 = \lVert AA^T - BB^T \rVert$?

Question

For matrices $A,B\in\mathbb{R}^{m\times n}$ and for any unit vector $x$, is the following true, and if so why?

$\lVert Ax \rVert^2 - \lVert Bx \rVert^2 = \lVert AA^T - BB^T \rVert$

Equivalently, is $x$ the eigenvector of $A^TA - B^TB$ corresponding to the largest eigenvalue. Why is this the same as saying that $x$ is a unit vector??

Answer 1

$$\lVert Ax\rVert^2-\lVert Bx\rVert^2=x^T(A^TA-B^TB)x\le \lVert A^TA-B^TB\rVert $$ if $x$ is a unite vector. This is true if $A^TA-B^TB$ is positive semidefinite.

Now, since $A^TA-B^TB$ is symmetric it is diagonalizable unitarily and hence $A^TA-B^TB=U\Lambda U^T$ for some orthogonal (unitary) matrix $U$. Then, the expression is $$y^T\Lambda y$$ where $y=Ux$. Now, $$y^T\Lambda y=\sum_i y_i^2 \lambda_i\le \lambda_{max}y^Ty=\lambda_{max}$$. So, in that case if $x$ is a unit eigenvector corresponding to the largest eigenvalue of $A^TA-B^TB$ the equality holds (again assuming $A^TA-B^TB$ is PSD).