Are $A^T B - I$ and $I - A B^T$ equal in Frobenius norm?

Question

Is $\text{frob}(A^T B - I) = \text{frob}(I - A B^T)$, where $\text{frob}$ is the Frobenius norm?

Can you show why?

I ask because, I am curious what is the intuitive meaning of Eq. 4, on page 3 here:https://arxiv.org/pdf/1908.07640.pdf

It seems to be just finding the $S$ matrix s.t. $S^{-1}$ is "closest" to $R$, because the Frobenius norm of the difference between two rotations is a rotation metric known as the deviation from the identity matrix: Eq 21, https://www.cs.cmu.edu/~cga/dynopt/readings/Rmetric.pdf

Answer 1

For the purposes of this question, I'll assume that 1) we're only interested in real-valued matrices, and 2) $A,B$ have the same dimensions (since otherwise $A^TB$ isn't square). As such, the Frobenius norm is defined as $\|A \|_F=\sqrt{\operatorname{tr}(A^T A)}$ where $\text{tr}$ is the trace operator. So two matrices $A,B$ are equal in Frobenius norm if $A^T A,B^T B$ have the same trace.

In particular, note that the cyclic property of trace ($\operatorname{tr}(AB)=\operatorname{tr}(BA)$) means that

$$\operatorname{tr}(A^T A)=\operatorname{tr}(AA^T)=\operatorname{tr}((A^T)^T (A^T)).$$ Hence $A^T$ has the same Frobenius norm as $A$ itself. In addition, we have $(-A)^T(-A)=A^T A$ and thus $-A$ also has the same Frobenius norm as $A$. Applying this to $A^TB-I$, we have

$$\|A^T B-I\|_F=\|-(B^T A-I)^T\|_F=\|I-B^T A^T\|_F$$ as asserted.

Answer 2

No, they aren't equal. You may use the classical example that $XY=0\not\to YX=0$ to construct a counterexample. Let $$ A=\pmatrix{0&1\\ 0&0},B=\pmatrix{1&0\\ 0&0}. $$ Then \begin{aligned} \|A^TB-I\|_F&=\left\|A^T-I\right\|_F=\sqrt{3},\\ \|I-AB^T\|_F&=\|I\|_F=\sqrt{2}. \end{aligned}