What does $\tan^{-2}\theta_i$ mean?

Question

In a paper the authors write ($A\in\mathbb{R}^{p\times p},B\in\mathbb{R}^{(n-p)\times p}$):

"Note as an aside that $\tan\theta_i, i=1,\ldots,p$, can also be expressed as the singular values of $BA^{-1}$. Hence, $\tan^{-2}\theta_i$ is the $i$th largest eigenvalue of $(A^\top A)(B^\top B)^{-1}$."

I tried to understand this and calculated ( $\sigma\hat{=}$ singular value, $\lambda\hat{=}$ eigenvalue):

$$\tan(\theta_i)=\sigma_i(BA^{-1})\\ \Leftrightarrow (\tan(\theta_i))^2=(\sigma_i(BA^{-1}))^2=\lambda_i((BA^{-1})^\top BA^{-1})=\lambda_i(A^{-\top}B^\top BA^{-1})\\ \Leftrightarrow \frac{1}{(\tan(\theta_i))^2}=\lambda_i((A^{-\top}B^\top BA^{-1})^{-1})=\lambda_i(AB^{-1}B^{-\top}A^\top)\neq\lambda_i(A^\top A(B^\top B)^{-1})$$

which is abviously different from what the authors claim. What is going wrong? Does $\tan^{-2}$ mean anything else?

Answer 1

If $A$ is invertable then the eigenvalues are the same. Let $$ AB^{-1}B^{-\top}A^\top x_i = \lambda_i x_i $$ then $$ A^\top AB^{-1}B^{-\top}A^\top x_i = \lambda_i A^\top x_i\\ A^\top AB^{-1}B^{-\top}z_i = \lambda_i z_i\\z_i \equiv A^\top x_i $$ so $$ \lambda(A^\top AB^{-1}B^{-\top}) = \lambda(AB^{-1}B^{-\top}A^\top) $$

Answer 2

The eigenvalues are equal, aren't they? After all,

$$AB^{-1}B^{-T}A^T= A^{-T}(A^TAB^{-1}B^{-T})A^T\sim A^TAB^{-1}B^{-T}=(A^TA)(B^TB)^{-1}$$