I am currently reading the paper by Hong, Yoo Pyo, and C-T. Pan. "Rank-revealing factorizations and the singular value decomposition." Mathematics of Computation 58.197 (1992): 213-232. and have a question concerning on argument in the proof of Theorem 2.3. More precisely, the authors state
$ diag(\sigma_{r+1}^2,\dots, \sigma_n^2) - W_{22}^T R_{22}^T R_{22} W_{22} \succeq 0 $ can be written as $\sigma_{r+1}^2 W_{22}^{-T}W_{22}^{-1} - R_{22}^T R_{22} \succeq 0$.
Here, $\sigma_{r+1}^2 \geq \dots \geq \sigma_n^2 \geq 0$ and $W_{22}$ is a real invertible $(n-r)\times (n-r)$ matrix and $R_{22}$ is a real $(n-r)\times (n-r)$ matrix. We use $\succeq$ to indicate positive semi-definiteness, i.e., $A \succeq B$ is $A-B$ is positive semi-definite for matrices $A,B$.
I don't really see how this follows directly and I think I am missing something quite obvious. Can someone please point out what I am missing? Thank you.
Like user1511 said, this is an implication. I don't think they meant to write "can be written as".
Start with $$ {\rm diag}(\sigma_{r+1}^2,\dots, \sigma_n^2) - W_{22}^T R_{22}^T R_{22} W_{22} \succeq 0, $$
Then multiply on the left by $W_{22}^{-T}$ and on the right by $W_{22}^{-1}$. You get
$$ W_{22}^{-T} {\rm diag}(\sigma_{r+1}^2,\dots, \sigma_n^2) W_{22}^{-1} - R_{22}^T R_{22} \succeq 0. $$
Then, use that $\sigma_{r+1}^2 \geq \dots \geq \sigma_n^2 \geq 0$, which implies
$$ W_{22}^{-T} {\rm diag}(\sigma_{r+1}^2,\dots, \sigma_{r+1}^2) W_{22}^{-1} - R_{22}^T R_{22} \succeq W_{22}^{-T} {\rm diag}(\sigma_{r+1}^2,\dots, \sigma_n^2) W_{22}^{-1} - R_{22}^T R_{22} \succeq 0. $$
Finally, you can rewrite $W_{22}^{-T} {\rm diag}(\sigma_{r+1}^2,\dots, \sigma_{r+1}^2) W_{22}^{-1} - R_{22}^T R_{22} = W_{22}^{-T} (\sigma_{r+1}^2 I )W_{22}^{-1} - R_{22}^T R_{22} = \sigma_{r+1}^2 W_{22}^{-T}W_{22}^{-1} - R_{22}^T R_{22}$.