Prove the norm-preserving matrix extension theorem $\min\Vert\binom{A~C}{B~W}\Vert_2=\max\{\Vert\binom{A}{B}\Vert_2,\Vert(A~~C)\Vert_2\}$ by symmetry? I want to prove the following norm-preserving matrix extension theorem:

Given matrices $A\in\mathbb C^{k\times k},B\in\mathbb C^{(n-k)\times k},C\in\mathbb C^{k\times(n-k)}$, then $$\min\bigg\Vert\begin{bmatrix}A&C\\B&W\end{bmatrix}\bigg\Vert_2=\max\bigg\{\bigg\Vert\begin{bmatrix}A\\B\end{bmatrix}\bigg\Vert_2,\Vert\begin{bmatrix}A&C\end{bmatrix}\Vert_2\bigg\}.$$
Here $\min$ means taking the minimum from all possible matrices $W$
of size $\mathbb C^{(−)×(−)}$.

I know there's a more "symmetry" version of this theorem, that is, given Hermitian matrix $A\in\mathbb C^{k\times k}$ and $B\in\mathbb C^{(n-k)\times k}$, then $\min_{W^H=W}\bigg\Vert\begin{bmatrix}A&B^H\\B&W\end{bmatrix}\bigg\Vert_2=\bigg\Vert\begin{bmatrix}A\\B\end{bmatrix}\bigg\Vert_2$. However the proof of this "symmetry" version is very lengthy and requires a lot computation. It's obvious that the original  theorem is stronger than the "symmetry" version. My question is, can we only use the "symmetry" version to obtain the original strong version? I tried to extend the original theorem by complementing its symmetry (e.g. $A$ to $\begin{bmatrix}&A\\A^H&\end{bmatrix}$) and failed, but I think this method might be working. Are there any ways of strengthening the "symmetry" version to the original version? Thanks!
 A: Consider the matrix $\begin{bmatrix}&A&&C\\A^H&&B^H&\\&B&X&W\\C^H&&W^H&Y\end{bmatrix}$ where $X^H=X,Y^H=Y$. From the construction of $W$ in the symmetry version of this theorem, we know that there exists $\lambda$ such that when
$$\begin{bmatrix}X&W\\W^H&Y\end{bmatrix}=-\begin{bmatrix}&B\\C^H&\end{bmatrix}\bigg(\lambda I-\begin{bmatrix}&A\\A^H&\end{bmatrix}^2\bigg)^{-1}\begin{bmatrix}&A^H\\A&\end{bmatrix}\begin{bmatrix}&B^H\\C&\end{bmatrix}$$
the $\bigg\Vert\begin{bmatrix}&A&&C\\A^H&&B^H&\\&B&X&W\\C^H&&W^H&Y\end{bmatrix}\bigg\Vert_2$ can take an minimum, in which we have $X=0,Y=0$. Therefore by the above theorem we have:
$$
\begin{align*}
\min_{W}\bigg\Vert\begin{bmatrix}&A&&C\\A^H&&B^H&\\&B&&W\\C^H&&W^H&\end{bmatrix}\bigg\Vert_2&=\min_{W,X^H=X,Y^H=Y}\bigg\Vert\begin{bmatrix}&A&&C\\A^H&&B^H&\\&B&X&W\\C^H&&W^H&Y\end{bmatrix}\bigg\Vert_2\\&=\bigg\Vert\begin{bmatrix}&A\\A^H&\\&B\\C^H&\end{bmatrix}\bigg\Vert_2=\rho\bigg(\begin{bmatrix}AA^H+CC^H&\\&A^HA+B^HB\end{bmatrix}\bigg)\\&=\max\bigg\{\bigg\Vert\begin{bmatrix}A\\B\end{bmatrix}\bigg\Vert_2,\Vert[A~~C]\Vert_2\bigg\}
\end{align*}
$$
On the other hand, since $$
\begin{align*}
\bigg\Vert\begin{bmatrix}&A&&C\\A^H&&B^H&\\&B&&W\\C^H&&W^H&\end{bmatrix}\bigg\Vert_2&=\rho\bigg(\begin{bmatrix}AA^H+CC^H&&AB^H+CW^H&\\&A^HA+B^HB&&A^HC+B^HW\\BA^H+WC^H&&BB^H+WW^H&\\&C^HA+W^HB&&C^HC+W^HW\end{bmatrix}\bigg)\\&=\max(\bigg\Vert\begin{bmatrix}A&C\\B&W\end{bmatrix}\bigg\Vert_2,\bigg\Vert\begin{bmatrix}A^H&B^H\\C^H&W^H\end{bmatrix}\bigg\Vert_2)=\bigg\Vert\begin{bmatrix}A&C\\B&W\end{bmatrix}\bigg\Vert_2
\end{align*}
$$
Therefore we have $\min_{W}\bigg\Vert\begin{bmatrix}A&C\\B&W\end{bmatrix}\bigg\Vert_2=\max\bigg\{\bigg\Vert\begin{bmatrix}A\\B\end{bmatrix}\bigg\Vert_2,\Vert[A~~C]\Vert_2\bigg\}$
