A question about one the properties of singular values of $A\in\mathbb{R}^{m\times n}$

Question

Let $A\in\mathbb{R}^{m\times n}$. Recall that $\sigma_i(A)$ denotes the $i$th singular values of $A$. Then for $k=1:\min\{m,n\}$, show that $$\sigma_k(A)=\displaystyle \max_{\substack{\dim(\mathcal{S})=k\\ \dim(\mathcal{T})=k}}\,\,\min_{\substack{0\neq x\in\mathcal{S}\\0\neq y\in\mathcal{T}}}\frac{y^TAx}{\|x\|_2\|y\|_2}=\max_{\substack{\dim(\mathcal{S})=k}}\,\,\min_{\substack{0\neq x\in\mathcal{S}}}\frac{\|Ax\|_2}{\|x\|_2}.$$ Note: based on Courant-Fischer theorem ([see https://en.wikipedia.org/wiki/Min-max_theorem), it is easy to see that $$\sigma_k(A)=\max_{\substack{\dim(\mathcal{S})=k}}\,\,\min_{\substack{0\neq x\in\mathcal{S}}}\frac{\|Ax\|_2}{\|x\|_2},$$ but I can't prove the middle equality, i.e., I can't prove $$\sigma_k(A)=\displaystyle \max_{\substack{\dim(\mathcal{S})=k\\ \dim(\mathcal{T})=k}}\,\,\min_{\substack{0\neq x\in\mathcal{S}\\0\neq y\in\mathcal{T}}}\frac{y^TAx}{\|x\|_2\|y\|_2}\quad\text{or}\quad \displaystyle \max_{\substack{\dim(\mathcal{S})=k\\ \dim(\mathcal{T})=k}}\,\,\min_{\substack{x\in\mathcal{S}\\y\in\mathcal{T}}}\frac{y^TAx}{\|x\|_2\|y\|_2}=\max_{\substack{\dim(\mathcal{S})=k}}\,\,\min_{\substack{0\neq x\in\mathcal{S}}}\frac{\|Ax\|_2}{\|x\|_2}.$$ Can somebody help me with this problem?