(Variational description of eigenvalues) Let $\boldsymbol{A}$ be a symmetric $n \times n$ matrix with eigenvalues $\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_n$. Let $\boldsymbol{S}:=\left(s_1, s_2, \ldots, s_n\right)$ be an orthogonal $n \times n$ matrix that diagonalizes $\boldsymbol{A}$, so that $\boldsymbol{S}^{\prime} \boldsymbol{A} \boldsymbol{S}=\operatorname{diag}\left(\lambda_1, \lambda_2, \ldots, \lambda_n\right)$. Show that
$$
\lambda_k=\max _{\boldsymbol{R}_{k-1}^{\prime} \boldsymbol{x}=\mathbf{0}} \frac{\boldsymbol{x}^{\prime} \boldsymbol{A} \boldsymbol{x}}{\boldsymbol{x}^{\prime} \boldsymbol{x}}=\min _{\boldsymbol{T}_{k+1}^{\prime} \boldsymbol{x}=0} \frac{\boldsymbol{x}^{\prime} \boldsymbol{A} \boldsymbol{x}}{\boldsymbol{x}^{\prime} \boldsymbol{x}} \quad(k=1, \ldots, n),
$$
where
$$
\boldsymbol{R}_k:=\left(\boldsymbol{s}_1, \boldsymbol{s}_2, \ldots, \boldsymbol{s}_k\right) \text { and } \boldsymbol{T}_k:=\left(\boldsymbol{s}_k, \boldsymbol{s}_{k+1}, \ldots, \boldsymbol{s}_n\right),
$$
and we agree to interpret $\boldsymbol{R}_0$ and $\boldsymbol{T}_{n+1}$ as "empty" in the sense that the restrictions $\boldsymbol{R}_0^{\prime} \boldsymbol{x}=$ $\mathbf{0}$ and $\boldsymbol{T}_{n+1}^{\prime} \boldsymbol{x}=\mathbf{0}$ do not impose a restriction on $\boldsymbol{x}$.