spectral radius vs. 2-norm [duplicate]

Solution 1:

Just to clarify the "tricky part," $\Sigma$ is a "diagonalized" matrix containing the singular values $\sigma_1$, $\sigma_2$, ... $\sigma_n$ where by the problem statement we take $\sigma_1$ to be the largest of these values $\left(\sigma_1\geq\sigma_2\geq\dots\geq\sigma_n\geq 0\right)$. The vector $V^T \bf{x}$ is then multiplied by the diagonal matrix $\Sigma$ such that \begin{align*} \left\Vert\Sigma V^T \textbf{x}\right\Vert &= \left\Vert\begin{pmatrix} \sigma_1 & 0 & \dots & 0 \\ 0 & \sigma_2 &\dots &0 \\ \vdots &\vdots & \ddots & \vdots \\ 0 &0& \dots & \sigma_n \end{pmatrix} V^T \textbf{x}\right\Vert \\ & \leq \left\Vert\begin{pmatrix} \sigma_1 & 0 & \dots & 0 \\ 0 & \sigma_1 &\dots &0 \\ \vdots &\vdots & \ddots & \vdots \\ 0 &0& \dots & \sigma_1 \end{pmatrix} V^T \textbf{x}\right\Vert \\ &= \left\Vert\sigma_1\begin{pmatrix} 1 & 0 & \dots & 0 \\ 0 & 1 &\dots &0 \\ \vdots &\vdots & \ddots & \vdots \\ 0 &0& \dots & 1 \end{pmatrix} V^T \textbf{x}\right\Vert \\ &= \vert\sigma_1\vert\left\Vert V^T\textbf{x}\right\Vert \\ &=\sigma_1 \left\Vert V^T\textbf{x}\right\Vert \end{align*}

I hope this helps you understand the proof.