Exercise 6.4.5

For a projection matrix $\mathrm{\Pi }$, we know that ${\mathrm{\Pi }}^T=\mathrm{\Pi }$ and ${\mathrm{\Pi }}^2=\mathrm{\Pi }$. So $\mathrm{\Pi }$ is symmetric, it can be written as $\mathrm{\Pi }=\mathit{Q\Lambda }{Q}^T$ and ${\mathrm{\Pi }}^2={\mathrm{\Pi }}^T\mathrm{\Pi }=Q{\Lambda }^2{Q}^T=\mathit{Q\Lambda }{Q}^T$, so we see that the eigevalues of $\mathrm{\Pi }$ are either 1 or zero, so ${\lambda }_{\mathit{max}}\le 1$.

Now let $y=x-z$, and consider the value of $\frac{y^T\mathrm{\Pi }y}{y^{T}y}$, its maximum value is the maximum eigenvalue of $\mathrm{\Pi }$, meaning $\frac{y^T\mathrm{\Pi }y}{y^{T}y}\le 1$, take norm on both sides, we have $\parallel y^T\mathrm{\Pi }y\parallel \le \parallel y^{T}y\parallel$, that is $\parallel y^T{\mathrm{\Pi }}^T\mathrm{\Pi }y\parallel \le \parallel y^{T}y\parallel$, i.e. $\parallel \mathrm{\Pi }y{\text{∥}}^2\le \parallel y{\text{∥}}^2$, we have $\parallel \mathrm{\Pi }y\parallel \le \parallel y\parallel$, i.e. $\parallel \mathrm{\Pi }(x-z)\parallel \le \parallel x-z\parallel$.