Let $Z \in \mathscr{M}_{d}(\mathbb{R})$ be an invertible matrix with $\mathrm{S} = Z^{T}Z$, and let $(\lambda_{i})_{1 \leqslant i \leqslant d}$ be the decreasing family of strictly positive eigenvalues of $S$ with associated orthonormal basis $(u_{1}, \ldots, u_{d})$. We consider $v_{i} = \frac{1}{\sqrt{\lambda_{i}}} Z u_{i}$ for all $1 \leqslant i \leqslant d$.
[(a)] Show that $(v_{1}, \ldots, v_{d})$ is an orthonormal basis of $\mathbb{R}^{d}$.
[(b)] Verify that if $U = (u_{1} | \ldots | u_{d})$, $V = (v_{1} | \ldots | v_{d})$ and $D = \operatorname{Diag}(\sqrt{\lambda_{1}}, \ldots, \sqrt{\lambda_{d}})$ then $Z = VDU^{T}$.
Let $Z \in \mathscr{M}_{d}(\mathbb{R})$ be an invertible matrix with $\mathrm{S} = Z^{T}Z$, and let $(\lambda_{i})_{1 \leqslant i \leqslant d}$ be the decreasing family of strictly positive eigenvalues of $S$ with associated orthonormal basis $(u_{1}, \ldots, u_{d})$. We consider $v_{i} = \frac{1}{\sqrt{\lambda_{i}}} Z u_{i}$ for all $1 \leqslant i \leqslant d$.
\begin{itemize}
\item[(a)] Show that $(v_{1}, \ldots, v_{d})$ is an orthonormal basis of $\mathbb{R}^{d}$.
\item[(b)] Verify that if $U = (u_{1} | \ldots | u_{d})$, $V = (v_{1} | \ldots | v_{d})$ and $D = \operatorname{Diag}(\sqrt{\lambda_{1}}, \ldots, \sqrt{\lambda_{d}})$ then $Z = VDU^{T}$.
\end{itemize}