Exercise 6.1.1

Proof. (The proof use the fact that the entries on the diagonal of $T$ are the eigenvalues of $A$, counting multiplicity, which seems not be explicitly stated in the book and can be found like in the Wiki.) $A=\mathit{UT}{U}^{\text{∗}}$. $\det <!--\; FUNCTION\; APPLICATION\; -->A=(\det <!--\; FUNCTION\; APPLICATION\; -->U)(\det <!--\; FUNCTION\; APPLICATION\; -->T)(\det <!--\; FUNCTION\; APPLICATION\; -->U^{\text{∗}})=\det <!--\; FUNCTION\; APPLICATION\; -->U=\underset{i=1}{\overset{n}{\mathrm{\Pi }}}{\lambda }_i$ because $U$ is unitary and $T$ is upper triangular with eigenvalues of $A$ on its diagonal.

To consider the trace, suppose $U=[u1,u2,...,\mathit{un}]$. Then $A$ can be represented by

where we exploit the orthogonality of $u1,u2,...\mathit{un}$. Then note that the trace of matrix $\mathit{uiu}{i}^{}=[u_{i1}\mathit{ui}{u}_{i2}\mathit{ui}...u_{\mathit{in}}\mathit{ui}]$ (outer product) is $u_{i1}^2+u_{i2}^2+...+u_{\mathit{in}}^2=\left|\right|\mathit{ui}|{\text{|}}^2=1$. Thus $\text{trace}A=\text{trace}({\lambda }_{1}u1u{1}^{})+\text{trace}({\lambda }_{2}u2u{2}^{})+...+\text{trace}({\lambda }_n\mathit{unu}{n}^{})={\lambda }_1+{\lambda }_2+...+{\lambda }_n={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n{\lambda }_i$. □