Exercise 10.A.18

Proof. Note that $\left|\right|T{e}_j|{\text{|}}^2$ is the sum of the the squares of the absolute values of the entries in the $j$-th column of the matrix of $T$ with respect to the basis $e_1,\dots ,e_n$. Thus, we’re essentially being asked to prove that $\mathrm{trace}<!--\; FUNCTION\; APPLICATION\; -->(T^{\text{∗}}T)$ equals the sum of the squares of the absolute values of the entries in the matrix of $T$ with respect to any orthonormal basis. We have

where these matrices are taken with respect to $e_1,\dots ,e_n$. The equation above gives the desired result, because $M(T^{\text{∗}})$ is the conjugate transpose of $M(T)$ and so the $j$-th diagonal entry of $M(T^{\text{∗}})M(T)$ equals the sum of the squares of the absolute values of the entries in the $j$-th column of $M(T)$. □