Exercise 7.D.17

Proof. (a) We have $T{e}_j=s_j{f}_j$. This implies that the matrix of $T$ if respect to the bases $e_1,\dots ,e_n$ and $f_1,\dots ,f_n$ is the diagonal matrix whose diagonal entries are the singular values of $T$. By 7.10, we have $T^{\text{∗}}{f}_j=s_j{e}_j$ for each $j$. If we replace $v$ with $f_j$ in the right hand side of the desired result we get the same thing, therefore

by the uniqueness of linear maps (see 3.5).

(b) Just apply the previous item to the formula given of the singular value decomposition of $T$.

(c) Note that the $e_j$’s are eigenvectors of $T^{\text{∗}}T$ with corresponding eigenvalue $s_j^2$. Thus $\sqrt{T^{\ast }T}{e}_j=s_j{e}_j$. Plugging $e_j$ in the place of $v$ in the right-hand side yields the same thing, so the result holds by uniqueness of linear maps again.

(d) The given formula satisfies $T{T}^{-1}=I$ and $T^{-1}T=I$ and it is well defined, because from Exercise 13 we see that none of $s_j$’s are $0$. □