Exercise 7.D.2

Answers

Proof. Define $T \in L (ℂ^{2})$ by

\begin{aligned} T e_{1} & = 0 \\ T e_{2} & = 5 e_{1} \end{aligned}

where $e_{1}, e_{2}$ is the standard basis of $ℂ^{2}$ . Then the matrix of $T$ with respect to this is basis is

M (T) = (\begin{matrix} 0 & 5 \\ 0 & 0 \end{matrix}),

which is upper triangular. Thus $0$ is the only eigenvalue of $T$ . We have

M (T^{∗} T) = M (T^{∗}) M (T) = (\begin{matrix} 0 & 0 \\ 5 & 0 \end{matrix}) (\begin{matrix} 0 & 5 \\ 0 & 0 \end{matrix}) = (\begin{matrix} 0 & 0 \\ 0 & 25 \end{matrix}) .

Hence the eigenvalues of $T^{∗} T$ are $0$ and $25$ . By 7.52, the singular values of $T$ are $0$ and $5$ . □

celiopassos

2017-10-06 00:00