Exercise 6.4.1

Answers

1.: Yes. Check $T T^{∗} = T^{2} = T^{∗} T$ .
2.: No. The two matrices $(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ and $(\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix})$ have $(\begin{matrix} 1 \\ 0 \end{matrix})$ and $(\begin{matrix} 0 \\ 1 \end{matrix})$ to be their unique normalized eigenvectors respectly.
3.: No. Consider $T (a, b) = (2 a, b)$ to be a mapping from $ℝ^{2}$ to $ℝ^{2}$ and $β$ to be the basis ${(1, 1), (1, 0)}$ . We have $T$ is normal with $T^{∗} = T$ . But ${[T]}_{β} = (\begin{matrix} 1 & 0 \\ 1 & 2 \end{matrix})$ is not normal. Furthermore, the converse is also not true. We may let $T (a, b), = (b, b)$ be a mapping from $ℝ^{2}$ to $ℝ^{2}$ and $β$ be the basis ${(1, - 1), (0, 1)}$ . In this time $T$ is not normal with $T^{∗} (a, b) = (0, a + b)$ . However, ${[T]}_{β} = (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix})$ is a normal matrix.
4.: Yes. This comes from Theorem 6.10.
5.: Yes. See the Lemma before Theorem 6.17.
6.: Yes. We have $I^{∗} = I$ and $O^{∗} = O$ , where $I$ and $O$ are the identity and zero operators.
7.: No. The mapping $T (a, b) = (- b, a)$ is normal since $T^{∗} (a, b) = (a, - b)$ . But it’s not diagonalizable since the characteristic polynomial of $T$ does not split.
8.: Yes. If it’s an operator on a real inner product space, use Theorem 6.17. If it’s an operator on a complex inner product space, use Theorem 6.16.

jephianlin

2011-06-27 00:00