Exercise 7.4.1

Answers

1.: Yes. See Theorem 7.17.
2.: No. Let $T (a, b) = (a, 2 b)$ be a transformation. Then the basis $β = {(1, 1), (1, 2)}$ is a $T$ -cyclic basis generated by $(1, 1)$ . But it is not a rational canonical basis.
3.: No. See Theorem 7.22.
4.: Yes. If $A$ is a square matrix with its rational canonical form $C$ with rational canonical basis $β$ , then we have $C = {[L_{A}]}_{β}$ .
5.: Yes. See Theorem 7.23(a).
6.: No. They are in general different. For example, the dot diagram of the matrix $(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ has only one dot. It could not form a basis.
7.: Yes. The matrix is similar to its Jordan canonical form and its rational canonical form. Hence the two forms should be similar.

jephianlin

2011-06-27 00:00