Exercise 7.3.1

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1.: No. If $p (t)$ is the polynomial of largest degree such that $p (T) = T_{0}$ , then $q (t) = t (p (t)$ is a polynomial of larger degree with the same property $q (T) = T_{0}$ .
2.: Yes. This is Theorem 7.12.
3.: No. The minimal polynomial divides the characteristic polynomial by Theorem 7.12. For example, the identity transformation $I$ from $ℝ^{2}$ to $ℝ^{2}$ has its characteristic polynomial ${(1 - t)}^{2}$ and its minimal polynomial $t - 1$ .
4.: No. The identity transformation $I$ from $ℝ^{2}$ to $ℝ^{2}$ has its characteristic polynomial ${(1 - t)}^{2}$ but its minimal polynomial $t - 1$ .
5.: Yes. Since $f$ splits, it consists of those factors ${(t - λ)}^{r}$ for some $r \leq n$ and for some eigenvalues $λ$ . By Theorem 7.13, the minimal polynomial $p$ also contains these factors. So $f$ divides $p^{n}$ .
6.: No. For example, the identity transformation $I$ from $ℝ^{2}$ to $ℝ^{2}$ has its characteristic polynomial ${(1 - t)}^{2}$ and its minimal polynomial $t - 1$ .
7.: No. For the matrix $(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ , its minimal polynomial is ${(t - 1)}^{2}$ but it is not diagonalizable.
8.: Yes. This is Theorem 7.15.
9.: Yes. By Theorem 7.14, the minimal polynomial contains at least $n$ zeroes. Hence the degree of the minimal polynomial of $T$ must be greater than or equal to $n$ . Also, by Cayley-Hamilton Theorem, the degree is no greater than $n$ . Hence the degree of the minimal polynomial of $T$ must be $n$ .

jephianlin

2011-06-27 00:00

Exercise 7.3.1

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