Exercise 7.4.6

Here we denote the degree of ${\varphi }_1$ and ${\varphi }_2$ by $a$ and $b$ respectively.

1.

By Theorem 7.23(b) we know the dimension of $K_{{\varphi }_1}$ and $K_{{\varphi }_2}$ are $a$ and $b$ respectly. Pick a nonzero element $v_1$ in $K_{{\varphi }_1}$. The $T$-annihilator of $v_1$ divides ${\varphi }_1$. Hence the $T$-annihilator of $v_1$ is ${\varphi }_1$. Find the nonzero vector $v_2$ in $K_{{\varphi }_2}$ similarly such that the $T$-annihilator of $v_2$ is ${\varphi }_2$. Thus ${\beta }_{v_1}\cup {\beta }_{v_2}$ is a basis of $V$ by Theorem 7.19 and the fact that $|{\beta }_{v_1}\cup {\beta }_{v_2}|=a+b=n$.

2.

Pick $v_3=v_1+v_2$, where $v_1$ and $v_2$ are the two vectors given in the previous question. Since ${\varphi }_1(T)(v_2)\ne 0$ and ${\varphi }_2(T)(v_1)\ne 0$ by Theorem 7.18. The $T$-annihilator of $v_3$ cannot be ${\varphi }_1$ and ${\varphi }_2$. So the final possibility of the $T$-annihilator is ${\varphi }_1{\varphi }_2$.

3.

The first one has two blocks but the second one has only one block.