2
$\begingroup$

Let $S$ be an operator on a finite-dimensional real vector space $U$ and assume that $$U = \langle S,u_1 \rangle \oplus \langle S,u_2 \rangle \oplus \cdots \oplus \langle S,u_6 \rangle$$ and $$\mu_{S,u_1}(x) = \mu_{S,u_2}(x) = (x^2+1)^5, \mu_{S,u_3} = (x^2+1)^4$$ $$\mu_{S,u_4}(x) = \mu_{S,u_5}(x) = (x^2+1)^2, \mu_{S,u_6} = x^2+1.$$ Set $U_i = \{u \in U:(s^2+I_U)^i(u)=0 \}$ for $i=1,2,3,4,5,6.$

How do we compute the dimension of each $U_i$?

$\endgroup$

1 Answer 1

3
$\begingroup$

We can write $U_1$ as

$$\langle S, (S^2 + I_V )^4(u_1) \rangle \oplus \langle S, (S^2 + I_V )^4(u_2) \rangle \oplus \langle S, (S^2 + I_V )^3(u_3) \rangle \oplus \langle S, (S^2 + I_V)(u_4) \rangle \oplus \langle S, (S^2 + I_V )(u_5) \rangle \oplus \langle S, u_6 \rangle.$$

Each of the vectors $(S^2+I_V )^4(u_1),(S^2+I_V )^4(u_2),(S^2+I_V )^3(u_3),(S^2+I_V )(u_4), (S^2 + I_V )(u_5), u_6$ has minimal polynomial $x^2 + 1.$ Therefore each of the six cyclic subspaces above have dimension $2$ and $U_1$ has dimension $12.$

We can write $U_2$ as $$\langle S, (S^2 + I_V )^3(u_1) \rangle \oplus \langle S, (S^2 + I_V )^3(u_2) \rangle \oplus \langle S, (S^2 + I_V )^2(u_3) \rangle \oplus \langle S, u_4 \rangle \oplus \langle S, u_5 \rangle \oplus \langle S, u_6 \rangle.$$ Each of the vectors $(S^2 + I_V )^3(u_1), (S^2 + I_V )^3(u_2), (S^2 + I_V )^2(u_3), u_4, u_5$ has minimal polynomial $(x^2 + 1)^2$ of degree $4.$ Therefore $U_2$ is the direct sum of five cyclic subspaces each with dimension $4$ and one with dimension $2,$ $\langle S,u6 \rangle.$ Therefore $\dim U_2 = 22.$

We can express $U_3$ as the direct sum $$\langle S, (S^2 + I_V )^2(u_1) \rangle \oplus \langle S, (S^2 + I_V )^2 (u_2) \rangle \oplus \langle S, (S^2 + I_V )(u_3) \rangle \oplus \langle S, u_4 \rangle \oplus \langle S, u_5 \rangle \oplus \langle S, u_6 \rangle.$$

Each of the vectors $(S^2 + I_V ^)2(u_1), (S^2 + I_V )^2(u_2), (S^2 + I_V )(u_3)$ has minimal polynomial $(x^2 + 1)^3$ of degree $6.$ The vectors $u_4$ and $u_5$ have minimal polynomial $(x^2 + 1)^2$ and $u_6$ has minimal polynomial $x^2 + 1.$ Consequently, $\dim U_3 = 3 \cdot 6 + 2 \cdot 4 + 2 = 28.$ In a similar fashion, $\dim U_4 = 2 \cdot 8+1 \cdot 6+2 \cdot 4+2 = 32.$ Finally, $\dim U_5 = \dim U_6 = \dim V = 34.$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .