Linear Operators, characteristic polyn.

[JFo]

Hi, I'm a little stuck on this problem. The question is:

Let $T$ be a linear operator on a two dimensional vector space $V$, and suppose that $T \neq cI$ for any scalar c. (here $I$ denotes the identity transformation). Show that if $U$ is any linear operator on $V$ such that $UT = TU$, then $U = g(T)$ for some polynomial $g(t)$

This is what I have so far:

Let f(t), g(t), h(t) denote the characteristic polynomial of T, U, and UT respectively. Since V is two dimensional, we know that f, g, and h are of degree 2 with leading coefficient (-1)^2 = 1. Therefore there exists scalars $a_0, a_1, b_0, b_1, c_0, c_1$ such that

$$f(t) = t^2 + a_1 t + a_0$$
$$g(t) = t^2 + b_1 t + b_0$$
$$h(t) = t^2 + c_1 t + c_0$$

from the Cayley-Hamilton Theorem, we know that

$$f(T) = T^2 + a_1 T + a_0 I = T_0 \ \ \ \ (1)$$
$$g(U) = U^2 + b_1 U + b_0 I = T_0 \ \ \ \ (2)$$
$$h(UT) = (UT)^2 + c_1 UT + c_0 I = T_0 \ \ \ \ (3)$$

where $T_0$ denotes the zero transformation

$$(1) \Rightarrow T^2 = -a_0 I - a_1 T \ \ \ \ (4)$$
$$(2) \Rightarrow U^2 = -b_0 I - b_1 U \ \ \ \ (5)$$
$$(3) \Rightarrow (UT)^2 = -c_0 I - c_1 UT \ \ \ \ (6)$$

composing (5) with (4) we get

$$U^2 T^2 = (UT)^2 = (-b_0 I - b_1 U)(-a_0 I - a_1 T)$$
$$\Rightarrow (UT)^2 = a_0b_0 I + a_0b_1 U + a_1b_0 T + a_1b_1 UT \ \ \ \ (7)$$

comparing (6) and (7) we have

$$-c_0 I - c_1 UT = a_0b_0 I + a_0b_1 U + a_1b_0 T + a_1b_1 UT \ \ \ \ (8)$$

This is where I'm stuck,,, I would like to be able to compare the c coefficients witht the a,b coefficients to come up with an equation for U in terms of I and T, but I'm not sure how I'm supposed to use the fact that $T \neq cI$ for any scalar c.

I realize that reading through this is a major pain in the butt, but thanks for your help (really!).

 

[kurt.physics]

Hint: You should be able to use the fact that if T ≠ cI for any scalar c then f(t) = t^2 + a_1 t + a_0 is not a perfect square. This implies that ab_0c_1 ≠ ab_1c_0.