# Prove the following; (vector spaces and linear operators)

#### crypt50

##### New member
a) V = U_1 ⊕ U_-1 where U_λ = {v in V | T(v) = λv}
b) if V = M_nn(R) and T(A) = A^t then what are U_1 and U_-1

When V is a vector space over R, and T : V -> V is a linear operator for which
T^2 = IV .

#### Deveno

##### Well-known member
MHB Math Scholar
Re: Prove the following;

Given $T^2 = I$, it's clear the only possible eigenvalues are 1 and -1, since the minimal polynomial for $T$ divides $x^2 - 1 = (x + 1)(x - 1)$.

Note that:

$\frac{1}{2}(T + I) - \frac{1}{2}(T - I) = I$

Thus:

$v = Iv = \frac{1}{2}(T + I)(v) - \frac{1}{2}(T - I)(v)$

I claim $u = \frac{1}{2}(T + I)(v) \in U_1$ and

$w = -\frac{1}{2}(T - I)(v) \in U_{-1}$.

To see this, observe that:

$T(u) = T(\frac{1}{2}(T + I)(v)) = \frac{1}{2}T(T + I)(v) = \frac{1}{2}(T^2 + T)(v)$

$= \frac{1}{2}(T + I)(v) = u$, while:

$T(w) = T(-\frac{1}{2}(T - I)(v)) = -\frac{1}{2}T(T - I)(v) = -\frac{1}{2}(T^2 - T)(v)$

$= \frac{1}{2}(T - I)(v) = -w$.

Thus $V = U_1 + U_{-1}$.

Your turn (prove this sum is direct).