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- #1

- Thread starter crypt50
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- Thread starter
- #1

- Feb 15, 2012

- 1,967

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).