Nov 22, 2013 Thread starter #1 C crypt50 New member Jun 29, 2013 21 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 .
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 .
Nov 22, 2013 #2 Deveno Well-known member MHB Math Scholar Feb 15, 2012 1,967 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).
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).