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- 1. Let T : V rightarrow V be a linear transformation sulci that T2 = 0 is the zero transformation. Show that itn T ker T If R: V rightarrow V is defined by R(v) = v + T(v) for all v in V. show that R is linear and invertible. Solution (a) Let x be in im(T). Then x = T(v) for some v in V. Now consider T(x): T(x) = T(T(v)) = T 2 (v) = 0 Since T(x) = 0, x must be in ker(T). This was true for a generic x in im(T), so we must have im(T) contained in ker(T). (b) R(v) = I(v) + T(v), where I is the identity transformation (which is linear). Since R is the sum of two linear transformations, it must itself be linear as well. If you want to write the proof out, it would look like this: Let v,w be in V, and k be a scalar. R(k*v+w) = (k*v+w) + T(k*v+w) = k*v + w + k*T(v) + T(w) [since T is linear] = k*[v+T(v)] + [w+T(w)] = k*R(v)+R(w) Therefore R is linear. Suppose R(v) = 0 0 = R(v) = v + T(v) T(v) = -v We know that T(T(v)) = 0 From the above, we have that T(v)=-v. Therefore we have:
- 2. T(-v) = 0 -T(v) = 0 [T is linear] Once again we know that T(v)=-v. Therefore we have: -(-v) = 0 v = 0 Therefore R(v)=0 --> v=0 Since R is linear, this means R is invertible.