Let V and W be vector spaces over the field F and let U be an isomorphism of V onto W. Prove T --> UTU^-1?

Question

is an isomorphism of L(V,V) onto L(W,W).

Answer 1

First, we need to show that this is a bijection. Note that S = UTU⁻¹ ⇔ T = U⁻¹SU, so this transformation has an inverse, namely the function S ↦ U⁻¹SU, therefore it is a bijection.

To see that it is a homomorphism, let S, T be any two elements of L(V, V). Then S+T ↦ U(S+T)U⁻¹ = (US + UT)U⁻¹ = USU⁻¹ + UTU⁻¹, so this transformation is linear. And if k∈F, we have that for any T∈L(V, V) that kT ↦ UkTU⁻¹ = kUTU⁻¹ (the scalar commutes because u is linear). Therefore, the transformation from L(V, V) to L(W, W) given by T ↦ UTU⁻¹ is a homomorphism, and since we've already shown it is bijective, it is an isomorphism. Q.E.D.

Answer 2

One won't be able to easily map factors no longer in W to 0. to be sure this, evaluate V to be R^2 and the subspace generated by potential of (a million,0). Map this to a million. then you particularly won't be able to map the two (a million,a million) and (0,a million) to 0. so which you will no longer basically map each and each ingredient no longer in W to 0 through fact the 1st one minus the 2d ought to map to a million. yet what you're able to do is right here: permit {w_i} be a foundation for W and we are able to finished that to a foundation for V, say with vectors {v_1}. all of us be attentive to how f is defined on each and each w_i. define g arbitrarily on the vectors {v_i}. that's needless to say linear, that's certainly defined and is of an identical opinion with f on W.