Help! Let V, W be (possibly different) vector spaces and let T : V rightarrow W

Question

Help!

Let V, W be (possibly different) vector spaces and let T : V rightarrow W be a linear transformation. Suppose it has an inverse transformation T-1 : W rightarrow V, i.e. TT-1 = Idv and T-1T = Idw, where Idv : V rightarrow V and Idw : W rightarrow W are the identity transformations1 on V and on W, respectively. Prove that T-1 is also a linear transformation. Prove that T maps a basis for V to a basis for W, i.e. if E = {v1, v2, ..., vn} is a basis for V, then F = {T(v1), T(v2), ..., T(vn)} is a basis for W. Conclude that dim(V) = dim(W).

Explanation / Answer

(a) Let a,b in W and k a scalar.

Since T^(-1) exists, then we note u,v such that T^(-1)(a)=u, T^(-1)(b)=v (and T(u)=a, T(v)=b)

Since T is linear T(u+kv) = T(u)+kT(v) = a+kb , Hence u+kv = T^(-1)(a+kb) by applying T^(-1) (and since TT^-1 = Id)

Which can again be written T^(-1)(a)+kT^(-1)(b) = T^(-1)(a+kb) : so T^(-1) is a linear transformation.

(b)

Suppose E ={v1,...,vn} is a basis for V.

Let's take the family F = { T(v1), ..., T(vn) }

Suppose there exist a1,...,an such that a1T(v1)+...+anT(vn) = 0

Then by linearity T(a1v1+...+anvn) = 0

And by applying T^-1 : a1v1+...+anvn = 0

Since E is a basis of V, then a1=...=an=0

So the vectors of F are linearly independant.

Let w in W , and let's note v in V such that T^(-1)(w)=v, so w = T(v)

Since E is a basis from V we can write : v = a1v1+...+anvn , so w = T(v) = a1T(v1)+...+anT(vn).

So F = { T(v1} , ..., T(vn) } is a base of W.

(c) Since card(F)=card(E)=n then dim(V)=dim(W)

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