View source for V and W are isomorphic if and only if dim(V ) = dim(W ). - Rhea

MA 35100 Question: 

Premise: Let V and W be two finite dimensional real vector spaces. We call that V and W are isomorphic if there is an isomorphism T : V → W . Show that V and W are isomorphic if and only if dim(V ) = dim(W ).

Proof: 

1) We begin with the if statement, by creating a basis of {v1, .... , vn} and {w1, .... , wn} for V and W respectively. Next, we define T: V -> W by T(Σ ai vi ) := Σ ai wi. Next, we're going to show that T is a well-defined function and an isomorphism. To show this, we're going to show that T is both injective and surjective. 

'''INJECTIVE''' T(v) = 0 for v ∈ V. Then, because {v1, .... , vn} is a basis of V, it spans V, and thus there exists constant a1, a2, .... , an such that v = a1v1 + a2v2 + ... + anvn = 0. 

Then, we get that T(v) = 0 ⇒ T(a1v1 + · · · + anvn) = 0
⇒ a1T(v1) + · · · + anT(vn) = 0
⇒ a1w1 + · · · + anwn = 0

Now think about the implication of T(vi) = wi fori = 1, · · · , n.
Because {w1, .... , wn) is a basis for W, it is linearly independent, and thus a1 = · · · = an = 0.
Thus, we will get that 
v = a1v1 + · · · + anvn = 0v1 + · · · + 0vn = 0
So v = 0, and hence T is injective.

'''SURJECTIVE''' Let w be an arbitrary vector in W.
Then, since {w1, .... , wn} is a basis for W, it spans W, and hence there exist constants a1, ... , an such that w = a1w1 + · · · + anwn.
Now we have v = a1v1 + · · · + anvn => 
T(v) =T(a1v1 + · · · + anvn)
=a1T(v1) + · · · + anT(vn)
=a1w1 + · · · + anwn = w
Where the third equality followed from the fact that T(vi) = wi for i =
1, · · · , n.
So we found v such that T(v) = w, and hence T is surjective.
Therefore, since T is an injective and surjective linear transformation, and
therefore by definition, V is isomorphic to W. 

2) Now we will begin with "only if". We will begin with the idea of T : V → W being an isomorphism and {v1,...,vn} a basis for V. Thus, we are going to show that {T(vi) : 1 ≤ i ≤ n} forms a basis for W.

We will assume that T(v) = w. As T: V->W and W = {w1, w2, ... , wn}, so there's a wi for i = 1, ... , n. Thus, when we map T: V->W, we let T(v) = T(a1v1 + ... + anvn) so we say a1w1 + .... + anwn. Now we claim that v = 0 so we give T(v) = 0. Thus, because v = 0, a1w1 + .... + anvn = 0, so {v1, ... , vn} spans V. Because [Σanvn = Σanwn], and because v is a basis of, w is a basis as well. Thus, {T(vi) : 1 ≤ i ≤ n} forms a basis for W.

(Jarrod Kopczynski)