Clarifying a theorem from Hoffman and Kunze, Linear algebra 
Theorem 9. Let $V$ and $W$ be finite dimensional vector spaces over the field $F$ such that $\dim V=\dim W.$ If $T$ is a linear transformation from $V$ into $W,$ the following are equivalent:
(i) $T$ is invertible.
(ii) $T$ is non-singular.
(iii) $T$ is onto.

And the book goes on to say : " We caution the reader not to apply Theorem $9$ except in the presence of finite dimensionality and with $\dim V=\dim W.$
If $T$ were invertible then $\dim V=\dim W$ would anyway hold right? My question is would anything happen to this theorem if drop this condition? Is it possible to have an invertible linear transformation with $\dim V\ne \dim W$?
 A: 
My question is would anything happen to this theorem if drop this condition? Is it possible to have an invertible linear transformation with $\dim V\ne \dim W$?

No. 
An invertible linear transformation between two vector spaces is an isomorphism, and since isomorphisms send bases to bases, and preserve their cardinality too, there's no way you could have $\dim(V)\neq\dim(W)$. That isomorphic spaces have equal dimensionality is true even for infinite dimensions.
Without finite dimensionality it is still true that an isomorphism is non-singular and onto. But the converses won't hold. You can have nonsingular transformations that aren't onto, and onto transformations that are singular.
A: It says the following conditions are equivalent. Which means that $T$ is onto will imply $T$ is invertible. You'll need dim $V$ = dim $W$ for that.
A: One way to see that the T is not invertible when $\dim V \neq \dim W$ is to recall that $$rank + nullity = \dim V$$ The rank of T is nothing but the dimension of the image of T and when T is onto, $\mathrm{rank (T)} = \dim W$
 Consequently, if $\dim V \neq \dim W$, then $nullity \neq 0$ and thus T is not $1:1$. 
