Okay, so the following actually isn't a theorem asserted by the book I'm using. It's actually an example given to illustrate the importance of isomorphisms. It was a specific example with specific numbers but I want to try and prove that it holds in general. So, here's the assertion:

Let $f:V \to W$ be an isomorphism. Let $U\_1$ and $U\_2$ be subspaces of V. Then, the following holds:

$$\dim(U\_1 \cap U\_2) = \dim(f(U\_1) \cap f(U\_2))$$

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Proof Attempt:

Let $(v\_1,v\_2,....,v\_n)$ be a basis for $U\_1 \cap U\_2$. Thus, that n-tuple is linearly independent and spans the intersection of the two subspaces. 

By a previously proven result, if $(v\_1,v\_2,....,v\_n)$ is a linearly independent list of vectors, then $\beta = (f(v\_1),f(v\_2),....,f(v\_n))$ is a linearly independent list of vectors. To show that this list forms a basis for $f(U\_1) \cap f(U\_2)$, we just need to show that it generates that set. 

Consider the following linear combination:

$\sum\_{k=1}^{n} \alpha\_k f(v\_k)$

Due to the linearity of f, we can rewrite the above as follows:

$f[\sum\_{k=1}^{n} \alpha\_k v\_k]$. 

$L(\beta) \subset f(U\_1) \cap f(U\_2)$. So, we have to prove containment in the other direction. Let $w \in f(U\_1) \cap f(U\_2)$. Then, $w \in f(U\_1)$ and $w \in f(U\_2)$. 

By the surjectivity of f, there exists a vector $u\_1 \in U\_1$ such that $f(u\_1) = w$ and there exists a vector $u\_2 \in U\_2$ such that $f(u\_2) = w$. By the injectivity of f, it's clear that $u\_1 = u\_2$. 

That also means that $u\_1 \in U\_1 \cap U\_2$ so it can be written as a linear combination of the vectors $(v\_1,v\_2,...,v\_n)$. Therefore, $f(u\_1)$ can be written as a linear combination of the vectors in $\beta$ and this proves that $f(U\_1) \cap f(U\_2) \subset L(\beta)$. This proves that $\beta$ generates $f(U\_1) \cap f(U\_2)$. 

So, $\beta$ forms a basis for $f(U\_1) \cap f(U\_2)$ and that basis has length $n$. Similarly, the basis for $U\_1 \cap U\_2$ also has length n. Thus, we conclude that:

$\dim(U\_1 \cap U\_2) = \dim(f(U\_1) \cap f(U\_2))$. 

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Once again, I need to know if my argument above is correct or not. I'm also pretty sure that there's a short and slick proof of this but I certainly can't see it (maybe with the use of set identities?).