# Prove $\dim(W_1 +W_2) =\dim W_1+\dim W_2 - \dim W_1\cap W_2$

Show that if $W_1$ and $W_2$ are finite-dimensional subspaces of $V$ , then there exists a natural exact sequence $0 \rightarrow W_1 \cap W_2 \rightarrow W_1 \oplus W_2 \rightarrow W_1 +W_2 \rightarrow 0$ and use it to get a proof of $\dim (W_1 +W_2) = \dim W_1 + \dim W_2 - \dim W_1 \cap W_2$.

Use the 1st isomorphism theorem?

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What is the question? – Jonas Meyer Feb 3 '13 at 0:27
This is nothing more than the definition of the direct sum of vector spaces. – Ehsan M. Kermani Feb 3 '13 at 0:36
@icurays1 You could have used \dim which is a native LaTeX (and MathJax) command. – Asaf Karagila Feb 3 '13 at 0:39
@ehsanmo How so? I don't know what you mean. – A Blumenthal Feb 3 '13 at 0:45
@AsafKaragila I always forget about that command... – icurays1 Feb 3 '13 at 1:54

The sequence has two "$\mapsto$", which I'll call $\Phi_1, \Phi_2$; the first should be the map $v \mapsto (v,v)$ and the second is $(v,w) \mapsto v - w$. That way the image of the first map is precisely the kernel of the second.
Now use the first isomorphism theorem, which gives an isomorphism $W_1 \oplus W_2/\ker \Phi_2 \mapsto W_1 + W_2$. So it suffices to compute the dimension of the domain of this map.
But $\ker \Phi_2 = \text{Im } \Phi_1$; the map $\Phi_1$ injects and so the dimension of its image is $\dim W_1 \cap W_2$ and (as you can check) the dimension of a quotient space is equal to the difference of the dimensions (in the finite dimensional case).
We conclude that $\dim (W_1 + W_2) = \dim W_1 + \dim W_2 - \dim(W_1 \cap W_2)$, as $\dim W_1 \oplus W_2 = \dim W_1 + \dim W_2$.