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Let $W_1$ and $W_2$ be finite dimensional subspace fo a vector space $V$. How should I start to prove that the subspaces $W_1 \cap W_2$ and $W_1+W_2$ are also finite dimensional and

\begin{eqnarray} dim(W_1 \cap W_2)+dim(W_1+W_2)=dim(W_1)+dim(W_2) \end{eqnarray}

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2 Answers 2

Well if $W_1,W_2$ have finite generating systems $E_1,E_2$, then $E_1 \cup E_2$ is a finite generating system of $W_1+W_2$. And $W_1 \cap W_2$ is a subspace of $W_1$, therefore also finite dimensional.

The formula follows from the isomorphism theorem $(W_1+W_2) / W_1 \cong W_2 / (W_1 \cap W_2)$ and the dimension formula $\dim(V/U)+\dim(U)=\dim(V)$.

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Here's a high-powered argument:

Consider the exact sequence $$0\to W_1\cap W_2\overset{f}{\to} W_1\times W_2\overset{g}{\to} W_1+W_2\to 0$$ with $f(w)=(w,-w)$ and $g(w_1,w_2)=w_1+w_2$. The alternating sum of the dimensions of an exact sequence of vector spaces is always $0$, so $\dim(W_1\cap W_2)-\dim(W_1\times W_2)+\dim(W_1+W_2)=0$. Observing that $\dim(W_1\times W_2)=\dim(W_1)+\dim(W_2)$ finishes the proof.

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This only works when we already know that $W_1 \cap W_2$ and $W_1+W_2$ are finite dimensional. –  Martin Brandenburg Apr 15 '13 at 18:21
    
@MartinBrandenburg True, although the first one is trivial (its a subspace of $W_1$, which is f.d.) and the second one is clearly spanned by the union of any bases for $W_1$ and $W_2$. –  Alex Becker Apr 15 '13 at 18:25
    
Yeah sure, this is also contained in my answer. I wanted to clarify this for the other readers. –  Martin Brandenburg Apr 16 '13 at 10:04

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