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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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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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  • $\begingroup$ That a subspace of a finite dimensional vector space is finite dimensional actually requires a bit more work (essentially it uses the fact that dimension is well defined, and the incomplete basis theorem) than the fact (which you explained in detail) that the sum is finite dimensional. $\endgroup$ – Marc van Leeuwen Apr 4 '16 at 13:51
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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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  • $\begingroup$ This only works when we already know that $W_1 \cap W_2$ and $W_1+W_2$ are finite dimensional. $\endgroup$ – Martin Brandenburg Apr 15 '13 at 18:21
  • $\begingroup$ @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 $W_2$. $\endgroup$ – Alex Becker Apr 15 '13 at 18:25
  • $\begingroup$ Yeah sure, this is also contained in my answer. I wanted to clarify this for the other readers. $\endgroup$ – Martin Brandenburg Apr 16 '13 at 10:04

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