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?

  • $\begingroup$ What is the question? $\endgroup$ – Jonas Meyer Feb 3 '13 at 0:27
  • $\begingroup$ This is nothing more than the definition of the direct sum of vector spaces. $\endgroup$ – Ehsan M. Kermani Feb 3 '13 at 0:36
  • $\begingroup$ @icurays1 You could have used \dim which is a native LaTeX (and MathJax) command. $\endgroup$ – Asaf Karagila Feb 3 '13 at 0:39
  • $\begingroup$ @ehsanmo How so? I don't know what you mean. $\endgroup$ – A Blumenthal Feb 3 '13 at 0:45
  • $\begingroup$ @AsafKaragila I always forget about that command... $\endgroup$ – 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 $\dim W_1 \oplus W_2 = \dim W_1 + \dim W_2$.

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  • $\begingroup$ Is there another way to prove this? $\endgroup$ – Fareed Abi Farraj Feb 11 '19 at 14:40
  • $\begingroup$ @FareedAF Sure. Let $e_1, \cdots, e_k$ be a basis for $W_1 \cap W_2$. Let $e_{k+1}, \cdots, e_n$ and $e_{k+1}', \cdots, e_n'$ be such that $\{ e_1, \cdots, e_k, e_{k+1}, \cdots, e_n\}$ is a basis for $W_1$ and $\{ e_1, \cdots, e_k, e_{k+1}', \cdots, e_m'\}$ is a basis for $W_2$. Check that $\{ e_1, \cdots, e_k, e_{k + 1}, \cdots, e_n, e_{k+1}', \cdots, e_m'\}$ is a basis for $W_1 + W_2$. We conclude $\dim (W_1 + W_2) = n + m - k = \dim(W_1) + \dim(W_2) - \dim(W_1 \cap W_2)$. $\endgroup$ – A Blumenthal Feb 12 '19 at 23:14
  • $\begingroup$ Thank you @A Blumenthal $\endgroup$ – Fareed Abi Farraj Feb 13 '19 at 7:29

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