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If $V_1,V_2$ are subspaces. The Dimension Identity says: $$\dim(V_1) + \dim(V_2) = \dim(V_1 + V_2) + \dim(V_1 \cap V_2)$$ But how to prove this? I know in the reference book, it says we should extend the basis. But could anyone give more details about this? To be specific in the proof given by the book, how to show $\{u_1, \cdots , u_k, v_{k+1}, \cdots , v_s, w_{k+1}, \cdots , w_t\}$ is a basis for $V_1 + V_2$

Reference

[1] https://link.springer.com/content/pdf/10.1007%2F978-1-4614-1099-7_1.pdf page 5, Theorem 1.1.

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  • $\begingroup$ I can't see any reference to $N([V_1V_2])$ in that theorem. Please clarify. $\endgroup$ – David Jan 22 '18 at 4:18
  • $\begingroup$ Dear @David, I'm sorry not say it clear. I think it's less related to our major problem. So I removed that and thus we can focus on the theorem given in the book. Thanks $\endgroup$ – maple Jan 22 '18 at 4:23
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I will outline the steps and leave you to fill in the algebra.

To show $\{u_1,\ldots,u_k,v_{k+1},\ldots,v_s,w_{k+1},\ldots,w_t\}$ is independent: let $$\alpha_1u_1+\cdots+\beta_1v_{k+1}+\cdots+\gamma_1w_{k+1}+\cdots=0\ .$$ This can be written $$\alpha_1u_1+\cdots+\beta_1v_{k+1}+\cdots=-\gamma_1w_{k+1}-\cdots\ .$$ The right hand side is in $V_2$ (should be obvious) and also in $V_1$ (why?), hence in $V_1\cap V_2$, hence is a linear combination of $u_1,\ldots\,$: $$-\gamma_1w_{k+1}+\cdots=\delta_1u_1+\cdots\ .$$ Explain why $\gamma_1,\ldots$ are all zero. Give a similar argument to explain why $\beta_1,\ldots$ are zero. Then explain why $\alpha_1,\ldots$ are zero.

To show $\{u_1,\ldots,u_k,v_{k+1},\ldots,v_s,w_{k+1},\ldots,w_t\}$ is a spanning set for $V_1+V_2$: let $x$ be in $V_1+V_2$. Then $x=x_1+x_2$ with $x_1\in V_1$ and $x_2\in V_2$ (why?). Use this to write $x$ as a linear combination of $u_1,\ldots,v_{k+1},\ldots,w_{k+1},\ldots\,$.

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  • $\begingroup$ Thanks, it really helpful $\endgroup$ – maple Jan 22 '18 at 11:41
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Consider the linear map $V_1\oplus V_2 \to V_1+V_2$ sending $(v,w)\mapsto v-w$. This is clearly surjective with kernel $(a,a), a \in V_1\cap V_2$. The result then follows from rank-nullity, if you like.

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@David has explained the reasoning about the bases extensions.

Another way to show it is to see that the following map is an isomorphism: \begin{align*} \dfrac{V_{1}}{V_{1}\cap V_{2}}\times\dfrac{V_{2}}{V_{1}\cap V_{2}}&\rightarrow\dfrac{V_{1}+V_{2}}{V_{1}\cap V_{2}}\\ (\overline{v},\overline{w})&\rightarrow\overline{v+w}. \end{align*}
The only concern is whether it is injective: For suppose $\overline{v+w}=0$, then $v+w\in V_{1}\cap V_{2}$, then $w=(v+w)-v\in V_{1}$, $v=(v+w)-w\in V_{2}$, so $\overline{v}=\overline{w}=0$, as claimed.

Then $\dim V_{1}-\dim(V_{1}\cap V_{2})+\dim V_{2}-\dim(V_{1}\cap V_{2})=\dim(V_{1}+V_{2})-\dim(V_{1}\cap V_{2})$.

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