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Let $V$ be a finite dimensional $k$-vector space and let $A,B,C$ three subspaces such that $A\supseteq B$.

Why is the following equation true?

$$\dim(A/B)=\operatorname{dim }\left(\frac{A+C}{B+C}\right)+\operatorname{dim }\left(\frac{A\cap C}{B\cap C}\right)$$

I was able to prove only that: $$\operatorname{dim }\left(\frac{A+C}{B+C}\right)=\operatorname{dim }\left(\frac{A}{A\cap(B+C)}\right)$$

$$\operatorname{dim }\left(\frac{A\cap C}{B\cap C}\right)=\operatorname{dim }\left(\frac{B+(A\cap C)}{B}\right)$$

  • 0
    What you mean by $A + B$? – 2017-02-20
  • 1
    the vector subspace generated by $A$ and $B$ – 2017-02-20

2 Answers 2

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That is because of the short exact sequence: $$0\to (B+C)/B\to (A+C)/B\to (A+C)/(B+C)\to 0$$ which implies the dimension of the middle term is the sum of the dimensions of the left and right terms, and the second isomorphism theorem: $$(B+C)/B\simeq C/B\cap C.$$

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Use:

  • If $V$ is a finite dimensional vector space and $S$ is a subspace of $V$, then $$\dim{V/S}=\dim{V}-\dim{S}$$

  • If $V$ is a finite dimensional vector space and $A$ and $B$ are subspaces of $V$, then $$\dim{(A+B)}=\dim{A}+\dim{B}-\dim{A\cap B}$$