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I know this question has been arlready asked, but as my reputation is too low I'm not allowed to post a comment, sorry for this second post.

I'm asked to prove :

  1. $(W_1+W_2)^0=W_1^0\cap W_2^0$.
  2. $(W_1\cap W_2)^0=W_1^0+W_2^0$

I managed to prove the first equation, but only "half" of the second one, i.e. to prove that $W_1^0+W_2^0$ is included in $(W_1\cap W_2)^0$.

I'm really stuck here, I don't know how to prove the reverse inclusion. I tried to use the dimension formula but it didn't lead me to anything satisfying.

Did I make a mistake? Or can anyone give me a hint here?

Thanks

Jeel Shah
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Jacques Ung
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1 Answers1

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As pointed out here, the relations \begin{gather*} (W_1 + W_2)^0 = W_1^0 \cap W_2^0 \tag{1}\\ (W_1 \cap W_2)^0 \supseteq W_1^0 + W_2^0\tag{2} \end{gather*} are valid for arbitrary vector spaces.

It's easy to check that $W_1^0 + W_2^0$ is a subspace of $(W_1 \cap W_2)^0$. If $W_1$ and $W_2$ are subspaces of a finite dimensional vector space $V$ then we have \begin{align*} \dim (W_1^0 + W_2^0) &= \dim W_1^0 + \dim W_2^0 - \dim(W_1^0 \cap W_2^0)\\ &= \dim V - \dim W_1 + \dim V - \dim W_2 - \dim (W_1 + W_2)^0 \\ &= 2\dim V - \dim W_1 - \dim W_2 - \left[\dim V -\dim(W_1 + W_2)\right] \\ &= \dim V - \dim W_1 - \dim W_2 + \dim W_1 + \dim W_2 - \dim (W_1\cap W_2)\\ &= \dim V - \dim (W_1\cap W_2)\\ &= \dim (W_1\cap W_2)^0. \end{align*} So $W_1^0 + W_2^0$ is a subspace of $(W_1 \cap W_2)^0$ with $\dim(W_1^0 + W_2^0) = \dim (W_1 \cap W_2)^0$, therefore $$W_1^0 + W_2^0 = (W_1 \cap W_2)^0.$$

leo
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  • Nice answer. How do you proof equation number 2 (one of the two that apply to arbitrary vector spaces)? – evaristegd May 21 '18 at 01:17
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    Get an element on the set on the left. Use the definitions of sum of vector spaces, annihilator space, and belonging to the annihilator of the intersection. – leo May 23 '18 at 01:38