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I have a problem as follows: $W_1$ and $W_2$ are subspaces of a finite-dimensional vector space $V$. $W^0$ is the annihilator of $W$.

(a) Prove $(W_1 + W_2)^0 = W_1^0 \cap W_2^0$.

(b) Prove $(W_1 \cap W_2)^0 = W_1^0 + W_2^0$.

Thoughts so far:

By definition of Annihilator, $W^0$ is the set of linear functionals that vanish on $W$. I feel that I also should use dual, but not sure how to put things together.

Gerry Myerson
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Megan
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  • Is this from Linear Algebra by Hoffman and Kunze? I think you can just write out the definitions of all four expressions, and the desired result follows immediately. – Divide1918 Nov 07 '19 at 14:21
  • @Divide1918: no, the second equality does not "follow immediately" from the definitions. Try to write down a proof and I'll show you where your mistake is. – Georges Elencwajg Aug 15 '20 at 15:19

1 Answers1

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Suppose $\mathbf{f}\in (W_1 + W_2)^0$. That means that $\mathbf{f}$ vanishes on all vectors of $W_1+W_2$. Since $W_1\subseteq W_1+W_2$, then $\mathbf{f}$ vanishes on all vectors of $W_1$, so $\mathbf{f}\in W_1^0$.

Argue similarly to show that $\mathbf{f}\in W_2^0$, to conclude that $(W_1+W_2)^0\subseteq W_1^0\cap W_2^0$.

For the converse inclusion, suppose that $\mathbf{g}\in W_1^0\cap W_2^0$. Then $\mathbf{g}$ vanishes on all vectors in $W_1$, and on all vectors on $W_2$. Let $w\in W_1+W_2$. Then we can express $w$ as $w=w_1+w_2$ with $w_1\in W_1$ and $w_2\in W_2$. Can you prove that $\mathbf{g}(w)=0$? If so, we will have shown that $\mathbf{g}$ vanishes at an arbitrary vector of $W_1+W_2$, so $\mathbf{g} \in(W_1+W_2)^0$.

Similar arguments can be used for (b).

Alternatively: $W_1+W_2$ is the smallest subspace that contains $W_1$ and $W_2$, so $(W_1+W_2)^0$ should be the largest subspace that is contained in both $W_1^0$ and $W_2^0$. Similarly for (b).

Arturo Magidin
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  • It's highly helpful! Thanks again!!!! – Megan Apr 23 '12 at 06:17
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    How can we prove the inclusion $(W_1 \cap W_2)^0 \subset W_1^0 + W_2^0$? I think I have to use the hypothesis of V being finite dimensional, but I don't see how. – Bias of Priene Jan 29 '19 at 12:48
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    @BiasofPriene: Here's what occurs to me now, almost 7 years later... Take a basis $\beta_1$ for $W_1\cap W_2$, and $\gamma_1,\gamma_2$ such that $\beta_1\cup\gamma_1$ is a basis for $W_1$, and $\beta_1\cup\gamma_2$ is a basis for $W_2$. If $\mathbf{f}\in (W_1\cap W_2)^0$, define functions $\mathbf{f}_1\in W_1^0$ and $\mathbf{f}_2\in W_2^0$ whose sum is $\mathbf{f}$ by specifying what they do in $\gamma_1$ and $\gamma_2$. – Arturo Magidin Jan 29 '19 at 15:51
  • Well, the problem is that I have to extend $\beta_1 \cup \gamma_1$ and $\beta_1 \cup \gamma_2$ to a basis for $V$ and then I don't know how to define $f_1$ and $f_2$ to "agree" in those elements with $f$. – Bias of Priene Jan 30 '19 at 16:26
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    @BiasofPriene Make one of them zero on those, and the other agree with $\mathbf{f}$. – Arturo Magidin Jan 30 '19 at 16:37
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    Referring to your answer (not your comments), can similar arguments really be used for part (b)? – user0 Dec 02 '19 at 18:53
  • Dear @user0: no, "similar arguments" cannot be used for part (b). One needs to do some real work, just like you did (and I upvoted your answer). – Georges Elencwajg Aug 15 '20 at 15:43