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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.

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1 Answer 1

up vote 3 down vote accepted

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).

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It's highly helpful! Thanks again!!!! –  Megan Apr 23 '12 at 6:17

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