Homework #3 for Math H110, fall '02: Perp walk
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If A is a subspace of V, define the subset A^perp of V^* as<br>
<center> A^perp = { f in V^*: f(a) = 0 for all a in A} </center> <br>
You should think of it as like "perpendicular", but without using 
an inner product.
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1. Show that A^perp is a subspace.

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2. Show that if B is a subspace of A, then B^perp contains A^perp.

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3. Show that there exists a basis of V such that some subset spans A.
Using this, write down some elements of V^* that give a basis of A^perp.

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4. Compute dim A^perp in terms of dim A and dim V.

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5. Let I : A -> V be the inclusion map - each vector a goes to itself.
Let I^*: V^* -> A^* be the backwards map on dual spaces.
What's the relation between A^perp and I^*?

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6. Show that (A^perp)^perp = A, once we identify A and (A^*)^*.