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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A. The zero functional in V^* is f(a)=0 for all of V. Definitely then
f(a)=0 for all of A. So A^perp contains the zero vector.
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If f,g are in A^perp, and r in F, then
(rf+g)(a) = rf(a) + g(a) = r0+0 = 0. So rf+g in A^perp.

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2. Show that if B is a subspace of A, then B^perp contains A^perp.
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A. If f vanishes on all of A, then it vanishes on all of B.

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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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A. Pick a basis of A and extend to a basis of V. Take the dual basis.
The dual basis elements <i>not</i> corresponding to the basis elements of A
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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A. We just found a basis, of size dim V minus dim A.

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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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A. A^perp is sitting inside V^*, and I^* maps out of V^*, so the
obvious way to connect them is to apply I^* to elements of A^perp.
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For any f in V (not necessarily in A^perp), 
to figure out what I^*(f) is we have to apply it
to elements a of A.
<center>I^*(f) (a) = f(I(a)) = f(a) </center>
<table>
<tr><td> f is in A^perp <td> if and only if this f(a)=0 for all a
<td><td> if and only f I^* f = 0 
<td><td> if and only if f is in the kernel of I^*.  </table>
So A^perp = ker I^*.
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6. Show that (A^perp)^perp = A, once we identify A and (A^*)^*.
(This uses V finite dimensional.)
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A. To be specific, we have A including into V, then mapping to (V^*)^*;
the question is to show that the image is (A^perp)^perp.
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If we let the dimensions of V and A be n and k, then A^perp is n-k dimensional,
and A^perp^perp is n-(n-k)=k dimensional again. So at least the dimensions
match.
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We proved in class that the "eat" map from V->V^*^* is 1:1, so A is injecting
into V^*^*. Therefore, if we can show that eat A is a subspace of 
A^perp^perp, we're done, using the dimensions matching. Here goes:
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If a in A, and f in A^perp, then (eat a)(f) = f(a) = 0. QED.