Homework #1 for Math H110, fall '02 answers
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1. A couple of people correctly pointed out that the proof of zero mapping
to zero in class wasn't complete. There are a number of ways to fix it
(by assuming additional axioms), but instead let's construct a
pathological example.
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Let F_2 be the field with two elements {0,1}, so 1+1=0. Find 
a "vector space" V over F_2, satisfying the axioms given in class:
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<li>vector addition is commutative associative, with an additive identity
<li>(e+f)v = ev + fv
<li>(ef)v = e(fv)
<li>1 times v = v for all v
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but with three elements (what would its dimension be, log_2 of 3?).

A1. Let V have three elements, {zero, small, big}.
Define + as "minimum", so zero+X = X, small+big = small, etc.
It's commutative and associative (though doesn't have additive inverses).
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Let 0,1 in F_2 <em>both</em> act as the identity. (There is an annoying
philosophical question here - does this mean 0=1, which wasn't supposed
to happen, according to the definition of field? What it really means is
that the map F_2 -> Functions(V,V) that takes a field element f to the
function "scale by f: V->V" isn't 1:1.)
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Then checking the above equations takes a finite amount of time.

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2. Curtis p15 #6. Suppose 0 has a multiplicative inverse. Show that 
the field only has one element.
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2A. First we show 1=1+1:
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1 = 0 * 0^{-1} = (0+0) * 0^{-1} = 0*0^{-1} + 0*0^{-1} = 1+1.
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Now multiply by any field element f to get f = f+f. Then subtract f,
to get 0=f.
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3. Curtis 2.4 #5,9,10
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A#5. (1,1,0), (1,0,-1) for example.
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A#9. If a = sum_i lambda_i b_i, and b_i = sum_j mu_ij c_j,
then 
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a = sum_i lambda_i sum_j mu_ij c_j <br>
  = sum_i sum_j lambda_i mu_ij c_j <br>
  = sum_j sum_i lambda_i mu_ij c_j <br>
  = sum_j (sum_i lambda_i mu_ij) c_j. 
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A#10. If S contains R, and R is dependent, then we can find a linear
dependence in S by taking one in R and putting the coefficient 0 in
front of the rest of the vectors in S. The analogous linear independence
has the set containment the other way.
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4. Let V be a fin. gen. vector space, and X,Y two subspaces of the
same dimension. 
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<li>Show that there is an invertible map T : V->V such that T(X) = Y.
<li>Show that there is a map T: V->V such that T(X) = T(V) = Y.
When is it invertible?
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A4.
Pick bases e,f of X and Y, and extend to bases e',f' of V.
Then define a linear map taking each e_i to f_i, and either
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<li>taking the rest of e' to the rest of f'
<li>taking the rest of e' to zero.
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This second guy is only 1:1 if there's no "rest", i.e. X=V. (In which
case Y=V too.)