1. Show that the kernel of a linear transformation is a subspace.
1A. Let K = ker T. Then T(0)=0, so K has the zero vector, and T(rv+w) = rT(v)+T(w) = r0+0 = 0 for any r in F, v,w in K.
2. Let T:V->W, and S:W->X. Find good bases of V,W,X so that each of T, S, and S o T all look nice. (Part of the problem is to decide what "nice" could mean.)
2A. I'm going to give this answer in terms of complementary subspaces, and only at the very end pick bases of all the spaces (which is sort of a distraction to the main argument).
The natural subspaces we have, without making any choices, are
(We could think about the intersection and span of ker T and ker ST, but it's not useful because ker ST contains ker T, which is trivial to prove. Similarly for im S and im ST.)Okay, now let's break our vector spaces into direct sums, starting with V. Pick a complement A to ker T inside ker ST, and a complement B to ker ST inside V. So A+B is a complement to ker T inside V, and therefore T : A+B -> im T is an isomorphism.(We proved this property of complements during the nullity + rank theorem.)
But even better: since A is inside ker ST, T actually gives an isomorphism of A with im T intersect ker S, and B with a complement of that intersection inside im T.
Inside W, then, so far, we have T(A) = im T intersect ker S, T(B) = a complement to that inside im T, and that's all of im T, so all we can get from V. Now we get started on W. Pick a complement C to T(A) inside ker S. (Casual observation: the subspace spanned by ker S and im T is therefore a direct sum of T(A), T(B), and C.) Pick a complement D to (ker S + im T) inside W.
Okay, W is now the direct sum of T(A), T(B), C, and D. If we hit these with S we get 0, S(T(B)), 0, and S(D), and also we know that S: T(B)+D -> im S is an isomorphism (since T(B)+D is a complement to ker S = T(A)+C).
Even better, ST : B -> im ST is an isomorphism. So inside X we've found our two subspaces, im ST = ST(B) and im S = ST(B) + S(D). Pick a complement E to im S inside W and we're done.
To recap: what do our maps look like?
| T: V->W | S: W->X | ST: V->X | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
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Now finally, we can be distracted by bases. Pick bases of everybody who isn't an image (i.e. their row is all 0s): that's ker T, A, B, C, D, E. From those construct bases of T(A), T(B), ST(B), S(D). Concatenate these bases together to get bases of V,W,X. Then with respect to those bases, the matrices expressing S, T, ST look like the three above (except much bigger, and each 1 has been replaced by an identity matrix).
If you still find it easier to think about matrices than linear transformations, it's worth trying the following. Take the above three matrices, and calculate all the "natural subspaces" listed before.
What theorems did we use to get all this to work? Only three:
Moreover, the first two are really from one theorem -- any linearly independent set can be extended to a basis. But in the above it was simpler to think of these two steps separately.
3. Show that any finite field has #elements = a power of a prime.
3A. Call our field F. Assume first that F has a subfield with p elements, p a prime (we'll prove this later), which we'll call G.
If so, then we can make F into a vector space over G: we know how to rescale elements of F by elements of G (even by F, but forget that), and we know how to add elements of F. Then all the vector space axioms follow from F's field axioms.
Since F is finite, it's finitely generated (we could just use all of F's elements to span F!). So by our theorem, it's got a basis as a G-vector space, and therefore there's a G-linear isomorphism tau : F -> G^n for some n.
Finally, the stupid thing: since it's an isomorphism, it's a 1:1 correspondence of finite sets, and they therefore have the same number of elements. So |F| = |G^n| = p^n.
Okay, now we've got to do the hard part - find G. (Confession: this part is not so obviously appropriate for a linear algebra class. But the first part definitely was.)
One property we know of the fields with p elements that we're used to is that every element is a sum of 1s. Zero, 1, 2 = 1+1, ... and eventually it wraps back around and p = 1+1+...+1 = 0.
So consider the subset G = {0, 1, 1+1, ...} inside F (it's usually not all of F). To emphasize: we're using F's notion of + to find these elements, and it's kind of cool that we're not making any choices -- 0 and 1 are God-given elements of a field, and + is specified. For short we'll refer to 1+1+1 as "three", etc. Put another way, we've got a natural map Z->F.
This subset G is obviously closed under +, and obviously finite (since it's a subset of F), so at some point it repeats: for some m less than n in Z we have m 1s = n 1s. But by subtracting m 1s (which we can do inside F, it has subtraction) we find out 0 = (n-m) 1s. Therefore the first number repeated is 0; let p be the smallest positive number such that p 1s = 0.
If p is not prime, so p = ab, then we find out that (a 1s)(b 1s) = 0, even though a 1s and b 1s aren't zero. But then we could divide by a 1s and reach a contradiction. So p is prime.
So far G is a p-element subset, closed under +. By distributivity, it's closed under times also. Since n 1s + (p-n) 1s = 0, it's closed under -. Finally since p is prime it's got division. So G is a field.
4. Show any 2x2 invertible is (uniquely) the product of an orthogonal matrix, a diagonal matrix with positive real entries, and a shear matrix (upper triangular with ones on the diagonal).
4A. We're trying to write M = ODS. Look at M^T M = S^T D^T O^T O D S = S^T D^2 S. If S and D are given by
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| d^2 | sd^2 |
| sd^2 | s^2 d^2 + e^2 |
| v.v | v.w |
| v.w | w.w |
So we can read off
| e = | sqrt(s^2 d^2 + e^2 - (sd^2)^2/d^2) |
| = sqrt(w.w - (v.w)^2/(v.v)) | |
| = sqrt((w.w)(v.v) - (v.w)^2) / |v| | |
| = sqrt((w.w)(v.v) - (|w||v|cos theta)^2) / |v| | |
| = |w| sqrt(1 - cos theta^2) | |
| = |w| sin theta |
where theta is the angle between v and w. It's nonzero, since M was invertible, and positive, since the angle is between 0 and 180 degrees. (Though it won't be necessary, let's simplify s further: it's |v||w|cos theta / |v|^2 = cos theta |w|/|v|.)
The argument so far: if M can be written as ODS, then the only possibility for D and S are as we've determined above. It remains to be shown that an O exists to complete the list.
But it's obvious what the O would have to be (if it exists): M S^-1 D^-1. Actually, let's think about what O^-1 would have to be: D S M^-1. It remains to show that D S M^-1 is orthogonal.
Look at (DSM^-1)^T (DSM^-1) = M^-T S^T D^T D S M^-1 = M^-T S^T D^2 S M^-1. (Here M^-T is some overly cute notation for (M^-1)^T = (M^T)^-1.) Recall we set up S,D so that this middle thing would be M^T M. So we get (DSM^-1)^T (DSM^-1) = M^-T M^T M M^-1 = 1. QED.
5. Curtis 2.7 #4,6
#4. Show that S,T two 2-d subspaces of R^3 must intersect in dim at least 1.
A. Pick bases for each, s_1, s_2, and t_1, t_2. All together they give a four element list. If they were linearly independent, they could be extended to a basis of size at least 4, but R^3 is 3-dimensional (and we proved that all bases are the same size).
So they're linearly dependent; there's some nontrivial combination As_1 + Bs_2 + Ct_1 + Dt_2 = 0. Therefore As_1 + Bs_2 = -Ct_1 - Dt_2. This is not the zero vector (since s_1 and s_2 are linearly independent, likewise t_1,t_2, and there's some nonzero coefficient used), but the left side is in S, and the right side in T, so we've found a nonzero vector in S intersect T.
#6. How many elements, 1-d spaces, bases in (F_2)^2?
A. Four, three, six (counting ordered bases).