Midterm #2 for Math H110, fall '02
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1. Let M be a square matrix in JCF with exactly two eigenvalues a,b.
What can M look like, if ker M plus Image(M) adds up to the whole space?
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A. First, let's name the "whole space" F^n.
Now count dimensions: <br>
dim (ker M + Im M) = dim ker M + dim Im M - dim (ker M intersect Im M)
= n - dim (ker M intersect Im M)<br>
(this second step using nullity + rank theorem). So it comes down to
whether ker M intersect Im M is nontrivial.
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To be in the bad case, we have some v = Mu 
a nonzero vector in ker M and Im M.
So ker M is nonzero, in particular one of a or b must be zero.
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Now let's look at a nilpotent block (which we only have because one of
a or b is zero). If it's of size 1 then it's zero, and doesn't contribute
to the image. If it's bigger, then notice that if we let u be its second
basis vector, we get v being its first.
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So far we've determined that if a,b are both nonzero, M is good.
And if one of them is zero, and there's a nilpotent block of size >1
somewhere, then M is bad. Finally, if one eigenvalue is zero and all
the nilpotent blocks are size 1, then we check that ker M and Im M
are spanned by different parts of the basis. 
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So the answer is "if a,b are both nonzero, then any JCF; if one of them
is zero, then all 0-blocks must be size 1".
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<hr>
2. What are the JCFs of these five matrices?
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i-iv. In these cases the upper triangularity makes it trivial to
compute the characteristic polynomials - the eigenvalues are already
down the diagonal.
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i, ii. In these two cases the matrix satisfies a squarefree polynomial.
Therefore it's diagonalizable, and you already know the eigenvalues.
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iii, iv. In these two cases look at the powers of M-1*Identity. 
The cube is nonzero; therefore we only have one line of suicides.
So there's just one JCF block.
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v. In this case, most people noticed the det is zero. That doesn't say
that <em>all</em> the eigenvalues are zero, just that some are. To figure
out how many, figure out the image -- it's vectors (a,a,a,a). Since there's
a 1-d image, there's a 3-d kernel, giving three 0-eigenvectors. And
(1,1,1,1) is an eigenvector; if we plug it in, we see it's a 4-eigenvector.
So the matrix is diagonalizable, with diagonal (4,0,0,0).
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<hr>
3. Let T: V->V have char poly p, and T': V/ker T -> V/ker T have char poly q,
where T' is the induced map on the kernel. What's p/q?
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A. Lots of people did it this way: pick a basis for ker T, extend to a 
basis of V, draw the matrices, and figure out that p = q * (char poly
of T restricted to ker T). So what is the map T: ker T -> ker T?
It's zero! So its characteristic polynomial is lambda^(dim ker T),
or maybe minus that, whatever.