Homework #5 for Math H110, fall '02

1. Let V_1,...,V_n be subspaces of a vector space V, such that V = sum_i V_i (not the direct sum, just the sum). Assume for each V_i, and b subspaces V_{j_1}...V_{j_b} not including V_i, that V_i intersect (V_{j_1} + ... V_{j_b}) = 0.

Show that V is the direct sum of V_1...V_n, i.e. every vector in V is uniquely the sum of a vector from each V_i.

A. We'll show that for any b+1 subspaces, that the sum of those subspaces is a direct sum, by induction on b, and eventually get to b=n-1.

The base case is b=1. Then we're told that each V_i intersects each V_j trivially. So any vector in their sum can indeed be written uniquely.

Now we want to show that V_i + V_{j_1} + V_{j_2} + ... + V_{j_b} is a direct sum. Equivalently, we want to know that there isn't a nontrivial way to write 0 = v + v_1 + v_2 + ... + v_b, where v \in V_i, and v_k \in V_{j_k} for each k=1...b.

If there were, we'd have -v = v_1 + v_2 + ... + v_b, so this vector would be in both V_i and in the sum of V_{j_1}+...+V_{j_b}. By our assumption (not the induction assumption), that vector -v must be zero.

Okay, now we have 0 = v_1 + ... + v_b. But by our induction assumption, those b spaces added together are a direct sum. So each v_i is zero.

That shows that V_i + V_{j_1} + V_{j_2} + ... + V_{j_b} is a direct sum, with b+1 terms. Repeat the induction until we get to b+1=n and we get the desired result.

2. Let V be a 5-dimensional space, and T : V->V a linear transformation such that T^3 = 0 and rank(T)=3. Determine the Jordan canonical form of T.

A. Since it's got T^3=0, none of the blocks can be larger than 3. That says that the blocks come as 3+2, or 3+1+1, or 2+2+1, or 2+1+1+1, or 1+1+1+1+1 (the zero matrix).

Then the rank is the number of 1s, i.e. 5 minus the number of zero rows. There's one zero row for each block, so the number of blocks is 5-rank = 2.

The only partition listed above with two blocks is 5 = 3+2. So the JCF is
010..
001..
000..
...01
...00

3. Same V,T as in problem 2. Consider the subspaces ker T, ker T^2, ker T^3, Im T, Im T^2, Im T^3. Which subspaces are contained in, or equal to, which others?

A. It's easiest to do this with a JCF basis, which I'll call e1,e2,e3,e4,e5, and the suicide diagram e3->e2->e1->0, e5->e4->0. Then each of our subspaces is a span of a subset of the JCF basis:

  • ker T contains e1, e4
  • ker T^2 contains e1, e2, e4, e5
  • ker T^3 contains e1, e2, e3, e4, e5
  • Im T contains e2, e1, e4
  • Im T^2 contains e1
  • Im T^3 contains no basis vectors.
    • So in fact none are equal to one another: ker T^3 > ker T^2 > Im T > ker T > Im T^2 > Im T^3.

    4. Let T be a diagonal matrix. Determine all the matrices that commute with T. (E.g. if T=0, it's _all_ matrices.)

    A. M commutes with T if the only nonzero entries M_ij are those such that T_ii = T_jj.

    The nicest case is when T has its eigenvalues all grouped together. Then M is block diagonal.

    5. Let T : F^n -> F^n be zero except for all 1s just above the diagonal, so T^n = 0, but T^{n-1} nonzero. Determine all the matrices that commute with T.

    A. Just the polynomials in T. These are the "banded" matrices that are upper triangular and constant along diagonals.

    6. Same question as 5, except now there are only some 1s above the diagonal (i.e. T is a nilpotent matrix in JCF).

    A. Let the blocks of T be of sizes a,b,c,d (maybe there are more than four, but whatever). Break up M into 16 rectangles, by separating the columns and also the rows, into these groups of sizes a,b,c,d. So above the diagonal blocks (which are square) we find an axb rectangle, etc.

    Then MT=TM says that each of these blocks must be banded and upper triangular, meaning only in the NE triangle.

    For example, if T is zero, then we're breaking M into n^2 1x1 blocks, and asking that they be banded upper triangular. Every 1x1 matrix is. So any M commutes with zero, hurrah.