Problem Set 8 (due Friday October 29)
MATH 110: Linear Algebra
Each problem is worth 10 points.
PART 1
1. Curtis p. 201 6.
2. Curtis p. 215 9.
1: Notice that if T2= 1 then the minimal polynomial of T m(x) divides
(x−1)(x+ 1). There are 3 cases: m(x) = x+ 1, m(x) = x−1, m(x) =
(x+ 1)(x−1). In each case we can apply the triangular form theorem to find
that V+and V−are exactly the nullspaces appearing in the direct sum.
2: a) It suffices to show that AB and BA have the same eigenvalues.
Suppose λis an eigenvalue of AB. Then ABx =λx so BABx =λBx so Bx
is an eigenvector of BA with eigenvalue λ. Similarly, if µis an eigenvalue
of BA then it is also an eigenvalue of AB. Therefore AB and BA have the
same eigenvalues. So their traces, which are the sum of their eigenvalues, are
equal. b) A=C−1BC or equivalently CA =BC for some C. Now
trace(B) = trace(BCC −1) = trace(C AC −1) = trace(AC C−1) = trace(A).
c) We think of trace as a linear transformation T r :Mn→F. We are in-
terested in the nullity (or dimensional of the nullspace of T r). This can be
computed from the rank and nullity theorem to be n2−rwhere ris the
dimension of the range of T r, which is 1. d) Matrices of the form AB −BA
have trace 0 from part a) and therefore the subspace generated by them is
a subspace of the nullspace of T r. Call this subspace Z. Let A, B simulta-
neously range over all matrices with only one 1 and all the rest zero, where
the 1 does not lie on the diagonal. Looking at AB −BA, we get matrices
that are either 0, or else have only one 1 in some position not on the di-
agonal, or else 1,−1 consecutively somewhere on the diagonal. There are
n2−n+n−1 = n2−1 such matrices, and they are clearly independent,
therefore Zhas dimension at least n2−1. It follows that Zis the nullspace
of T r.
Note: part d) should have read: “is generated by matrices of the form
AB −BA where A, B ∈Mn(F)”.
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