Problem Set 9 (due November 12)
MATH 110: Linear Algebra
Each problem is worth 10 points.
PART 1
1. Curtis p. 161 10.
2. Curtis p. 226 7.
3. Curtis p. 243 6.
Solutions: 1. Vandermonde Determinant: Let Dbe the Vandermonde
determinant and let A=Qi<j(ζi−ζj). Let x=ζi. Then by expanding across
the ith row of the Vandermonde matrix in the determinant calculation, we
see that D=p(x) where pis a polynomial of degree n−1. In fact, since ζj
is a root of pfor all j6=i(since if x=ζjwe have two rows of the matrix
equal and hence its determinant is 0) we have that D=c(x−ζ1· · · (x−ζn)
with the term (x−ζi) missing in the product. Since iis arbitrary in the
argument above, we see that D=kA where kis a polynomial in ζ1,...ζn.
But examining the coefficient of ζn−1
1ζn−2
2· · · 1 in D, we see that it is either
1 or −1, and therefore kmust be either 1 or −1.
2. Similarity of Diagonal matrices: Two similar matrices have the same
eigenvalues, and diagonal matrices have as their diagonal entries their eigen-
values. Using these two facts, we see that if the matrices are diagonal and
similar, then certainly their diagonals are rearrangements of each other (since
they must have the same elements, namely the eigenvalues). Conversely, if
the diagonal elements are rearrangements of each other, then the matrices are
similar because they represent the same linear transformation, with respect
to the same basis except with the basis elements rearranged.
3. We are assuming that Vis finite dimensional. Suppose that λv=λw.
Then for any linear functional f, we have that f(v) = f(w). Therefore
f(v−w) = 0. Let B={v1, . . . , vn}be a basis for Vand C={fi}be the
dual basis of B. Then in particular, fi(v−w) = 0 for every i, and so if we
write v−w=Pn
i=1 civi, we have that ci= 0 for every i. Therefore v−w= 0
and v=w. Thus we have that our map is 1 −1, and it follows that it is an
isomorphism, since dim(V∗)∗) = dim(V). To see this, let D={µi}be the
dual basis of C, that is µi(fj) = 0 if j6=iand 1 if j=i. Notice that in fact,
1
