Problem Set 6 Solutions
MATH 110: Linear Algebra
Each problem is worth 5 points.
PART 1
1.
2.
3.
4.
Curtis
Curtis
Curtis
Curtis
p.
p.
p.
p.
150
150
182
182
2.
6.
1.
2.
Solution to 2: Suppose u1 , u2 , . . . , un are orthonormal vectors in Rn .
Write them as row vectors in a matrix A. Notice that AAt = I (At denotes the transpose of A, i.e. the matrix obtained by flipping the entries
of A across the diagonal). Furthermore, det(A) = det(At ) because the determinant can be calculated by expanding by rows or by columns. Thus,
det(AAt ) = det(A)2 = 1, hence the result. Note that we have only shown
that the determinant can be 1 or −1. It is clear that the determinant can be
1 (for example if ui = ei , the ith unit vector). It is not hard to see that there
are examples with the determinant equal to −1.
Remember that the starred problem is non collaborative.
Problem 1(10)
Two matrices A and B are said to be similar if there exists a matrix C
such that B = C −1 AC. Show that two matrices that are similar have the
same determinant.
Proof: det(B) = det(C −1 AC) = det(C −1 )det(A)det(C) = det(C −1 )det(C)det(A) =
det(C −1 C)det(A) = det(I)det(A) = det(A).
Problem 2(10)
For each of the following statements about square matrices, give a proof
or find a counter example:
a) det(A + B) = det(A) + det(B).
False. Counterexample: A = B = I are 2 × 2 matrices.
b) det((A + B)2 ) = (det(A + B))2 .
Proof: det(A + B)2 ) = det(A + B)det(A + B) = (det(A + B))2 .
c) det((A + B)2 ) = det(A2 + 2AB + B 2 ).
False, because A and B do not necessarily commute. Counterexample:
Let A, B be 2 × 2 matrices where A is the identity with an extra 1 in the
1
upper right hand corner and B has two ones in the first column and two
zeros in the second. Then the left hand side of c) is 1 and the right hand
side is 0.
d) det((A + B)2 ) = det(A2 + B 2 ).
False. Counterexample: A = B = I are 2 × 2 matrices.
Problem 3 (15)
Given n2 functions fij on an interval (a, b) define F (x) = det[fij (x)] for
each x in (a, b). Prove that the derivative F 0 (x) is a sum of n determinants
F 0 (x) =
n
X
detAi (x).
i=1
where Ai (x) is the matrix obtained by differentiating the functions in the ith
row of [fij (x)].
P
Q
Solution: F (x) = σ∈Pn (σ) ni=1 fi,σ(i) . Here σ is a permutation of
1, 2, . . . , n and Pn is the set of all permutations. (σ) is the sign of σ, which
is either plus or minus one. Taking the derivative of F (x) we find that
F 0 (x) =
X
σ∈Pn
(σ)
n
X
0
fj,σ(j)
j=1
n
Y
fi,σ(i) .
i=1,i6=j
This is equal to
n X
X
0
(σ)fj,σ(j)
j=1 σ∈Pn
n
Y
fi,σ(i)
i=1,i6=j
which is simply nj=1 detAj .
Problem 4 (10)
Prove Demoivre’s theorem:
P
(cos θ + i sin θ)n = cos nθ + i sin nθ.
Proof:
(cos θ + i sin θ)n = (eiθ )n = ei(nθ) = cos nθ + i sin nθ.
PART 3 - Optional Problem
Let f, g, h be relatively prime nonzero polynomials (not all constant) with
f + g = h.
Define n0 (f ) to be the number of distinct roots of a polynomials f . Show
that
max(deg(f ), deg(g), deg(h)) ≤ n0 (f gh) − 1.
2