SELECTED SOLUTIONS FROM THE HOMEWORK
ANDREW J. BLUMBERG
1. Solutions
(1.1, 14) Show that every unit vector in R2 is of the form [cos θ1 , cos θ2 ], where θ1 is
the angle the vector makes with the positive x axis and θ2 is the angle the
vector makes with the positive y axis.
Proof. A direct argument for this can be made using trigonometric considerations; a unit vector specifies a point on the circle of radius 1, and such
a point can be expressed as (cos θ1 , sin θ1 ). Since sin θ1 = cos θ2 , we can
conclude the desired result.
Alternatively, we can use the dot product to handle this easily as follows.
A unit vector v ∈ R2 can be written v = v · e1 + v · e2 . Since v · e1 = cos θ1
and v · e2 = cos θ2 , the result follows.
(1.1, 25) If x is a vector in Rn and c1 6= c2 , show that c1 x = c2 x implies that x = 0.
Proof. Consider the ith coordinate xi of x. By the definition of scalar
multiplication, c1 xi = c2 xi . If c1 = 0, then c2 xi = 0 and since c2 6= 0,
xi = 0. Similarly, if c2 = 0 then we can conclude that xi = 0. If both c1
and c2 are nonzero, then xi = cc12 xi ; since c1 6= c2 , this is true only if xi = 0.
Since i was arbitrary, x = 0.
(1.2, 3) (a) Show that [a, b] and [−b, a] are orthogonal.
(b) Show that the lines given by the equations ax + by + c = 0 and bx −
ay + d = 0 are perpendicular.
Proof. For the first part, we compute the dot product and find that
3[a, b] · [−b, a] = a(−b) + ba = −ba + ba = 0,
so the two vectors are orthogonal.
For the second part, we proceed (as the text suggests) by finding vectors
in the direction of the lines and then taking the dot product. Let us first
assume that a and b are nonzero. Then we can find points on each line by
setting x and y to zero respectively; for ax + by + c = 0, we get (0, − cb ) by
setting x = 0 and (− ac , 0) by setting y = 0. Subtracting these, we get the
vector (− ac , cb ). Carrying out the analogous calculation on bx − ay + d = 0
leads to (0, ad ) and (− db , 0) and hence the vector ( db , ad ). Taking the dot
product of these two vectors yields 0. If a = b = 0, then the question is
vacuous, and if one is zero then we have to plug in a different value for
either x or y in order to avoid dividing by 0 — but otherwise the procedure
is analogous.
Date: February 9, 2015.
1
2
ANDREW J. BLUMBERG
(1.2, 9) Prove that if (x + y)(x − y) = 0, then ||x|| = ||y||.
Proof. Expanding, we find that
(x + y)(x − y) = x · x − x · y + y · x − y · y = ||x||2 − ||y||2 ,
and so ||x||2 = ||y||2 . Since the norm of a vector is always nonnegative, we
can take the square root of both sides and deduce that ||x|| = ||y||.
(1.2, 11) (a) Prove that ||x + y||2 = ||x||2 + ||y||2 if and only if x · y = 0.
(b) Prove that if x, y, z are mutually orthogonal then ||x + y + z||2 =
||x||2 + ||y||2 + ||z||2 .
(c) Prove that x · y = 41 (||x + y||2 − ||x − y||2 ).
Proof. (a) Expanding, we find that
||x + y||2 = (x + y) · (x + y) = x · x + 2x · y + y · y = ||x||2 + ||y||2 + 2x · y.
Clearly, if x · y = 0, then 2x · y = 0 and the desired conclusion holds.
On the other hand, if ||x + y||2 = ||x||2 + ||y||2 , then subtracting from
both sides we obtain 0 = 2x · y and so x · y = 0.
(b) Since ||x + y + z + ||2 = (x + y + z) · (x + y + z), the result follows by
expanding and observing that by hypothesis x · y = x · z = y · z = 0.
(c) This again follows by expanding out the right-hand side.
(1.2, 12) Given x, y, z with x orthogonal to both y and z, prove that x is orthogonal
to c1 y + c2 z.
Proof. Computing, we have that
x · (c1 y + c2 z) = x · (c1 y) + x · (c2 z) = c1 (x · y) + c2 (x · z).
By hypothesis, (x·y) = (x·z) = 0, and so the desired conclusion follows. (1.4, 4) Prove that if AT = B T , then A = B.
Proof. By definition AT = B T implies that (AT )ij = (B T )ij for all i and
j, which implies that Aji = Bji for all i and j, and so A = B.
(1.4, 6) (a) Prove that if A and B are diagonal, then A + B is diagonal.
(b) Prove that if A and B are symmetric, then A + B is symmetric.
Proof.
Since A and B are diagonal, Aij and Bij are zero for off-diagonal elements.
By definition, (A + B)ij = Aij + Bij , and so (A + B)ij = 0 is i 6= j.
If A and B are symmetric, Aij = Aji and Bij = Bji for all i and j. By
definition,
(A + B)ij = Aij + Bij = Aji + Bji = (A + B)ji ,
so we conclude that (A + B) is symmetric.
(1.4, 7) Use induction to prove that if A1 , A2 , . . . , An are upper triangular matrices
of the same size, then Σni=1 Ai is upper triangular.
SELECTED SOLUTIONS FROM THE HOMEWORK
3
Proof. We start with the base case of n = 2. A matrix A is upper triangular
if Aij = 0 for i > j. Since (A1 + A2 )ij = (A1 )ij + (A2 )ij , we see that this
quantity is 0 for i > j and so A1 + A2 is upper triangular. Now assume the
result is true for n ≤ m and consider the sum Σm+1
i=1 Ai . Rewriting this as
(Σm
i=1 Ai ) + Am+1 ,
applying the induction hypothesis for n = m we see that Σm
i=1 Ai is upper
triangular and applying the induction hypothesis for n = 2 we conclude
that the sum above is upper triangular.
E-mail address: [email protected]