Math 317: Linear Algebra
HW 1 Solutions
Fall 2015
1. Given 4ABC, let M and N be the midpoints of AB and AC, respectively. Prove
−−→
−−→
that M N = 12 BC.
Figure 1: 4ABC
Proof: Consulting the Figure 1 above, we let x = AB and y = AC. Via
−−→
the parallelogram law, we observe that BC = x − y. Now, since M is the
midpoint of AB, then AM = 12 AB = 12 x. In a similar manner, AN = 21 y.
−−→
Once again from the parallelogram law, we note that M N = 21 x − 12 y =
−−→
1
(x − y) = 12 BC.
2
2. Suppose that v, w ∈ Rn and c is a scalar. Prove that Span(v+cw, w) = Span(v, w).
Proof: To prove set equality, we show that Span(v + cw, w) ⊂ Span(v, w)
and Span(v, w) ⊂ Span(v + cw, w). Suppose that x ∈ Span(v + cw, w).
Then x can be written as a linear combination of v + cw and w. That
is, x = c1 (v + cw) + c2 w for some scalars c1 and c2 . This implies that
x = c1 v + (c1 c + c2 )w, so that x is a linear combination of v and w.
Thus x ∈ Span(v, w). Now to show containment on the other side, we
let x ∈ Span(v, w). Then x = c1 v + c2 w for some scalars c1 and c2 .
We add and subtract c1 cw to get that x = c1 v + c2 w + c1 cw − c1 cw =
c1 (v + cw) + (c2 − c1 c)w. Thus, we now have now expressed x as a linear
combination of v + cw and w. Thus x ∈ Span(v + cw, w). This shows that
Span(v, w) ⊂ Span(v + cw, w). Since we have shown containment on both
sides, we have that Span(v + cw, w) = Span(v, w).
3. Suppose the vectors v and w are both linear combinations of v1 , . . . , vk .
(a) Prove that for any scalar c, cv is a linear combination of v1 , . . . , vk .
1
Math 317: Linear Algebra
HW 1 Solutions
Fall 2015
Proof : Suppose that v is a linear combination of v1 , . . . , vk . Then
there are scalars c1 , c2 , . . . , ck so that v = c1 v1 + c2 v2 + . . . + ck vk .
Thus we have that cv = c(c1 v1 + c2 v2 + . . . + ck vk ) = cc1 v1 + cc2 v2 +
. . . + cck vk = ĉ1 v1 + ĉ2 v2 + . . . ĉk vk where ĉi = cci are scalars. Thus
cv is a linear combination of v1 , v2 , . . . , vk .
(b) Prove that v + w is a linear combination of v1 , . . . , vk .
Proof : Suppose that v and w are linear combinations of v1 , . . . , vk .
Then v = c1 v1 +. . .+ck vk and w = d1 v1 +. . .+dk vk where the c0i s and
d0i s are scalars. Thus v +w = (c1 v1 +. . .+ck vk )+(d1 v1 +. . .+dk vk ) =
(c1 + d1 )v1 + . . . + (ck + dk )vk which shows that v + w is a linear
combination of v1 , . . . , vk .
4. (a) Using only the properties listed in Exercise 28, prove that for any x ∈ Rn , we
have 0x = 0.
Proof: We begin by observing that by 28(c) (applying this both for the
scalar 0 and the vector 0) we have that 0x = (0 + 0) x and 0x+0 = 0x.
By the distributive property (i.e. 28g), we note that 0x = (0 + 0) x =
0x + 0x. We see that we have two different representations for 0x and
thus we may equate them to obtain: 0x + 0 = 0x + 0x. Now, by
28d, 0x has an additive inverse, that is, there is a term, call it −0x
such that 0x + (−0x) = 0. Using the associativity property in 28b to
reorder terms, we obtain: (−0x + 0x) + 0 = (−0x + 0x) + 0x which
then gives (using 28c) 0x = 0.
(b) Using the result of part a, prove that (−1)x = −x.
Proof: We begin by noting that −x is the element such that x+(−x) =
0. Thus, we begin by proving that x + (−1x) = 0. Using 28h and
28f, we obtain x + (−1x) = 1x + (−1x) = (1 − 1) x = 0x = 0, where
the last equation is true from part (a) of this exercise. Using 28c,
28d and the associativity property in 28b to reorder terms, we obtain
(−x+x)+(−x) = (−x+x)+−1x =⇒ 0−x = 0−1x =⇒ −x = −1x.
5. Starting with the vectors a = (a1 , a2 ) and b = (b1 , b2 ), draw a triangle (in Quadrant
I) whose sides are the magnitude of a and b. Write the third side of the triangle in
terms of a and b. Let φ be the angle opposite of the side of the triangle that you
have written in terms of a and b. The law of cosines asserts that
c2 = kak2 + kbk2 − 2kakkbk cos φ,
(1)
2
Math 317: Linear Algebra
HW 1 Solutions
Fall 2015
where c is the third side of the triangle that you have written in terms of a and b.
Using (1) and the geometric form of the definition of dot product, show that
a · b = a1 b 1 + a2 b 2
(2)
Figure 2: Labelled Triangle
Proof : Using the parallelogram law for the triangle given above in Figure
2, we find that c = ka − bk and thus from the law of cosines we have that
ka − bk2 = kak2 + kbk2 − 2kakkbk cos φ,
(3)
where φ is the angle between vectors a and b. Now a − b = (a1 − b1 , a2 − b2 )
if a = (a1 , a2 ) and b = (b1 , b2 ). Thus, by definition of the magnitude of a
vector, we have that ka − bk2 = (a1 − b1 )2 + (a2 − b2 )2 = a21 − 2a1 b1 + b21 +
a22 − 2a2 b2 + b22 . By the geometric definition of the dot product, we have that
kakkbk cos φ = a·b, and thus (using the fact that kak2 = a21 +a22 and kbk2 =
b21 +b22 ), we find that kak2 +kbk2 −2kakkbk cos φ = a21 +a22 +b21 +b22 −2(a·b).
Thus we can rewrite (3) as
a21 − 2a1 b1 + b21 + a22 − 2a2 b2 + b22 = a21 + a22 + b21 + b22 − 2(a · b),
(4)
which simplifies precisely to
a · b = a1 b 1 + a2 b 2 .
√
6. Suppose that x, y ∈ Rn , kxk = 2, kyk = 1, and the angle between x and y is
3π/4. Show that the vectors 2x + 3y and x − y are orthogonal.
3
Math 317: Linear Algebra
HW 1 Solutions
Fall 2015
Proof : We wish to show that 2x + 3y · x − y = 0. Using associated
properties of dot product, as well as the geometric definition of the dot
product: x · y = kxkkyk cos θ, we obtain
(2x + 3y) · (x − y)
2x · x − 2x · y + 3x · y − 3y · y
2kxk2 + x · y − 3kyk2
2kxk2 + kxkkyk cos θ − 3kyk2
√ 2
√
2( 2) + ( 2)(1) cos(3π/4) − 3(1)
4−1−3
=
=
=
=
=
= 0
7. Prove the triangle inequality: For any vectors x, y ∈ Rn , kx + yk ≤ kxk + kyk.
Scratch work : Before I prove this claim, I need to do some scratch work to determine
where I should begin in my proof. Following the hint, we may compute kx + yk2
using a proposition proved in class to obtain:
kx + yk2 = kxk2 + 2(x · y) + kyk2 .
Similarly, we can square the right hand side of the triangle inequality to obtain
(kxk + kyk)2 = kxk2 + 2kxkkyk + kyk2 .
Now, if the triangle inequality were true, then kx+yk ≤ kxk+kyk =⇒ kx+yk2 ≤
(kxk + kyk)2 =⇒ kxk2 + 2(x · y) + kyk2 ≤ kxk2 + 2kxkkyk + kyk2 =⇒ x · y ≤
kxkkyk. The last inequality is exactly the CS (or CBS) inequality that we proved
in class. Thus my scratch seems to suggest that if I start off with the CS inequality,
which I know is true, I should be able to deduce the triangle inequality.
Proof: Suppose that x, y ∈ Rn . By the CS (or CBS) inequality, we have
the following:
x · y ≤ kxkkyk
kxk + 2 (x · y) + kyk2 ≤ kxk2 + 2kxkkyk + kyk2
kx + yk2 ≤ (kxk + kyk)2
kx + yk ≤ kxk + kyk.
2
=⇒
=⇒
=⇒
4