Math 317: Linear Algebra
Worksheet 1: 1.1,1.2,1.4
Fall 2015
Name:
1. (a) Suppose that x, y ∈ Rn are nonparallel vectors. Prove that if sx + ty = 0 then
s = t = 0. (Hint: It may be useful to use contradiction here. That is, assume
that either s 6= 0 or t 6= 0, and show that this is impossible by contradicting
the original hypothesis.)
Proof: Suppose by way of contradiction that s 6= 0. (We will later see
that the same argument works if we assume that t 6= 0 instead.) There
are two cases to consider. First, if t = 0, then sx + ty = sx = x = 0,
since s 6= 0. Since one of the vectors is 0, then x and y are neither
parallel nor nonparallel, contradicting the hypothesis that x, y ∈ Rn
are nonparallel vectors. Now let us suppose that t 6= 0. Then sx+ty =
0 =⇒ sx = −ty =⇒ x = − st y, which says that x is parallel to
y, (since x is a constant multiple of y, i.e x = cy with c = − st . This
is a contradiction to the hypothesis which states that x, y ∈ Rn are
nonparallel vectors. Thus s = 0. The argument works the same way
if t 6= 0. Thus, s = t = 0.
(b) Use (a) to prove that if ax + by = cx + dy, then a = c and b = d.
Proof: Once again, we assume that x, y ∈ Rn are nonparallel vectors.
Then ax + by = cx + dy =⇒ (a − c)x + (b − d)y = 0. But from part
(a), we know that a − c = 0 =⇒ a = c and b − d = 0 =⇒ b = d.
2. Using only those properties in Section 1.1, Problem 28, prove that if a + c = b + c
for some vectors a, b, c ∈ Rn , then a = b. This is the cancellation property that
some of you used (without proof) in your homework assignment.
Proof: To prove that a + c = b + c for some vectors a, b, c ∈ Rn =⇒
a = b, we use properties 28(b) and 28(d). 28(d) tells us that there is an
element, call it −c such that c + (−c) = 0. Using this fact, along with the
associativity property in 28(b), we obtain: a + (c − c) = b + (c − c) =⇒
a + 0 = b + 0 =⇒ a = b by 28(c).
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Math 317: Linear Algebra
Worksheet 1: 1.1,1.2,1.4
Fall 2015
3. Let u, v ∈ Rn and suppose that u − v is orthogonal to u + v. Prove that u and v
have the same magnitude.
Proof : Suppose that u−v is orthogonal to u+v. Then (u − v)·(u + v) = 0.
Using the associated properties of dot product, we obtain:
(u − v) · (u + v) =
=
=
=
u·u+u·v−u·v−v·v
u·u−v·v
kuk2 − kvk2
0.
Thus, kuk2 = kvk2 =⇒ kuk = ±kvk =⇒ kuk = kvk since kuk ≥
0, kvk ≥ 0. Thus u and v have the same magnitude.
4. (a) For some x ∈ Rn , show that x · y = x · z does not necessarily imply that
y = z by coming up with a suitable example. (Suggestion: Look for examples
in R2 . Is it possible for a vector x to be orthogonal to two different vectors? )
Following the suggestion above, we look for examples in R2 . Let x =
(1, 2). We look for a vector y that is orthogonal to x. (You certainly
don’t have to use this approach, but this is a nice way to come up with
an example for the problem above.) If y is orthogonal to x = (1, 2)
then (1, 2) · y = 0 =⇒ (1, 2) · (y1 , y2 ) = 0 =⇒ y1 + 2y2 = 0. From
here there are many values for y1 and y2 that we can pick which will
satisfy this equation. One set of values will serve as a vector for y and
another set of values will serve as a vector for z so that x·y = x·z = 0.
We see here that y = (2, −1) and z = (4, −2) will get the job done.
(b) Suppose that x · y = x · z for all x ∈ Rn . Show that y = z.
Proof: Suppose that x · y = x · z for all x ∈ Rn . We want to show
that y = (y1 , y2 , . . . , yn ) = z = (z1 , z2 , . . . , zn ) which is equivalent to
show that y1 = z1 , y2 = z2 , . . . yn = zn . Let x1 = (1, 0, . . . , 0). Then
x1 ·y = x1 ·z =⇒ 1y1 +0y2 +. . .+0yn = 1z1 +0z2 +. . .+0zn =⇒ y1 =
z1 . We see that we can do this for each component of y and z. For
i = 1, 2, . . . , n, let xi = (0, 0, . . . , 1, . . . , 0) where 1 is the ith element of
the vector xi . Then xi · y = xi · z =⇒ 0y1 + 0y2 + . . . + 1yi + . . . 0yn =
0z1 + 0z2 + . . . + 1zi + . . . + 0zn =⇒ yi = zi . Since this true for each
i = 1, 2, . . . , n, then y = z.
5. Determine if (3, 4, −1, 6) lies in span {(1, 2, −1, 2), (−2, 3, 1, −1), (−1, 3, 2, 1)}.
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Math 317: Linear Algebra
Worksheet 1: 1.1,1.2,1.4
Fall 2015
Recalling that span {(1, 2, −1, 2), (−2, 3, 1, −1), (−1, 3, 2, 1)} is the set of
all linear combinations of (1, 2, −1, 2), (−2, 3, 1, −1), (−1, 3, 2, 1), asking if
(3, 4, −1, 6) lies in span {(1, 2, −1, 2), (−2, 3, 1, −1), (−1, 3, 2, 1)} is equilvalent to asking if there are constants c1 , c2 and c3 such that
 
 
 
 
3
1
−2
−1
4
2
3
3
  = c1   + c2   + c3   .
1
−1
1
2
6
2
−1
1
This is equivalent to the following system of equations:
3
4
1
6
=
=
=
=
c1 − 2c2 − c3
2c1 + 3c2 + 3c3
−c1 + c2 + 2c3
2c1 − c2 + c3 .
We can write this system of equations in matrix form as follows:


 
1 −2 −1  
3
c1
 2


3
3    4

c = 
 −1 1
2  2
1
c3
2 −1 1
6
To solve this system of equations, we can put this system in augmented
matrix form and find its corresponding reduced row echelon form. The
augmented matrix [A|b] is given as

1 −2 −1 3
 2
3
3 4 
.

 −1 1
2 1 
2 −1 1 6

We row reduce the augmented matrix, starting with 1 (boxed in) as our
pivot entry and making each entry (in the same column) below 1 a zero.


[1] −2 −1 3
 2
3
3 4 
3 →R3

 R1 +R
−→
 −1 1
2 1  −2R1 +R2 →R2
−2R1 +R4 →R4
2 −1 1 6


1 −2 −1 3
 0 [7] 5 −2  (1/7)R2 →R2


−→
 0 −1 1
4 
0
0 3
3
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Math 317: Linear Algebra
Worksheet 1: 1.1,1.2,1.4

1 −2 −1
3
 0 [1] 5/7 −2/7  R2 +R3 →R2


−→
 0 −1 1
4  −3R2 +R4 →R4
0
0 3
3

Fall 2015


1 −2 −1
3
 0 1
5/7 −2/7 
3 +R4 →R4

 −(1/2)R−→
 0 0 [12/7] 26/7 
6/7
0 0
6/7


1 −2 −1
3
 0 1 5/7 −2/7 


 0 0 12/7 26/7 
0 0
0
−1
The last row in our augmented matrix is equlivalent to 0c1 + 0c2 + 0c3 = −1
which doesn’t make any sense. Thus, this system is inconsistent and so
there is not a set of scalars c1 , c2 and c3 so that b is a linear combination of
(1, 2, −1, 2), (−2, 3, 1, −1), (−1, 3, 2, 1).
6. Determine the reduced row echelon form of A and give the solution to Ax = 0 in
standard form.


2 −2 4
A = −1 1 −2
3 −3 6
We can simultaneously solve Ax = 0 and find the reduced row echelon form
of A by starting with the coorsponding augmented matrix [A|0].



[2] −2 4 0
1 →R1
 −1 1 −2 0  (1/2)R

−→
3 −3 6 0

[1] −1
 0 0
0 0

[1] −1 2 0
R1 +R2 →R2
−1 1 −2 0 
−→
−3R1 +R3 →R3
3 −3 6 0

2 0
0 0 
0 0
From the reduced row echelon form of A we see that x1 is our pivot variable
and x2 , x3 are our free variables. From the first equation, we have that
x1 − x2 + 2x3 = 0 =⇒ x1 = x2 − 2x3 . In standard form, the solution
becomes
  
   

 
 
x1
x2 − 2x3
x2
−2x3
1
−2
x2  =  x2  = x2  +  0  = x2 1 + x3  0 
x3
x3
0
x3
0
1
Note that this says that the solution set to Ax = 0 is precisely
span {(1, 1, 0), (−2, 0, 1)}.
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