Math 416 Midterm 1. Solutions.
Question 1, Version 1. Let us define an addition and a scalar multiplication on R2 as
(x1 , x2 ) + (y1 , y2 ) = (x1 + y1 , x2 + y2 ),
c(x1 , x2 ) = (cx1 , x2 ).
Why do these operations not make R2 into a vector space?
Solution: The simplest way to see this is that we know that for any vector v, 0v = 0. But
notice that since this is “regular” addition for vectors, the zero vector is then 0 = (0, 0). But
we also have 0(v1 , v2 ) = (0, v2 ) which is not the zero vector for v2 6= 0. Thus we must not
have a vector space.
If we are checking axioms, we see that axiom #8 fails:
(a + b) · (x1 , x2 ) = ((a + b)x1 , x2 ),
while
a · (x1 , x2 ) + b · (x1 , x2 ) = (ax1 , x2 ) + (bx1 , x2 ) = ((a + b)x1 , 2x2 ),
and these are not the same!
Question 1, Version 2. Let us define an addition and a scalar multiplication on R2 as
(
(cx1 , x2 /c), c 6= 0,
(x1 , x2 ) + (y1 , y2 ) = (x1 + y1 , x2 y2 ), c(x1 , x2 ) =
(0, 0),
c = 0.
Why do these operations not make R2 into a vector space?
Solution: We first note that the zero vector in this space must be (0, 1), since
(x, y) + (0, 1) = (x + 0, y · 1) = (x, y).
But then we must have 0 · (x, y) = (0, 1) for any x, y (since 0 · v = 0). But according to the
definition, 0 · (x, y) = (0, 0).
If we are checking axioms, again axiom #8 fails: we write
(a + b) · (x1 , x2 ) = ((a + b)x1 , x2 /(a + b)),
while
a · (x1 , x2 ) + b(x1 , x2 ) = (ax1 , x2 /a) + (bx1 , x2 /b),
and in general
x2
x 2 x2
6=
+ .
a+b
a
b
Question 1, Version 3. Let us define an addition and a scalar multiplication on R2 as
(x1 , x2 ) + (y1 , y2 ) = (x1 + y1 , x2 − y2 ),
c(x1 , x2 ) = (cx1 , −cx2 ).
Why do these operations not make R2 into a vector space?
Solution: Recall that Axiom #5 says that
1x = x,
but here we have 1(2, 3) = (2, −3) 6= (2, 3).
Question 2, Version 1. Write down the solution set of the system
2x1 + 3x2 + 4x3 = −2,
x1 + 2x2 + 3x3 = −2,
2x1 + x2 + 4x3 = −2.
Solution: We write this as an augmented matrix and then do row reductions.
 
2 3 4 −2
1
 1 2 3 −2  ,  2
2 1 4 −2
2


1 2
3 −2
 0 −1 −2 2  
0 −3 −2 2

 
1 2 3 1
−2
 0 1 2 −2  ,  0
0 0 1 −1
0

 

−2
2 3 −2
1
2
3
 
,
3 4 −2 , 0 −1 −2 2
1 4 −2
2 1
4 −2

 
1 2 3 1 2
3 −2
−2

 
0 1
2 −2 , 0 1 2 −2 ,
0 0 4 −4
0 −3 −2 2
 

2 3 1 0 0 −2
1
, 0 1 0 0 .
1 0 0
0 1 −1
0 0 1 −1
So there is a unique solution, and it is x1 = 1, x2 = 0, and x3 = −1.
Question 2, Version 2. Write down the solution set of the system
x1 + 3x2 + 4x3 = 8,
x1 + 2x2 − 3x3 = 0,
2x1 − x2 + 4x3 = 5.
Solution: We write this as an augmented matrix and then do row reductions.
 
8
1 3
4 8
1
3
4
 1 2 −3 0  ,  0 −1 −7 −8
2 −1 4 5
2 −1 4 5

 
8
1 3
4 8
1
3
4
 0 1
7 8 , 0 1 7 8
0 −7 −4
−11
0 0 45 45
 
 

5
1 3 4 8
1
0
4
1
 0 1 0 1 , 0 1 0 1 , 0
0 0 1 1
0 0 1 1
0


1 3
4 8
 ,  0 −1 −7 −8  ,
0 −7 −4 −11

 
1 3 4 8
, 0 1 7 8 ,
0 0 1 1

0 0 1

1 0 1 .
0 1 1
 
So there is a unique solution, and it is x1 = 1, x2 = 1, and x3 = 1.
Question 2, Version 3. Write down the solution set of the system
x1 + 3x2 + 4x3 = 10,
x1 − 2x2 + 3x3 = −1,
x1 + x2 + 4x3 = 6.
Solution: We write this as an augmented matrix and then do row reductions.
 
1
1 3 4 10
 1 −2 3 −1  ,  0
1
1 1 4 6

 
1 3
4 10
 0 −5 −1 −11  , 
0 1
0 2

 
1 3 4 10
1 0
 0 1 0 2 , 0 1
0 0 1 1
0 0

3
4 10
−5 −1 −11
1
4 6
1 3 4 10
0 0 −1 −1
0 1 0 2
 
4 1
4
, 0
0 2
1 1
0

1 3
4 10
 ,  0 −5 −1 −11  ,
0 −2 0 −4
 

1 3 4 10
, 0 0 1 1 ,
0 1 0 2

0 0 0

1 0 2 .
0 1
1
 
So there is a unique solution, and it is x1 = 0, x2 = 2, and x3 = 1.
Question 3, Version 1. Show that S = {1, x − 1, x + 1, x2 , x2 + 2} spans P2 (R). Write down
a subset of S that is a basis for P2 (R).
Solution: A really quick way to show that S spans P2 (R) is to note that 1 and x2 are both in
S, and clearly x ∈ Span(S), since (x + 1) + (x − 1) = x. Since {1, x, x2 } is a basis for P2 (R),
it certainly spans, so S is a spanning set.
More generally, we write down the linear combination
a1 (1) + a2 (x − 1) + a3 (x + 1) + a4 (x2 ) + a5 (x2 + 2) = α + βx + γx2 .
If we can always find ai to solve this for any α, β, γ, then these vectors span.
This is the same as asking if we can solve the system
a1 − a2 + a3 + 2a5 = α,
a2 + a3 = β,
a4 + a5 = γ.
This gives the augmented matrix

1 −1 1 0 2 α
 0 1 1 0 0 β .
0 0 0 1 1 γ

But when we write it this way, it is clear that we can: adding the second row to the first puts
this in RREF, and we will have three pivots. (We could work harder for the actual solution,
but that is not needed here.)
So now we need to find a subset that is a basis. Since we know that any basis of P2 (R) contains
exactly 3 vectors, we need to throw out two.
Let’s keep 1 and x2 , we like those. So now we just need one more vector that is independent of
those two. But of course, any vector with an x term will be independent, so pick either one,
say x − 1.
Then our basis is {1, x − 1, x2 }.
Question 3, Version 2. Show that S = {1, x − 2, x + 1, x2 − 1, x2 } spans P2 (R). Write down
a subset of S that is a basis for P2 (R).
Solution: A really quick way to show that S spans P2 (R) is to note that 1 and x2 are both in
S, and clearly x ∈ Span(S), since (x + 1) + (x − 2) + 1 = x. Since {1, x, x2 } is a basis for
P2 (R), it certainly spans, so S is a spanning set.
More generally, we write down the linear combination
a1 (1) + a2 (x − 2) + a3 (x + 1) + a4 (x2 − 1) + a5 (x2 ) = α + βx + γx2 .
If we can always find ai to solve this for any α, β, γ, then these vectors span.
This is the same as asking if we can solve the system
a1 − 2a2 + a3 − a4 = α,
a2 + a3 = β,
a4 + a5 = γ.
This gives the augmented matrix


1 −2 1 −1 0 α
 0 1 1 0 0 β .
0 0 0 1 1 γ
But when we write it this way, it is clear that we can: adding twice the second row to the
first puts this in RREF, and we will have three pivots. (We could work harder for the actual
solution, but that is not needed here.)
So now we need to find a subset that is a basis. Since we know that any basis of P2 (R) contains
exactly 3 vectors, we need to throw out two.
Let’s keep 1 and x2 , we like those. So now we just need one more vector that is independent of
those two. But of course, any vector with an x term will be independent, so pick either one,
say x + 1.
Then our basis is {1, x + 1, x2 }.
Question 3, Version 3. Show that S = {1, 4, x + 1, x2 , x2 − 2} spans P2 (R). Write down a
subset of S that is a basis for P2 (R).
Solution:
Question 4, Version 1. Let T : R3 → R2 be the linear transformation defined by
T (x, y, z) = (2x − 3y, x + y + z).
Compute a basis for N (T ) and for R(T ). Show how this matches the Rank–Nullity Theorem.
Solution: First we compute N (T ). We have
(2x − 3y, x + y + z) = (0, 0),
or the system
2x − 3y = 0,
x + y + z = 0.
This gives a homogeneous system, which after row reduction is
2 −3 0
1 1 1
1 1
1
1 1 1
1 0 3/5
,
,
,
,
.
1 1 1
2 −3 0
0 −5 −2
0 1 2/5
0 1 2/5
This gives the equations
x1 = −3/5x3 ,
x2 = −2/5x3 .
Clearly x3 is free, and we can then write the nullspace as


−3/5
x3  −2/5  .
1
This one vector is a basis for N (T ).
We want to compute R(T ). One way to do this is to compute
T (1, 0, 0) = (2, 1),
T (0, 1, 0) = (−3, 1),
T (0, 0, 1) = (0, 1).
Clearly there are two independent vectors in that list (in fact, any pair will do). Therefore
dim(R(T )) = 2, and thus the range of T is all of R2 .
Finally, we notice that 1 + 2 = 3, which confirms The Rank–Nullity Theorem.
Question 4, Version 2. Let T : R3 → R2 be the linear transformation defined by
T (x, y, z) = (x + 3y, x − y + z).
Compute a basis for N (T ) and for R(T ). Show how this matches the Rank–Nullity Theorem.
Solution: First we compute N (T ). We have
(x + 3y, x − y + z) = (0, 0),
or the system
x + 3y = 0,
x − y + z = 0.
This gives a homogeneous system, which after row reduction is
1 3 0
1 3 0
1 3
0
1 0 3/4
,
,
,
.
1 −1 1
0 −4 1
0 1 −1/4
0 1 −1/4
This gives the equations
1
3
x1 = − x3 , x2 = x3 .
4
4
Clearly x3 is free, and we can then write the nullspace as


−3/4
x3  1/4  .
1
This one vector is a basis for N (T ).
We want to compute R(T ). One way to do this is to compute
T (1, 0, 0) = (1, 1),
T (0, 1, 0) = (3, −1),
T (0, 0, 1) = (0, 1).
Clearly there are two independent vectors in that list (in fact, any pair will do). Therefore
dim(R(T )) = 2, and thus the range of T is all of R2 .
Finally, we notice that 1 + 2 = 3, which confirms The Rank–Nullity Theorem.
Question 4, Version 3. Let T : R3 → R2 be the linear transformation defined by
T (x, y, z) = (x, y + z).
Compute a basis for N (T ) and for R(T ). Show how this matches the Rank–Nullity Theorem.
Solution: First we compute N (T ). We have
(x, y + z) = (0, 0),
or the system
x = 0,
y + z = 0.
This is already in reduced form, so we write x = 0, y = −z, and then the nullspace is given by


0
z 1 .
−1
This one vector is a basis for N (T ).
We want to compute R(T ). One way to do this is to compute
T (1, 0, 0) = (1, 0),
T (0, 1, 0) = (0, 1),
T (0, 0, 1) = (0, 1).
Clearly there are two independent vectors in that list. The last two are clearly dependent, but
if we throw one copy out, then we are left with {(0, 1), (1, 0)} which we know well as a basis
for R2 . Therefore dim(R(T )) = 2, and thus the range of T is all of R2 .
Finally, we notice that 1 + 2 = 3, which confirms The Rank–Nullity Theorem.
Question 5, Version 1. Let V be a vector space, and assume that {x, y} is a basis for V .
Show that {x, x + y} is also a basis for V .
Solution: First we need to show that this pair of vectors spans V , and then show that it is
independent.
We first write
ax + by = (a − b)x + b(x + y)
So any combination of x, y is also a combination of x, x − y, and thus Span(x, x + y) ⊇
Span(x, y) = V .
To show that they are independent, we write down the general combination
ax + b(x + y) = (a + b)x + by.
Since x, y are independent, this means a+b = b = 0, but the only solution to that is a = b = 0.
Question 5, Version 2. Let V be a vector space, and assume that {x, y} is a basis for V .
Show that {x, 2y} is also a basis for V .
Solution: First we need to show that this pair of vectors spans V , and then show that it is
independent.
We first write
b
ax + by = ax + (2y)
2
So any combination of x, y is also a combination of x, 2y, and thus Span(x, 2y) ⊇ Span(x, y) =
V.
To show that they are independent, we write down the general combination
ax + b(2x) = ax + 2by.
Since x, y are independent, this means a = 2b = 0, but the only solution to that is a = b = 0.
Question 5, Version 3. Let V be a vector space, and assume that {x, y} is a basis for V .
Show that {x, x − y} is also a basis for V .
Solution: First we need to show that this pair of vectors spans V , and then show that it is
independent.
We first write
ax + by = (a + b)x − b(x − y)
So any combination of x, y is also a combination of x, x − y, and thus Span(x, x − y) ⊇
Span(x, y) = V .
To show that they are independent, we write down the general combination
ax + b(x − y) = (a + b)x − by.
Since x, y are independent, this means a + b = −b = 0, but the only solution to that is
a = b = 0.
Question 6, Version 1. Prove that the zero vector in any vector space is unique.
Solution:
Let 0 and 00 both be a zero for the vector space V .
Since 0 is a zero,
0 + 00 = 00 .
Since 00 is a zero,
0 + 00 = 0.
Therefore
0 = 0 + 00 = 00 .