Math 369 Final Exam Practice Problem Solutions
1. Consider the linear transformation T : R2 → R3 given by


x+y
x
T
= 2x + y  .
y
3x + y
What is the nullspace of T ?
 
0
x
x
Answer: By definition, we want to find the vectors
that solve T
= 0, which means
y
y
0
solving the system of equations
x+y =0
2x + y = 0
3x + y = 0
This gives the augmented matrix

1 1
2 1
3 1
Subtracting twice row 1 from row 2 and three times

1 1
0 −1
0 −2
and then subtracting twice row 2 from row 3 yields

1 1
0 −1
0 0

0
0 .
0
row 1 from row 3 yields

0
0 ,
0

0
0 .
0
Therefore, the reduced system of equations is
x+y =0
−y = 0,
so in fact y = 0 and x = 0. Therefore, the nullspace consists only of the zero vector:
0
null(T ) =
.
0
2. Let
     
3
4 
 1
B = 2 , 2 ,  4  .


4
2
x
(a) Find a value for x which makes B linearly dependent and prove that the result really is linearly
dependent.
Answer: If x = 6, then we see that
       
1
3
4
0
2 + 2 − 4 = 0 ,
4
2
6
0
so this set is linearly dependent.
1
(b) Find a value for x which makes B linearly independent and prove that the result really is linearly
independent.
Answer: It turns out that any choice other than x = 6 will work, so we could choose, say, x = 0.
Suppose there were a linear combination giving zero:
 
 
   
1
3
4
0
a 2 + b 2 + c 4 = 0 .
4
2
0
0
This corresponds to the augmented matrix

1 3 4
2 2 4
4 2 0

0
0 .
0
Now row-reduce: subtract twice row 1 from row 2 and

1
3
4
0 −4 −4
0 −10 −16
four times row 1 from row 3:

0
0 .
0
Subtract 5/2 times row 2 from row 3:

1
0
0
3
−4
0
4
−4
−6

0
0 .
0
This gives the system of equations
a + 3b + 4c = 0
−4b − 4c = 0
−6c = 0
The third equation says c = 0; plugging that into the second equation gives b = 0, and then
plugging both into the first equation gives a = 0. SO we see that the only linear combination of
the vectors giving the zero vector is the trivial linear combination, and hence the three vectors
are linearly independent.
3. Find the minimal polynomial of the matrix

−4
A= 2
1
−2
1
2

−3
6 .
0
Answer: Notice that

−4
A2 =  2
1
−2
1
2

−3
−4
6  2
0
1
−2
1
2
 
−3
9
6  = 0
0
0
0
9
0

0
0 .
9
Therefore A2 − 13 = 0, so the minimal polynomial must divide x2 − 9. Since A itself is not a multiple
of the identity matrix, the minimal polynomial cannot be degree 1, so we conclude that the minimal
polynomial is exactly x2 − 9.
4. Let F : V → V be a linear transformation.
2
(a) Let λ ∈ R be a number. Give the definition of the eigenspace Vλ of F associated to λ.
Answer: The eigenspace is defined to be
*
*
*
Vλ = { v ∈ V : F ( v ) = λ v }.
*
(b) Show that 0 ∈ Vλ (this is true regardles of what λ is).
*
*
*
*
Proof. Since F is linear, F ( 0 ) = 0 = λ 0 , so 0 ∈ Vλ .
*
*
*
*
(c) Suppose that λ 6= τ , but that v ∈ Vλ and v ∈ Vτ . Prove that this means v = 0 .
*
*
Proof. Since v ∈ Vλ and v ∈ Vτ , we know that
*
*
*
F (v ) = λv
*
*
and F ( v ) = τ v .
*
But since F ( v ) = F ( v ), this means that
*
*
λv = τ v .
*
In particular, if vi is the ith entry in v , this tells us that λvi = τ vi ; since λ 6= τ , this can only
*
*
happen if vi = 0. Since the same argument works for all i, this implies that v = 0 .
*
5. Let V be a vector space with an inner product. Let v ∈ V be some particular vector and define
*
*
*
W = {w ∈ V : hw, v i = 0}.
Prove that W is a subspace of V .
Proof. We have to prove the usual 3 axioms.
*
First, we want to show that 0 ∈ W . To do so, note that
* *
h 0 , v i = 0,
*
*
since we’ve shown before that the inner product of 0 with anything must be zero. Therefore, 0 ∈ W .
*
*
*
*
*
*
Next, suppose w1 , w2 ∈ W . The goal is to show that w1 + w2 ∈ W . Note that, since w1 , w2 ∈ W , we
know that
*
*
*
*
hw1 , v i = 0 and hw2 , v i = 0.
Therefore, using the additivity of the inner product,
*
*
*
*
*
*
*
hw1 + w2 , v i = hw1 , v i + hw2 , v i = 0 + 0 = 0,
*
*
so indeed w1 + w2 ∈ W .
*
*
*
Finally, if w ∈ V and λ ∈ R, then we know that w, v i = 0, so it follows from homogeneity of the inner
product that
* *
* *
hλw, v i = λhw, v i = λ · 0 = 0,
*
so λw ∈ W .
Therefore, since all three properties hold, we conclude that W is a subspace of V .
3
6. Alice and Bob are each given the same 7 × 11 matrix B. Alice computes that the nullspace of B is
3-dimensional, while Bob computes that the column space of B is 8-dimensional. Can they both be
right? Why or why not?
Answer: No, this is not possible. Although the rank-nullity theorem says that
#(columns of B) = dim null(B) + dim col(B)
and these numbers don’t violate that, there is no way that the column space of B can be 8-dimensional.
Since B has only 7 rows, the columns of B are 7-dimensional vectors, and there is no way that a
collection of 7-dimensional vectors can span an 8-dimensional space.
7. Consider R3 with the slightly unusual inner product
1
1
1
u1 v1 + u2 v2 + u3 v3 .
6
8
27
* *
h u, v i =
 
 
1
1
*
Let e 1 = 2 and e 2 = −2.
3
3
*
*
*
(a) Show that { e 1 , e 2 } is an orthonormal set with respect to this weird inner product.
*
*
*
*
*
*
Proof. We just need to compute h e 1 , e1 i, h e 2 , e 2 i, and h e 1 , e 2 i:
1
·1·1+
6
1
*
*
h e 2, e 2i = · 1 · 1 +
6
1
*
*
h e 1, e 2i = · 1 · 1 +
6
*
*
h e 1, e 1i =
*
1
1
1 1 1
·2·2+
·3·3= + + =1
8
27
6 2 3
1
1
1 1 1
· (−2) · (−2) +
·3·3= + + =1
8
27
6 2 3
1
1
1 1 1
· 2 · (−2) +
· 3 · 3 = − + = 0,
8
27
6 2 3
*
so e 1 and e 2 have norm 1 and are orthogonal with respect to this weird inner product.
 
0
*
*
*
*
(b) Let v 3 =  8 . Apply the Gram-Schmidt prodcedure to the set { e 1 , e 2 , v 3 } to get an orthonor18
*
*
*
mal set. (Hint: you already know from part (a) that { e 1 , e 2 } is orthonormal, so v 3 is the only
vector that needs adjustment.)
*
*
*
Answer: Since we already have e 1 and e 2 , we only need to figure out e 3 . Recall that we find
*
e 3 in a two-step process: first, we compute
*
*
*
*
*
*
*
*
w3 = v 3 − h v 3 , e 1 i e 1 − h v 3 , e 2 i e 2
 
 
 
0
1
1
=  8  − 4 2 + 0 −2
18
3
3
   
0
4
=8−8
18
12
 
−4
=  0 .
6
Then, we let
 
   
−4
−4
−2
1
1
1
*
*
e 3 = * w3 = √  0  =  0  =  0 
2
kw3 k
4
6
6
3
4
Therefore, the desired orthonormal set is
     
1
−2 
 1
2 , −2 ,  0  .


3
3
3
8. Shown below left is a picture before applying an unknown matrix C, and below right is the result after
applying the matrix. What is the absolute value of det(C)? (You do not need to determine the matrix
C to solve this problem, though of course that is one approach.)
-4
-3
-2
4
4
3
3
2
2
1
1
1
-1
2
3
4
-4
-3
-2
1
-1
-1
-1
-2
-2
-3
-3
-4
2
3
4
-4
Answer: With some work, you can show that the matrix C =
0 1/2
, so | det(C)| = |0 − 1 · 12 | =
1 0
| − 12 | = 12 .
However, you could also just figure this out directly from the picture. Remember that the determinant
of the matrix is supposed to be the volume of the image of a unit box under the action of the matrix.
Since each square in the original picture gets sent to a rectangle of the same height and half the width,
the image of each box will have area equal to half the original area (indeed, the area of the new house
is half the area of the old house). So the absolute value of the determinant must be 1/2.
9. Prove or disprove the following claim: If A and B are n × n matrices and A is invertible, then
det(ABA−1 ) = det(B).
1
Proof. First, recall that det(A−1 ) = det(A)
. Next, remember that the determinant of a product is the
product of the determinants. Therefore,
det(ABA−1 ) = det(A) det(B) det(A−1 ) = det(A) det(B)
1
= det(B).
det(A)
10. Let C 0 ([−1, 1]) be the vector space of continuous functions on the closed interval [−1, 1]. Let E be the
set of even continuous functions on [−1, 1]; in other words,
E = {f ∈ C 0 ([−1, 1]) : f (x) = f (−x) for all x ∈ [−1, 1]}.
Prove that E is a subspace of C 0 ([−1, 1]).
5
Proof. As usual, we have to show that E satisfies the three subspace axioms.
First, if z(x) = 0 is the zero function, then z(x) = 0 = z(−x) for all x ∈ [−1, 1], so z ∈ E.
Next, if f, g ∈ E, then we know that f (x) = f (−x) and g(x) = g(−x) for all x ∈ [−1, 1]. Therefore, for
any x ∈ [−1, 1], we have that
(f + g)(x) = f (x) + g(x) = f (−x) + g(−x) = (f + g)(−x),
so f + g ∈ E.
Finally, if h ∈ E and λ ∈ R, then we know that h(x) = h(−x) for all x, so
(λh)(x) = λh(x) = λh(−x) = (λh)(−x)
for all x ∈ [−1, 1], so we see that λh ∈ E.
Therefore, since it satisfies the three conditions, we conclude that E is a subspace of C 0 ([−1, 1]).
6