Math 369 Exam #2 Practice Problem Solutions
1. Consider the function f : R2 → R3 given by


2x − 3y
x
f
=  x + y .
y
2x
(a) Show that f is a linear transformation.
Proof. We need to show that f is additive and homogenous. For additivity, suppose
x2
are two vectors in R2 . Then
y2
f
x1
y1
+
x2
x1 + x2
=f
y2
y1 + y2


2(x1 + x2 ) − 3(y1 + y2 )
=  (x1 + x2 ) + (y1 + y2 ) 
2(x1 + x2 )

 

2x1 − 3y1
2x2 − 3y2
=  x1 + y1  +  x2 + y2 
2x1
2x2
x1
x2
=f
+f
,
y1
y2
so f is additive.
x
For homogeneity, suppose
is a vector in R2 and λ is a number. Then
y
f
x
λx
λ
=f
y
λy


2λx − 3λy
=  λx + λy 
2λx


2x − 3y
= λ x + y 
2x
x
= λf
,
y
so f is homogeneous.
 
−1
(b) Is  2  in the image of f ?
2
Answer: Yes. We need to solve the equation

  
2x − 3y
−1
 x + y  =  2 .
2x
2
1
x1
y1
and
x
1
=
. We can check that this
y
1
 
−1
x
1
f
=f
=  2 .
y
1
2
If you form the augmented matrix and solve, you see that
is correct:
2. Consider the matrix

−3
A= 2
0
−4
3
0

0
0 .
2
Find the characteristic polynomial of A.
Answer: By definition, the characteristic polynomial is

−3 − x
pA (x) = (−1)3 det(A − x13 ) = − det  2
0

0
0 .
2−x
−4
3−x
0
To compute the determinant, expand on the bottom row to get


−3 − x
−4
0
−3 − x
3−x
0  = − 0 + 0 + (−1)3+3 (2 − x) det
pA (x) = − det  2
2
0
0
2−x
−4
3−x
= −(2 − x) [(−3 − x)(3 − x) − 2(−4)]
= (x − 2)(x2 − 9 + 8)
= (x − 2)(x2 − 1)
or, after factoring
pA (x) = (x − 2)(x − 1)(x + 1).
3. Logan and Terry are both computing with the same 5 × 3 matrix. Logan determines that the nullspace
of the matrix is 2-dimensional, while Terry computes that the column space is 2-dimensional. Can
they both be right? Justify your answer.
Answer: No, they cannot both be correct. The 5 × 3 matrix is the matrix of a linear transformation
T : R3 → R5 . By the rank-nullity theorem,
dim(R3 ) = dim(null(T )) + dim(im(T )).
Since the left hand side is 3, there’s no way that the two terms on the right hand side can both be 2.
4. Suppose V is a vector space and that f : V → V is a linear transformation. Let
Vλ = {v ∈ V : f (v) = λv}.
Show that Vλ is a subspace of V .
Proof. We prove the usual three facts:
*
*
*
*
*
(i). If 0 is the zero vector in V , then f ( 0 ) = 0 = λ 0 since f is linear, so 0 ∈ Vλ .
(ii). If u, w ∈ Vλ , then we know that f (u) = λu and f (w) = λw. But then
f (u + w) = f (u) + f (w) = λu + λw = λ(u + w),
where the first equality is because f is additive. Therefore, u + w ∈ Vλ .
2
(iii). If u ∈ Vλ and r is a number, then f (u) = λu, so we see that
f (ru) = rf (u) = rλu = λ(ru),
where the first equality is because f is homogeneous. Therefore, ru ∈ Vλ .
Having proved the three conditions, we conclude the Vλ is a subspace of V .
5. Consider the following images:
-3
-2
3
3
2
2
1
1
1
-1
2
3
-3
-2
1
-1
-1
-1
-2
-2
-3
-3
2
3
The figure on the right shows the image of the figure on the left under the action of the linear transformation T : R2 → R2 .
(a) What is the matrix for T with respect to the standard basis on R2 ? In other words, what is
[T ]std2 →std2 ?
Answer:
We know
that the columns of [T ]std2 →std2 must be the images of the standard basis
1
0
vectors
and
under the action of T . Just from the picture, we can see that
0
1
1
1/2
T
=
0
3/2
so
[T ]std2 →std2
0
3/2
and T
=
,
1
1/2
1/2 3/2
=
.
3/2 1/2
1
−1
(b) What are T
and T
?
1
1
Answer: By multiplying, we see that
1
1/2 3/2
1
2
T
=
=
1
3/2 1/2
1
2
and
T
−1
1/2 3/2
−1
1
=
=
,
1
3/2 1/2
1
−1
both of which look reasonable based on the picture.
3
1
−1
,
. What is [T ]B→B ? (Hint: Use (b)...)
1
1
1
1
−1
−1
Answer: Notice that T
=2
and T
= −1
, so the matrix of T with
1
1
1
1
respect to the basis B is just the diagonal matrix
2 0
[T ]B→B =
.
0 −1
(c) Let B =
4