Math 317: Linear Algebra
Practice Exam 1
Fall 2015
Name:
*Please be aware that this practice test is longer than the test you will see on September
28, 2015. Also, this test does not cover every possible topic that you are responsible for
on the exam. For a comprehensive list of all topics covered on the exam, please see the
exam topics document on the website.
I have purposefully strayed away from computation type problems (except for one or two)
to give you some additional insight and practice on the types of proofs you could see on
the exam. The exam is roughly 40% theory and 60% computational, so please be sure
that you know how to compute answers as well! (See Exam Guide topics).
1. Suppose that a, b, c ∈ Rn such that a is parallel to b and b is parallel to c. Prove
that span(a, b, c) = span(a).
Proof: Suppose that a is parallel to b. Then there is a constant c1 ∈ R
such that a = c1 b. Similarly since b is parallel to c then there is a constant
c2 ∈ R such that b = c2 c. Now if w ∈ span(a, b, c) then by definition, w =
d1 a + d2 b + d3 c for some d1 , d2 , d3 ∈ R. But b = c11 a and c = c12 b = c12 c11 a.
So,
w = d1 a + d2
1 1
1
a + d3
a = ca,
c1
c2 c1
where c = d1 + d2 c11 + d3 c12 c11 ∈ R. So w ∈ span(a), and thus span(a, b, c) ⊂
span(a). But, we always have that span(a) ⊂ span(a, b, c) since if v ∈
span(a), then v = c1 a = c1 a + 0b + 0c for c1 ∈ R which implies that
v ∈ span(a, b, c). Thus span(a, b, c) = span(a).
2. Find a parametric equation for the line through points A = (1, 2, 1) and B =
(2, 1, 0).
Let x0 = (1, 2, 1) and v = (2, 1, 0) − (1, 2, 1) = (1, −1, −1). Then the parametric equation for the line l is given by l = {(1, 2, 1) + t(1, −1, −1), t ∈ R}.
3. Determine if the following property is true or false. If it is true, prove it. If it false,
provide a counterexample or explain why the statement cannot be true.
For any vectors x, y, z ∈ Rn , x · (y · z) = (x · y) · z.
This is a false statement. If we compute (y · z), then we are left with a
scalar, call it c. So x · (y · z) = x · c which doesn’t make sense as we can
not take the dot product of a vector with a scalar.
4. Prove that for any vectors x, y ∈ Rn , we have that
1
kx + yk2 + kx − yk2 = kxk2 + kyk2 .
2
1
Math 317: Linear Algebra
Practice Exam 1
Fall 2015
Proof: Using properties of the dot product we have that:
1
kx + yk2 + kx − yk2
2
1
1
((x + y) · (x + y)) + ((x − y) · (x − y))
2
2
1
1
2
2
2
kxk + 2 (x · y) + kyk +
kxk − 2 (x · y) + kyk2
2
2
1
1
2
kxk + kyk2 +
kxk2 + kyk2
2
2
kxk2 + kyk2
=
=
=
=
5. Does there exist a system of linear equations with exactly 3 solutions? If yes, provide
an example; if not, explain why not.
No. We recall from class that a system of linear equations can have exactly
one solution, no solution, or infinitely many solutions. Suppose that a
system Ax = b has exactly 3 solutions, call them x1 , x2 and x3 . Then
Ax1 = b, Ax2 = b, and Ax3 = b. I claim that
a1 fourth solution is
x1 +x2 +x3
x1 +x2 +x3
. This is so because A
= 3 A (x1 + x2 + x3 ) =
given by
3
3
1
1
(Ax
+
Ax
+
Ax
)
=
(3b)
=
b.
1
2
3
3
3
Suppose that A is an n×n matrix that is both upper triangular and lower triangular.
Prove that A is a diagonal matrix.
Proof: Suppose that A is an n × n matrix that is both upper triangular
and lower triangular. Since A is upper triangular, then Aij = 0 for i > j.
However, A is also lower triangular, so Aij = 0 for i < j. This is equivalent
to saying that Aij = 0 whenever i 6= j which is precisely what it means for
A to be diagonal.
6. If A2 = B, prove that both A and B must be square matrices.
Proof: Suppose that A is an m × n matrix. Then in order for the multiplication of A and itself to be defined, we need the inner dimensions (of m × n
m × n) to agree. Thus m = n which forces A to be a square matrix. Note
that the outer dimensions will tell us the resulting size of the product A2
which is n × n. Thus B is also a square matrix.
7. Explain why the following claim is false:
If A is a 2 × 2 matrix such that A2 = 0 where 0 denotes the zero matrix, then A is
nonsingular.
2
Math 317: Linear Algebra
Practice Exam 1
Fall 2015
This is a false statement. If A is nonsingular, then so is A2 since the product
of two nonsingular matrices is still nonsingular. Since A is nonsingular, the
rank of A is 2 (as it is a 2 × 2 matrix). Thus the rank of A2 would also have
to be 2. Since A2 = 0, there are no nonzero rows in the row echelon form
of A2 which tells us that the rank of A2 is 0. This contradicts the fact that
A is nonsingular.
8. Prove that if T is a linear transformation from Rn to Rn , then T (0) = 0.
Proof: We observe that T (0) = T (0 + 0) = T (0) + T (0) since T is a linear
transformation. So T (0) = 2T (0) =⇒ T (0) = 0.
3