Last name:
name:
1
Quiz 5 (Notes, books, and calculators are not authorized)
Show all your work in the blank space you are given on the exam sheet. Always justify your
answer. Answers with no justification will not be graded.
Question 1: Let A = [c1 . . . cn ] be a real-valued n × n matrix with column vectors c1 , . . . , cn .
Assume that there is a positive number α > 0 such that X T AX ≥ αkXk2`2 for all X ∈ Rn .
Show that {c1 , . . . , cn } is a basis of Rn .
Let us show that the columns vectors are linearly independent. Let x1 , . . . , xn ∈ R be such that
x1 c1 + . . . xn cn = 0. Then denoting X = (x1 , . . . , xn )T , this means that AX = 0. Then
0 = X T AX ≥ αkXk2`2 . This implies that kXk`2 = 0, i.e., X = 0. Hence {c1 , . . . , cn } is linearly
independent. This is also a basis of Rn since there are exactly n vectors. Note in passing that this
also proves that null space of A reduces to {0}, which implies that A is invertible (since A is square)
a b
Question 2: Let V be the vector space over R of the matrices of the following form
,
c 0
1 1
2 2
1 3
where a, b, c ∈ R. Are the vectors A1 =
, A2 =
, A3 =
linearly
2 0
2 0
3 0
independent in V ?
Let X = (x1 , x2 , x3 ) ∈ R3 be so that x1 A1 + x2 A2 + x3 A3 = 0. Then X solves


1 2 1
1 2 3 X = 0.
2 2 3
Let us reduce the matrix of the linear system in echelon form


 

1
1 2 1
1 2 1
1 2 3 ∼ 0 0 2 ∼= 0
0
0 −2 1
2 2 3
2
−2
0

1
1 .
2
There are only three pivots. This means that the solution set of the above linear system is {0}.
This means that the set of vectors {A1 , A2 , A3 } is linearly independent.
2
Quiz 5, October 8, 2014
Question 3: Is the set S := {(1, 2, 3), (2, 2, 0), (3, 0, 0)} a basis of R3 ? Why?
R3 is three-dimensional. The bases of R3 have three vectors. Since S contains three vectors; it
suffice to show that S is linear independent to conclude that the set is a basis. Consider the linear
combination
λ1 (1, 2, 3) + λ2 (2, 2, 0) + λ3 (3, 0, 0) = 0.
This means that
λ1 + 2λ2 + 3λ3 = 0,
2λ1 + 2λ2 = 0,
3λ1 = 0.
We immediately infer that λ1 = λ2 = λ3 = 0. The set is linearly independent and therefore it is a
basis of R3
Question 4: Express the function v(t) = e2t + 4et − 3 as a linear combination of z1 (t) =
e2t − 2et + 5, z2 (t) = 2e2t − 3et , z3 (t) = et + 3.
We look for (x1 , x2 , x3 ) ∈ R3 so that v = x1 z1 + x2 z2 + x3 z3 . This means that the following holds
for all t ∈ R:
e2t + 4et − 3 = x1 (e2t − 2et + 5) + x2 (2e2t − 3et ) + x3 (et + 3)
= e2t (x1 + 2x2 ) + et (−2x1 − 3x2 + x3 ) + 1(5x1 + 3x3 ).
This implies
x1 + 2x2 = 1,
We can write this
form:

1
2
−2 −3
5
0
−2x1 − 3x2 + x3 = 4
5x1 + 3x3 = −3.
system in the form of an augmented matrix and compute the reduced echelon
 
1
0 1
1 4  ∼ 0
0
3 −3
2
1
−10
This means that v = −3z1 + 2z2 + 4z3 .
 
0 1
1
1 6  ∼ 0
3 −8
0
2
1
0
0
1
13
 
1
1
6  ∼ 0
52
0
0
1
0
0
0
1

−3
2
4