# Last name: name: 1 Quiz 5 (Notes, books, and calculators are not authorized) ```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
Question 1: Let A = [c1 . . . cn ] be a real-valued n &times; n matrix with column vectors c1 , . . . , cn .
Assume that there is a positive number α &gt; 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
```