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1
Quiz 4 (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.


3 5 6
Question 1: Compute the LU factorization of A = 6 12 16. Give all the details.
9 23 35



1 0 0
3 5 6
A = 2 1 0 0 2 4
3 4 1
0 0 1
Question 2: (a) Consider C as a vector space over the field R. Consider the vectors u = 1 and
v = i, where recall that i2 = −1. Is the set {u, v} linearly independent? Give all the details.
Letpx1 , x2 ∈ R be so that x1 u + x2 v = 0, i.e., x1 + ix2 = 0. The modulus of this complex number
is x21 + x22 = 0. This means that x1 = 0 and x2 = 0. In conclusion the set {1, i} is linearly
independent.
(b) Consider C as a vector space over the field C. Consider the vectors u = 1 and v = i. Is the
set {u, v} linearly independent? Give all the details.
Let x1 , x2 ∈ C be so that x1 u + x2 v = 0, i.e., x1 + ix2 = 0. Then there two ways to solve this
problem
(i) Observe that x1 = −i and x2 = 1 is a nonzero solution to this problem, thereby proving that
the set {1, i} is linearly dependent.
(ii) Let x1 = a + ib, a, b ∈ R, and x2 = c + id, c, d ∈ R. Then
a + ib + i(c + id) = 0.
This implies that
a − d = 0,
b + c = 0.
There are 4 unknowns and two equations. This means there are at least two free variables. In
conclusion there is a non-zero solution. This proves that the set {1, i} is linearly dependent. (Note
that a = d = 0 and b = 1, c = −1 is a solution, i.e., x1 = −i, x2 = 1.)
2
Quiz 4, October 2, 2014
Question 3: Is the function p(x) = −3 + 4 cos(x) + sin(x) a linear combination of u1 (x) =
5 − 2 cos(x) + sin(x), u2 (x) = −3 cos(x) + 2 sin(x), u3 (x) = 3 + cos(x)? If yes give the linear
combination. Give all the details.
Let us assume that p(x) is a linear combination of u1 , u2 , u3 , i.e., there are t1 , t2 , t3 such that
p(x) = t1 u1 (x) + t2 u2 (x) + t3 u3 (x). Then
−3 + 4 cos(x) + sin(x) = t1 (5 − 2 cos(x) + sin(x)) + t2 (−3 cos(x) + 2 sin(x)) + t3 (3 + cos(x)),
= 5t1 + 3t3 + (−2t1 − 3t2 + t3 ) cos(x) + (t1 + 2t2 ) sin(x).
This implies that
5t1 + 3t3 = −3;
We can write this
form:

1
2
−2 −3
5
0
−2t1 − 3t2 + t3 = 4,
t1 + 2t2 = 1.
system in the form of an augmented matrix and compute the reduced echelon
 
0 1
1
1 4  ∼ 0
3 −3
0
2
1
−10
 
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
This means that p(x) = −3u1 (x) + 2u2 (x) + 4u3 (x).
Question 4: Let {v1 , v2 , v3 , v4 , v5 } be five non-zero (column) vectors in R5 . Assume that this
set is orthogonal, i.e., vi ·vj = 0 is i 6= j. Determine whether this set is linearly independent or
linearly dependent. Justify your answer with all the details.
Consider X = (x1 , x2 , x3 , x4 , x5 )T ∈ R5 so that
x1 v1 + x2 v2 + x3 v3 + x4 v4 + x5 v5 = 0.
This equation is equivalent to the linear system
[v1 v2 v3 v4 v5 ]X = 0.
There are at least two ways to answer the question.
(1) Since the columns of the matrix are orthogonal, we know from class that [v1 v2 v3 v4 v5 ] is
invertible, thereby implying that X = 0 is the only solution. Hence the set {v1 , v2 , v3 , v4 , v5 } is
linearly independent.
(2) We have forgotten the above result and we establish it again. Upon taking the dot product of
the equation x1 v1 + x2 v2 + x3 v3 + x4 v4 + x5 v5 = 0 with v1 we obtain x1 kv1 k2`2 = 0, which implies
that x1 = 0 since v1 is non-zero. Applying the same argument with all the vectors, v2 , . . . , v5 we
infer that x2 = . . . = x5 = 0. In conclusion X = 0 is the only solution. This means that the set
{v1 , v2 , v3 , v4 , v5 } is linearly independent.