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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.


2
4
3
7
7
Question 1: Compute the LU factorization of A = 6
4 −12 2



1 0 0
2 4
3
A = 3 1 0 0 −5 −2
2 4 1
0 0
4
Question 2: Let {u1 , u2 , u3 } be three linearly independent vectors in a vector space V over
R. Are the vectors p1 = u1 − 3u2 + 2u3 , p2 = 2u1 − 4u2 − u3 , p3 = u1 − 5u2 + 7u3 linearly
independent in V ?
Let X = (x1 , x2 , x3 ) ∈ R3 be so that x1 p1 + x2 p2 + x3 p3 = 0. Then (x1 + 2x2 + x3 )u1 + (−3x1 −
4x2 −5x3 )u2 +(2x1 −x2 +7x3 )u2 = 0. Since the three vectors {u1 , u2 , u3 } are linearly independent
this is equivalent to saying that X = (x1 , x2 , x3 )T solves
x1 + 2x2 + x3 = 0
−3x1 − 4x2 − 5x3 = 0
2x1 − x2 + 7x3 = 0.
We can put this system into the following matrix form:


1
2
1
−3 −4 −5 X = 0.
2 −1 7
Let us reduce the matrix of the linear system in echelon form
 



1 2
1
1
1
2
1
−3 −4 −5 ∼ 0 2 −2 ∼= 0
0 −5 5
0
2 −1 7
2
2
0

1
−2 .
0
There are only two pivots. There is one free variable. This means that the solution set of the above
linear system is not {0}. There is some nonzero vector X so that x1 p1 + x2 p2 + x3 p3 = 0. This
means that the vectors p1 , p2 , p3 are linearly dependent.
2
Quiz 4, September 27, 2012
Question 3: Is W = {at2 + bt + c : a, b, c ∈ R, a + b + |c| = 0} a subspace of the vector space
of univariate polynomials of degree at most 2 over the field R? Why?
Consider p1 (t) = −t2 − 1 and p2 (t) = −t2 + 1. Then p1 (t) + p2 (t) = −2t2 and clearly p1 + p2 is
not a member of W . This proves that W is not a subspace of P2 (t).
Question 4: Let {v1 , v2 , v3 , v4 , v5 } be an arbitrary set of five vectors in R4 . Show that they
are necessarily linearly dependent.
Consider X = (x1 , x2 , x3 , x4 , x − 5)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
The matrix [v1 v2 v3 v4 v5 ] has at least one free variable column since there are at most 4 pivots
(since there are four lines) and there are 5 columns. This means that the null space of A is not {0}.
There is a nonzero vector X so that
[v1 v2 v3 v4 v5 ]X = 0 = x1 v1 + x2 v2 + x3 v3 + x4 v4 + x5 v5 ,
thereby proving that {v1 , v2 , v3 , v4 , v5 } must be linearly dependent.