# Last name: name: 1 Quiz 4 (Notes, books, and calculators are not authorized) ```Last name:
name:
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


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.
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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.
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