MATH 240 QUIZ 5 NAME Solutions
Prof. J. Beachy Friday, 10/5/2007 Score / 20
1. (4.6 #2; 10 pts) Which of the following sets of vectors are bases for R3? Explain your answer.
(a) {(1,2,0),(0,1,−1)}
Since dim(R3) = 3, there aren’t enough vectors.
(b) {(1,1,−1),(2,3,4),(4,1,−1),(0,1,−1)}
There are too many vectors.
(c) {(1,0,0),(0,2,−1),(3,4,1),(0,1,0)}
There are too many vectors.
(d) {(3,2,2),(−1,2,1),(0,1,0)}
3−1 0
2 2 1
2 1 0
1−1/3 0
2 2 1
2 1 0
1−1/3 0
0 8/3 1
0 5/3 0
1−1/3 0
0 5/3 0
0 8/3 1
At this stage you can see that it will reduce to the identity, so the vectors are linearly independent.
2. (4.6 #13; 10 pts) Find a basis for the subspace P3spanned by the polynomials t3+t2−2t+ 1, t2+ 1, t3−2t,
2t3+ 3t2−4t+ 3. What is the dimension of this subspace?
Set up the equation c1(t3+t2−2t+ 1) + c2(t2+ 1) + c3(t3−2t) + c4(2t3+ 3t2−4t+ 3) = 0. Remember that
in this equation the right hand side is the zero polynomial. If you rearrange the terms as coefficients of t3,t2,tand
1, the each coefficient must be zero. This gives a system of 4 equations in the unknowns c1,c2,c3and c4, which has
the following matrix.
1 0 1 2
1 1 0 3
−2 0 −2−4
1 1 0 3
1 0 1 2
0 1 −1 1
0 0 0 0
0 1 −1 1
1 0 1 2
0 1 −1 1
0 0 0 0
0 0 0 0
This shows that c1=−c3−2c4and c2=c3−c4. Choosing first c3= 1 and c4= 0 and then c3= 0 and c4= 1
shows that the 3rd and 4th polynomials can be written in terms of the first two.
Conclusion: t3+t2−2t+ 1 and t2+ 1 form a basis for the subspace.
The algorithm in the text is this: put the coefficients of the polynomials in as columns of a matrix and row
reduce. The columns that contain the leading 1’s form the basis.
A second method of solution is to put the coefficients in as rows of a matrix. Since elementary row operations
do not change the subspace generated by a set of vectors, after you row reduce the matrix you have a basis.
1 1 −2 1
0 1 0 1
1 0 −2 0
2 3 −4 3
1 1 −2 1
0 1 0 1
0−1 0 −1
0−1 0 −1
1 0 −2 0
0 1 0 1
0 0 0 0
0 0 0 0
The first two rows correspond to t3−2tand t2+ 1. These two polynomials form a basis for the subspace.
Either method shows that the dimension of the subspace is 2.