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Problems #3,33 pg.213-215
#3, pg.213: Determine whether the following set is a basis for R3×1:
{[10−3],[31−4],[−2−11]}.
Solution: We will test for independent of these vectors by the definition of linear independence: consider the vector equation
x1[10−3]+x2[31−4]+x3[−2−11]=0
We can solve this vector equation by considering its augmented matrix and reducing it to reduced echelon form:
[13−2001−10−3−410]∼[101001−100000].
We can easily see there are nontrivial solutions to this vector equation by rewriting this matrix as a system of equations:
{x1+x3=0x2−x3=0
and hence we see that there are infinitely many solutions that can be found by choosing values for the free variable x3:
→x=[x1x2x3]=[−x3x3x3]=x3[−111].
This means that the set is a dependent set of vectors (hence not a basis). We can still check to see if it spans R3×1, though. Let →b=[b1b2b3]∈R3×1. We must determine if there are weights α1,α2,α3 such that
α1[10−3]+α2[31−4]+α3[−2−11]=[b1b2b3]. We can set this up as an augmented matrix and row reduce to reduced echelon form:
[13−2b101−1b2−3−41b3]∼[101b1−3b201−1b20003b1−5b2+b3].
This solution is equivalent to the system of equations
{α1+α3=b1−3b2α2−α3=b20=3b1−5b2+b3
Hence this system only have a solution whenever 0=3b1−5b2+b3 (not all things in R3×1 have this property!). Thus these vectors do not span R3.
#33, pg.215: Consider the polynomials →p1(t)=1+t2 and →p2(t)=1−t2. Is {→p1,→p2} a linearly independent set in P3? Why or why not?
Solution: Yes it is independent. Try to solve the vector equation
x1→p1+x2→p2=0
or equivalently
x1(1+t2)+x2(1−t2)=0
yielding
(x1+x2)+(x1−x2)t2=0.
Equating coefficients yields the following system of equations:
{x1+x2=0x1−x2=0
which is equivalent to
{x1=−x2x1=x2
So x1=x2=−x2. Thus the only solution here is x2=0 and hence x1=0, a trivial solution. Hence the vectors →p1 and →p2 are linearly independent.
