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Discover Linear Algebra: A first course in linear algebra

Jeremy Sylvestre

Discovery guide 18.1 Discovery guide
A4US

Suppose V is a vector space and S is a finite spanning set for V (i.e. V=SpanS). In the previous chapter, we saw that if S is linearly dependent, then (at least) one vector can be removed from S, and the resulting smaller set will still be a spanning set. You can imagine repeating this process until finally you are left with a spanning set that is linearly independent.
This leads to the following definition.
basis for a vector space
a linearly independent spanning set for the space

Discovery 18.1.

In each of the following, determine whether S is a basis for V. If it is not a basis, make sure you know which property S violates, independence or spanning.

(a)

V=R3, S={(1,0,0),(1,1,0),(1,1,1),(0,0,2)}.

(b)

V=R3, S={(1,0,0),(1,1,0),(0,0,2)}.

(c)

V=M2(R), S={[2010],[100−1],[0011]}.

(d)

V= the space of 2×2 upper triangular matrices,
S={[1001],[1101],[1201],[1301]}.

(e)

V= the space of 3×3 lower triangular matrices,
S={[100000000],[000100000],[000010000],[000000100],[000000010],[000000001]}.

(f)

V=P3(R), the space of all polynomials of degree 3 or less, S={1,x,x2}.

(g)

V=P3(R), S={1,x,x2,x3}.
As discussed in the introduction to this discovery guide above, a spanning set that is not linearly independent contains redundant information in the form of vectors that are not actually needed to form a spanning set. This redundancy manifests itself in other ways, as the next discovery activity will demonstrate.

Discovery 18.2.

Consider the set S={(1,0),(1,1),(1,−1)} of vectors in R2. This set spans R2 but is not linearly independent.

(a)

Since S spans R2, it is possible to express vector (3,−3) as a linear combination of the vectors in S.
Demonstrate a way to do this:
(3,−3)=XX(1,0)+XX(1,1)+XX(1,−1).

(b)

Here is the redundant part. Demonstrate a different way to express (3,−3) as a linear combination of the vectors in S:
(3,−3)=XX(1,0)+XX(1,1)+XX(1,−1).

(c)

How many different ways do you think there are to do this?
The next discovery activity will demonstrate that the redundancies of Discovery 18.2 cannot happen for a basis.

Discovery 18.3.

Suppose V is a vector space, S={v1,v2,v3} is a basis for V, and w is a vector in V.
Since S is a spanning set, there is a way to express w as linear combinations of the vectors in S:
w=a1v1+a2v2+a3v3.
Suppose there were a different such expression:
w=b1v1+b2v2+b3v3.
Use the vector identity
w−w=0
and the two different expressions for w above to show that having these two different expressions violates the linear independence of S.
Discovery 18.3 shows that when we have a basis S={v1,v2,…,vn} for a vector space V, each vector in V has one unique expression as a linear combination of the vectors in S. For w=c1v1+c2v2+⋯+cnvn, the coefficients c1,c2,…,cn are called the coordinates of w relative to S. Since these coordinates consist of n coefficients, we sometimes relate w to a vector in Rn by collecting its coordinates into an n-tuple:
(w)S=(c1,c2,…,cn).
This is called the coordinate vector of w relative to S.

Discovery 18.4.

In each of the following, determine the coordinate vector of w relative to the provided basis S for V.

(a)

V=M2(R), S={[1000],[0100],[0010],[0001]}, w=[−153−2].

(b)

V=M2(R), S={[1000],[1100],[0010],[0011]}, w=[−153−2].

(c)

V=P3(R), S={1,x,x2,x3}, w=3+4x−5x3.

(d)

V=R3, S={(−1,0,1),(0,2,0),(1,1,0)}, w=(1,1,1).

(e)

V=R3, S={(1,0,0),(0,1,0),(0,0,1)}, w=(−2,3,−5).

Discovery 18.5.

In each of the following, determine which vector w in V has the given coordinate vector (w)S.

(a)

V=M2(R), S={[1000],[0100],[0010],[0001]}, (w)S=(3,−5,1,1).

(b)

V=M2(R), S={[1000],[1100],[0010],[0011]}, (w)S=(3,−5,1,1).

(c)

V=P3, S={1,x,x2,x3}, (w)S=(−3,1,0,3).

(d)

V=R3, S={(−1,0,1),(0,2,0),(1,1,0)}, (w)S=(1,1,1).

(e)

V=R3, S={(1,0,0),(0,1,0),(0,0,1)}, (w)S=(−2,3,−5).

Discovery 18.6.

Coordinate vectors let us transfer vector algebra in a space V to the familiar space Rn.
For example, consider the basis
S={[1000],[1100],[0010],[0011]}
for the space M2(R) from Task 18.5.b.

(a)

In Task 18.5.b you have already determined the vector w in M2(R) that has coordinate vector (w)S=(3,−5,1,1). Now do the same to determine the vector v in M2(R) that has coordinate vector (v)S=(−1,2,0,3).

(b)

Do some algebra in M2(R):
Using your vectors v from Task a, and w from Task 18.5.b compute the linear combination 2v+w.
Note: Vectors v and w “live” in the space M2(R), so your computation in this task should involve 2×2 matrices, and should also result in a 2×2 matrix.

(c)

Do the same algebra in R4:
Compute 2(v)S+(w)S, using the coordinate vectors (v)S and (w)S provided to you in Task 18.6.a.
Note: These coordinate vectors “live” in the space R4, so your computation in this task should involve four-dimensional vectors, and should also result in a four-dimensional vector.

(d)

Compare your results:
Consider your four-dimensional result vector from Task c as a coordinate vector for some vector in M2(R) relative to S. Similarly to your computations in Task 18.5.b and Task a, determine the matrix in M2(R) that has coordinate vector equal to your result vector from Task c. Then compare with your result matrix from Task 18.6.b. Surprised?
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