Bases and Dimension

Recall that a basis of a subspace of is a subset of that is linearly independent and spans . A basis allows us to uniquely express every element of as a linear combination of the elements of the basis. Several questions may come to mind at this time. Does every subspace of have a basis? We know that bases are not unique. If there is more than one basis, what, if anything, do they have in common?

Exploring Dimension

How would you describe If you answered that is a line in , you are correct. While the two vectors span the line, it is not necessary to have both of them in the spanning set to describe the line.

What is the minimum number of vectors needed to span a line?

Answer: .

Observe also that the vectors in the given spanning set are not linearly independent, so they do not form a basis for . How many vectors would a basis for have?

Answer: .

Now consider another subspace of : Geometrically, is a plane in . Note that the vectors in the spanning set are linearly independent. Can we remove one of the vectors and have the remaining vector span the plane?

What is the minimum number of vectors needed to span a plane?

Answer: .

How many vectors would a basis for a plane have?

Answer: .

Our observations in Exploration init:dimensionintro hint at the idea of dimension. We know that a line is a one-dimensional object, a plane is a two-dimensional object, and the space we reside in is three-dimensional.

Based on our observations in Exploration init:dimensionintro, it makes sense for us to define dimension of a vector space (or a subspace) as the minimum number of vectors required to span the space (subspace). We can accomplish this by defining dimension as the number of elements in a basis.

We have to proceed carefully because we don’t want the dimension to depend on our choice of a basis. So, before we state our definition, we need to make sure that every basis for a given vector space (or subspace) has the same number of elements.

Proof
Suppose . Without loss of generality, assume that . Because spans , every of can be written as a linear combination of elements of :

Consider the vector equation \begin{align} \label{eq:sizeofbases} b_1\vec{w}_1+b_2\vec{w}_2+\ldots +b_s\vec{w}_s=\vec{0} \end{align}

By substitution, we have: \begin{align*} &b_1\vec{w}_1+b_2\vec{w}_2+\ldots +b_s\vec{w}_s=\\ \\ =&b_1(a_{11}\vec{v}_1+a_{21}\vec{v}_{2}+\ldots +a_{t1}\vec{v}_t)+b_2(a_{12}\vec{v}_1+a_{22}\vec{v}_{2}+\ldots +a_{t2}\vec{v}_t)+\ldots \\ &+b_s(a_{1s}\vec{v}_1+a_{2s}\vec{v}_{2}+\ldots +a_{ts}\vec{v}_t)\\ \\ =&(b_1a_{11}+b_2a_{12}+\ldots +b_sa_{1s})\vec{v}_1 +(b_1a_{21}+b_2a_{22}+\ldots +b_sa_{2s})\vec{v}_2+ \ldots \\ &+(b_1a_{t1}+b_2a_{t2}+\ldots +b_sa_{ts})\vec{v}_t\\ \\ =&\vec{0} \end{align*}

Because ’s are linearly independent, we must have For all . This gives us a system of equations and unknowns. We can write the system as a matrix equation.

Recall our assumption that . By Theorem th:rankandsolutions, we know that the system has infinitely many solutions. This shows that equation (eq:sizeofbases) has a nontrivial solution (in fact, infinitely many of them). But this shows that is linearly dependent and contradicts our assumption that is a basis of . We conclude that .

The following section will guarantee that dimension is defined for every subspace of .

Every Subspace of has a Basis

Proof
See Practice Problem prob:atmostnlinindinrnproof.

Proof
Consider the equation \begin{align} \label{eq:expandinglinindset}a\vec{u}+a_1\vec{v}_1+\ldots +a_k\vec{v}_k=\vec{0} \end{align}

We need to show that . Suppose , then . But this contradicts the assumption that is not in the span of . So, . But because are linearly independent.

This means that (eq:expandinglinindset) has only the trivial solution, and is linearly independent.

Proof
Suppose that is a linearly independent subset of . If then is already a basis of . If , choose in such that is not in . The set is linearly independent by Lemma lemma:expandinglinindset.

If we are done; otherwise choose such that is not in . Then is linearly independent, and the process continues. We claim that a basis of will be reached eventually. If no basis of is ever reached, the process creates arbitrarily large independent sets in . But this is impossible by Lemma lemma:atmostnlinindinrn.

Practice Problems

Problems prob:finddimension1-prob:finddimension3

For each given set of vectors, find .

Use Theorem th:linindandrank of VEC-0110.
Answer:
Answer:
Answer:
Prove Lemma lemma:atmostnlinindinrn.
Look at the proof of Theorem th:dimwelldefined.
Let be a basis of . Suppose is a nonsingular matrix. Show that is also a basis of .
To show that spans , express as a linear combination of , and .

Exercise Source

Practice Problem prob:matrixtimesbasisvectors is adopted from Keith Nicholson’s Example 5.2.12 Linear Algebra with Applications. (CC-BY-NC-SA)

W. Keith Nicholson, Linear Algebra with Applications, Lyryx 2018, Open Edition, p 277.

2024-09-26 22:13:00