Lesson: Dimension

Question 2

1 point

Transcript

Before we look at extending a linearly independent set into a basis, we will need a few more facts. In our technique for shrinking a spanning set to a basis, we depend on the fact that, if we get small enough, we will eventually be linearly independent. Now we want to know that if we get big enough, we will eventually be a spanning set. And the first step towards understanding why this happens is to realize that every basis for a vector space has the same number of vectors.

Consider the following. Lemma 4.3.2: Suppose that V is a vector space, and the span of {v1 through vn} equals V. If {u1 through uk} is a linearly independent set in V, then k is less than or equal to n. In other words, this says that the number of vectors in a linearly independent set will be less than or equal to the number of vectors in a spanning set.

The proof of this lemma is available in the text. Instead, I’d like to focus on the main result, which follows easily from the lemma.

Theorem 4.3.3: If the set B of vectors {v1 through vn} and the set C of vectors {u1 through uk} are both bases of a vector space V, then k = n.

Note that this theorem does not say that a vector space will have a unique basis, merely that the number of vectors in a basis is unique. We call this unique number of basis vectors the dimension of the vector space.

Definition: If a vector space V has a basis with n vectors, then we say that the dimension of V is n, and write that dim V = n.

The dimension of the trivial vector space O = {the 0 vector} is defined to be 0, consistent with our definition that the basis for O is the empty set. If a vector space V does not have a basis with finitely many elements, then V is called infinite-dimensional. We will not be studying the properties of infinite-dimensional spaces in this course, however.

Let’s look at our standard examples. Rn has standard basis {e1 through en}, so the dimension of Rn equals n. M(m, n) has a standard basis consisting of the mn matrices with a 1 in one entry and a 0 in the other entries. As such, the dimension of M(m, n) equals mn. Pn has standard basis {1, x, x-squared, through x-to-the-n}, so the dimension of Pn equals n + 1. Count carefully. The function space, functions on (a, b), even continuous functions on (a, b) are all infinite-dimensional spaces.

So one last example: Note that in the previous lecture, we found that this set T2 was a basis for the span of T, where T was this set. Since T2 contains 3 matrices, this means that the dimension of the span of T is 3.

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