## Transcript

Now that we know that the vector spaces in this course have a finite number of vectors in their basis, we can proceed to extend any linearly independent subset to a basis, and the way we do so is easy. Just pick a vector not already in the span and add it.

Theorem b: If the set T of vectors {v1 through vk} is a linearly independent set, and if w is not in the span of T, then {v1 through vk, w} is also a linearly independent set.

Here’s how we prove my theorem. Suppose that T is the set of vectors {v1 through vk} and is linearly independent, and that w is not in the span of T. To see that our bigger set {v1 through vk, w} is linearly independent, we will look for solutions to this equation. Suppose by way of contradiction that t-sub-(k+1) does not equal 0. Well, then we can divide by t-sub-(k+1) and get that (t1/t(k+1))v1 through to (tk/t(k+1))vk + w = 0. Well, that means that we can write w = –(t1/t(k+1))v1 – through to – (tk/t(k+1))vk. But this means that we can write w as a linear combination of the vectors in T. As this contradicts our choice of w as not being in the span of T, from this contradiction, we know that we must get that t(k+1) = 0, and this turns our linear independence equation into t1v1 + all the way through tkvk + 0w = 0, which, of course, is the same as the equation t1v1 + all the way through tkvk = 0. Now since T is linearly independent, we know that the only solution to this equation is t1 through tk = 0. As such, we’ve shown that the only solution to our larger equation t1v1 + all the way through tkvk through t(k+1)w = 0 is that t1 through tk through t(k+1) = 0, and so this means that our set {v1 through vk, with w} is linearly independent.

Now, the fact that our vector spaces have a finite basis does more than guarantee that our expansion process will come to an end. It actually tells us when our process will end because our basis will need to have exactly dim V elements in it. We see how to use this fact in the following example.

So let’s produce a basis B for the plane P in R3 with equation 2x1 + 4x2 – x3 = 0, and then we will extend the basis B to a basis C for R3. Well, we already know that the dimension of any plane in R3 is 2, so to find a basis B for P, we simply need to find two linearly independent vectors on our plane. But a set of two vectors is linearly independent whenever they are not a scalar multiple of each other. So we can quickly note that v1 = [1; 0; -2] and v2 = [2; -1; 0] are two vectors that satisfy 2x1 + 4x2 – x3 = 0 that are not scalar multiples of each other. And thus, our set B = {[1; 0; -2], [2; -1; 0]} is a basis for P.

Now, we know that the dimension of R3 is 3, so to extend B to a basis C of R3, we simply need to find one vector that is not in the span of B. But the span of B is precisely the vectors on the plane P, so this means we are looking for any vector not on the plane. That is, we are looking for any vector that does not satisfy the equation 2x1 + 4x2 – x3 = 0. The vector [1; 1; 1] quickly comes to mind as such a vector. And so we have that C = {[1; 0; -2], [2; -1; 0], [1; 1; 1]} is a basis for R3.

Now, during the process of shrinking spanning sets and expanding linearly independent sets, we have discovered the following. Theorem 4.3.4: Let V be an n-dimensional vector space.

- A set of more than n vectors in V must be linearly dependent.
- A set of fewer than n vectors cannot span V.
- A set of n elements of V is a spanning set for V if and only if it is linearly independent.

As a personal note, I would like to refer to part 3 as the two-out-of-three rule for proving that something is a basis. By definition, a basis must be a spanning set and linearly independent, but we also know that our basis must have the correct number of elements. So really, there are three features that our basis must have: spanning, linear independence, and n elements. Between the definition of a basis and part 3, we see that showing that any two of these three features will guarantee that our set is a basis.