 So what happens if we don't like the basis vectors that we have? We can construct a new one. Or can we? So let's consider the simplest case where our vector space has two basis vectors. Since the vector space itself consists of all linear combinations of our basis vectors, this means that the basis vectors themselves have to be in the vector space. And so we might take for our new basis some linear combination of the old basis vectors. So let's take some arbitrary linear combination of v1 and v2 and form a new set of vectors w1 and w2. And here's the challenge. Let's make sure that w1, w2 is still a basis for our vector space. And so that means we need to check two things. First, that these two vectors are independent. And second, that they actually span our original vector space. In order to span v, any linear combination of the vectors in v must be expressible as a linear combination of the vectors in w. So we must be able to solve linear combination of v vectors equals some linear combination of the w vectors. So let's set this up as a system of equations. So it's important to keep in mind that the things that we don't know are the values of x, the things that we do know are the values of v. Since I have w1 and w2 defined in terms of the vectors v1 and v2, let's go ahead and substitute those in and collect our coefficients. And if we compare the coefficients of v1 and v2, we'll get a system of equations. And we can then row reduce the augmented coefficient matrix. And remember our goal is to try and find solutions for x1 and x2. And we see that if a11, a22, minus a12, a21 is not equal to zero, this system will have a solution for any values of v1 and v2. Now this is a good opportunity to introduce some terms that are commonly used by mathematicians. The thing to notice here is that in order to solve this system of equations, all you need is this condition. And here's a fun alternative fact. The Beatles tried to write a song about this. But they couldn't find a proper rhyme for a11, a22, minus a12, a21 not equal to zero. So they used the word love instead. But I digress. Because meeting this condition is all we need, a mathematician would say that this condition is sufficient for w to span v. So this guarantees that the set of vectors w spans our vector space. But in order to be a basis, we also need to make sure that the vectors are independent. So how can we do that? Remember there's a theorem that says that a set of vectors will be independent if and only if the only linear combination equal to the zero vector is the trivial linear combination. And so we want to make sure that the only solution to linear combination of the w vectors equals the zero vectors is to have x1 and x2 both equal to zero. So we'll set up a system of equations. And since we know w1 and w2 in terms of v1 and v2, we'll substitute those in and collect our coefficients. Now it's worth noting that right now we have linear combination of the v vectors equal to the zero vector. Since we know that v is a basis, then this means that the only solution to this system of equations occurs when these coefficients are both equal to zero. And that gives us a nice system of equations. We can re-reduce our system. And remember that when we augment this with our column of constants, that column of constants is just going to be zero. And so provided that a11a22-a21a12 is not equal to zero, our system will have only one solution, x1 equal to zero, x2 equal to zero. And that's going to be the only solution to linear combination of the w vectors equal to the zero vector. And so our vectors will be independent. And since we've already established that they span our vector space v, then w will also form a basis for our vector space v.