00:01
We've got some vectors and their bases of rn written as columns.
00:07
So in general, vi is going to equal aij, or say vj, so i can do this like this.
00:16
Vj is a1j, a2j, dot dot dot, anj, like that.
00:26
And a is an n by n matrix.
00:28
That's cool.
00:28
We want to show two things.
00:30
First, if a is invertible, then av1, av2, through avn is the basis of rn.
00:37
Let's prove that real quick.
00:40
So first of all, let's let v be an arbitrary vector of rn.
00:48
What that means, it's an arbitrary vector of rn, and what we can do now is take a inverse times v, and let's call this u.
01:04
We know that a is invertible, so a inverse exists, and we have this u, it's some element of rn.
01:12
That's cool.
01:13
And because our vi are a basis, we can express u is equal to a sum, this i goes from, do we start at 1, or is it, we're starting at 1, from 1 to n, of some alpha i times vi.
01:30
We can express u uniquely as a linear combination of our vi.
01:39
Now if we apply a to both of these sides, we get au is equal to a sum, i goes to 1, 1 to n, of a times alpha i times vi, which is precisely a sum, by linearity, of alpha i times avi.
02:05
So we've expressed au as a sum of the avi, but remember au is a, a inverse v, which is equal to v.
02:15
So that means that we've expressed v, which you'll recall was an arbitrary vector of rn, as a linear combination of avi, which means that these guys are, these guys span rn.
02:29
In particular, as well, we can show that if v is a sum, i equals 1 to n, for alpha i avi, we can apply a inverse to it, and get a inverse v, which is u, is equal to a sum of, let's see, alpha i sum scalars, times a inverse times a times vi, which cancel out, and we get vi.
03:06
Now there's only one way to do this, because vi are a basis, you'll recall, which means this, since this goes both, because this goes both ways, because a is invertible, that means there's only one way to express v as a linear combination of the avi's, which means not only do they span, but they're also linearly independent, this linear combination is unique, which means they are a basis, q, e, d.
03:35
Next up, we want to show that if these are a basis of rn, show that a is invertible.
03:40
So let's see, suppose that avi, as a set, as i goes from 1 to n, is a basis.
03:52
In particular, that means that the set avi is linearly independent.
04:05
Of course it also means that it spans the space, spans rn.
04:12
But i'm actually going to go back through that and just let u be an arbitrary vector in rn, and express u as a sum of our avi, for i equals 1 to n.
04:35
We can do this uniquely, so this is defined uniquely, or excuse me, not just a sum of that, we need our alpha times i.
04:43
These alpha i are unique, so now we go from here, define v to be a similar sum, but without the matrix here.
04:59
Now i want to make a subtle difference, make clear a distinction, subtle, from what i was doing earlier.
05:07
Here i let v be arbitrary, and then defined u in terms of v, and then found our alpha i, and went back to our i's here.
05:20
Because a was invertible, we knew that u could also be any vector in rn...