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# Showing that an eigenbasis makes for good coordinate systems

## Video transcript

I've talked a lot about the idea that eigenvectors could make for good bases or good basis vectors. So let's explore that idea a little bit more. Let's say I have some transformation. Let's say it's a transformation from Rn to Rn, and it can be represented by the matrix, A. So the transformation of x is equal to the n-by-n matrix, A times x. Now let's say that we have n linearly independent eigenvectors of A. And this isn't always going to be the case, but it can often be the case. It's definitely possible. Let's assume that A has n linearly independent eigenvectors. So I'm going to call them v1, v2, all the way through vn. Now, n linearly independent vectors in Rn can definitely be a basis for Rn. We've seen that multiple times. And what I want to show you in this video is that this makes a particularly good basis for this transformation. So let's explore that. So the transformation of each of these vectors-- I'll write it over here. The transformation of vector 1 is equal to A times vector 1 and since vector 1 is an eigenvector of A, that's going to be equal to some eigenvalue lambda 1 times vector 1. We could do that for all of them. The transformation of vector 2 is equal to A times v2, which is equal to some eigenvalue lambda 2 times v2. And I'm just going to skip all of them and just go straight to the nth one. We have n of these eigenvectors. You might have a lot more. We're just assuming that A has at least n linearly independent eigenvectors. In general, you could take scaled up versions of these and they'll also be eigenvectors. Let's see, so the transformation of vn is going to be equal to A times vn. And because these are all eigenvectors, A times vn is just going to be lambda n, some eigenvalue times the vector, vn. Now, what are these also equal to? Well, this is equal to, and this is probably going to be unbelievably obvious to you, but this is the same thing as lambda 1 times vn plus 0 times v2 plus all the way to 0 times vn. And this right here is going to be 0 times v1 plus lambda 2 times v2 plus all the way, 0 times all of the other vectors vn. And then this guy down here, this is going to be 0 times v1 plus 0 times v2 plus 0 times all of these basis vectors, these eigenvectors, but lambda n times vn. This is almost stunningly obvious, right? I just rewrote this as this plus a bunch of zero vectors. But the reason why I wrote that is, because in a second, we're going to take this as a basis and we're going to find coordinates with respect to that basis, and so this guy's coordinates will be lambda 1, 0, 0, because that's the coefficients on our basis vectors. So let's do that. So let's say that we define this as some basis. So B is equal to the set of-- actually, I don't even have to write it that way. Let's say I say that B, I have some basis B, that's equal to that. What I want to show you is that when I do a change of basis-- we've seen this before-- in my standard coordinates or in coordinates with respect to the standard basis, you give me some vector in Rn, I'm going to multiply it times A, and you're going to have the transformation of it. It's also going to be in Rn. Now, we know we can do a change of basis. And in a change of basis, if you want to go that way, you multiply by C inverse, which is-- remember, the change of basis matrix C, if you want to go in this direction, you multiply by C. The change of basis matrix is just a matrix with all of these vectors as columns. It's very easy to construct. But if you change your basis from x to our new basis, you multiply it by the inverse of that. We've seen that multiple times. If they're all orthonormal, then this is the same thing as a transpose. We can't assume that, though. And so this is going to be x in our new basis. And if we want to find some transformation, if we want to find the transformation matrix for T with respect to our new basis, it's going to be some matrix D. And if you multiply D times x, you're going to get this guy, but you're going to get the B representation of that guy. The transformation of the vector x is B representation. And if we want to go back and forth between that guy and that guy, if we want to go in this direction, you can multiply this times C, and you'll just get the transformation of x. And if you want to go in that direction, you could multiply by the inverse of your change of basis matrix. We've seen this multiple times already. But what I've claimed or I've kind of hinted at is that if I have a basis that's defined by eigenvectors of A, that this will be a very nice matrix, that this might be the coordinate system that you want to operate in, especially if you're going to apply this matrix a lot. If you're going to do this transformation on a lot of different things, you're going to do it over and over and over again, maybe to the same set, then it maybe is worth the overhead to do the conversion and just use this as your coordinate system. So let's see that this will actually be a nice-looking, easy-to-compute-with and actually diagonal matrix. So we know that the transformation-- what is the transformation of-- let's write this in a bunch of different formats. Let me scroll down a little bit. So if I wanted to write the transformation of v1 in B coordinates, what would it be? It's just going to be equal to-- well, these are the basis vectors, right? So it's the coefficient on the basis vectors. So it's going to be equal to lambda 1, and then there's a bunch of zeroes. It's lambda 1 times v1 plus 0 times v2 plus 0 times v3, all the way to 0 times vn. That's what it's equal to. But it's also equal to D, and we can write D like this. D is also a transformation between Rn and Rn, just a different coordinate system. So D is going to just be a bunch of column vectors d1, d2, all the way through dn times-- this is the same thing as D times our B representation of the vector v1. But what is our B representation of the vector v1? Well, the vector, v1 is just 1 times v1 plus 0 times v2 plus 0 times v3 all the way to 0 times vn. v1 is a basis vector. That's just 1 times itself plus 0 times everything else. So this is what its representation is in the B coordinate system. Now, what is this going to be equal to? And we've seen this before. This is all a bit of review. I might even be boring you. This is just equal to 1 times d1 plus 0 times d2 plus 0 times all the other columns. This is just equal to d1. So just like that, we have our first column of our matrix D. We could just keep doing that. I'll do it multiple times. The transformation of v2 in our new coordinate system with respect to our new basis is going to be equal to-- well, we know what the transformation of v2 is. It's 0 times v1 plus lambda 2 times v2 and then plus 0 times everything else. And that's the same thing as D, d1, d2, all the way through dn times our B representation of vector 2. Well, vector 2 is one of the basis vectors. It's just 0 times v1 plus 1 times v2 plus 0 times v3 all the way, the rest is 0. So what's this going to be equal to? This is 0 times d1 plus 1 times d2 and 0 times everything else, so it's equal to d2. I think you get the general idea. I'll do it one more time just to really hammer the point home. The transformation of the nth basis vector, which is also an eigenvector of our original matrix A or of our transformation in standard coordinates, in B coordinates, is going to be equal to what? Well, we wrote it right up here. It's going to be a bunch of zeroes. It's 0 times all of these guys plus lambda n times vn. And this is going to be this guy d1, d2, all the way to dn times the B representation of the nth basis vector, which is just 0 times v1, 0 times v2 and 0 times all of them, except for 1 times vn. And so this is going to be equal to 0 times d1 plus 0 times d2 plus 0 times all of these guys all the way to 1 times dn. So that's going to be equal to dn. And just like that, we know what our transformation matrix is going to look like with respect to this new basis, where this basis was defined or it's made up of n linearly independent eigenvectors of our original matrix A. So what does D look like? Our matrix D is going to look like-- its first column is right there. We figured that one out. Lambda 1, and then we just have a bunch of zeroes. Its second column is right here. d2 is equal to this. It's 0, lambda 2, and then a bunch of zeroes. And then this is in general the case. The nth column is going to have a zero everywhere except along the diagonal. It's going to be lambda n. It's going to be the eigenvalue for the nth eigenvector. And so the diagonal is going to look-- you're going to have lambda 3 all the way down to lambda n. And our nth column is lambda n with just a bunch of zeroes everywhere. So D, when we picked-- this is a neat result. If A has n linearly independent eigenvectors, and this isn't always the case, but we can figure out that eigenvectors and say, hey, I can take a collection of n of these that are linearly independent, then those will be a basis for Rn. n linearly independent vectors in Rn are a basis for Rn. But when you use that basis, when you use the linearly independent eigenvectors of A as a basis, we call this an eigenbasis. The transformation matrix with respect to that eigenbasis, it becomes a very, very nice matrix. This is super easy to multiply. It's super easy to invert. It's super easy to take the determinant of. We've seen it multiple times. It just has a ton of neat properties. It's just a good basis to be dealing with. So that's kind of the big takeaway. In all of linear algebra, we did all this stuff with spaces and vectors and all of that, but in general, vectors are abstract representations of real world things. You could represent a vector as the stock returns or it could be a vector of weather in a certain part of the country, and you can create these spaces based on the number of dimensions and all of that. And then you're going to have transformations. Sometimes, like when we learn about Markov chains, your transformations are essentially what's the probability after one time increment that something state will change to something else, then you'll want to apply that matrix many, many, many, many times to see what the stable state is for a lot of things. And I know I'm not explaining any of this to you well, but I wanted to tell you that all of linear algebra is really just a very general way to solve a whole universe of problems. And what's useful about this is you can have transformation matrices that define these functions essentially on data sets. And what we've learned now is that when you look at the eigenvectors and the eigenvalues, you can change your bases so that you can solve your problems in much simpler ways. And I know it's all very abstract right now, but you now have the toolkit, and the rest of your life, you have to figure out how to apply this toolkit to specific problems in probability or statistics or finance or modeling weather systems or who knows what else.