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.