And so from the perspective of our new coordinate frame

v will be a different vector.

How can we find the representation of v in our new coordinate frame?

Well this is not too complicated.

What we want to do is find out how much v goes over in the x1 direction and

how much v goes over in the x2 direction and how can we do that?

What we're going to do is we're going to project v onto a unit vector x1,

hat, x1 new hat for the new unit vector in the direction of x1.

And the projection of v onto x1 new hat,

the unit vector pointing in the direction x1 new,

will give us the amount that v lines up with the x1 new direction.

And in the same way the projection of v on to x2 new hat,

the unit vector pointing in the x2 new direction

will give us the amount that v points in the x2 new direction, so v2 new.

So, how do we actually write out the math?

So, if we were to write, v1 new, what does that equal?

That just equals, the projection of v

onto the unit vector, pointing in our x1 new direction.

And as we learned in earlier weeks, that's just the dot product between v and

x1 new hat.

It's important that x1 new hat is a unit vector so

that v1 new isn't scaled by an unnecessary amount.

And likewise,

v2 new = v x2 new.

And so how can we write out this equation as a matrix?

Well, dot products are very easy things to stick into matrix equations.

So we can write v1 new,

the vector v1 new is equal to some matrix times

our v old, sorry that should be v new not v1 new.

And this matrix is our change of basis matrix.

And in each row of our change of basis matrix we have

a row vector corresponding to one of the new basis vectors.

So the top row of our change of basis matrix

is the row vector for x1 new hat and

the bottom row is the row vector for x2 new hat.

We write the little t to indicate the transpose,

meaning that we’re taking a column vector to a row vector.

So in this case x is a 2 x 2 matrix.

So let's go through an example,

we'll draw out our original x1 old as one axis.

And x2 old as the other and

let's draw the vector so

v old will be, how about 1, 4.

So it goes over 1 and that's one that's 4,

not perfectly proportioned but that's okay.

How would we represent this vector in a coordinate frame

that has been rotated 45 degrees?

Well, the first thing we need is to make our change of basis matrix.

So, we need to find the unit vectors x1 hat new and

x2 hat new that point in the directions of our new axes.

Well, this is just a trigonometry problem, so we can right the x1 hat

new is equal to 1 over the square root of 2,

1 over the square root of 2.

Because if it's at 45 degrees,

it has to go up as much as it goes over and the square root of 2 term

comes in because the magnitude of this vector has to be equal to 1.

Similarly, x2 hat new is equal to,

while we go over minus 1 over square root of 2 and up 1 over square root of 2.

This allows us to write our change of basis matrix x.

So how did we do that?

What we did was we brought x1 hat new to the top row.

So we have 1 over square root of 2, 1 over square root of 2 and

we have x hat 2 new in the bottom row.

So that's minus 1 over square root of 2, 1 over square root of 2.

So then how do we write V new?

Well from our equation,

that was equal to x times v old.

And remember when you multiply a matrix by a column vector, what you're doing is

taking the dot product of each row of the matrix with the column vector.

So this equals 1 over root 2,

1 over root 2 minus 1 over root 2,

1 over root 2 times 1, 4.

So now doing this matrix multiplication, gives us 1 over square root of 2,

plus four4 over square root of 2, in the top.

So that's 5 over square root of 2 and minus 1 over square root of 2 + 4

over square root of 2 in the bottom so that's 3 over square root of 2.

And that equals about 3.5 and 2.1.

So in our new coordinate frame we go over 3.5 and

up 2.1 and that's all she wrote.

All right, so here's some intuition.

If you have a vector v, it's really the same vector in kind of a deeper way,

regardless of what basis you represent it in.

I could represent v in the old basis, or the new basis,

or maybe even a newer basis, newer x2, newer.

But regardless of what basis I represent v itself doesn't actually change.

The only thing that changes is the numbers we use to describe v.

And so the change in basis formula tells you how to find the numbers

that describe v in one basis, given the numbers that describe v in another basis.

As well as your change of basis matrix,

which relates the old basis and the new basis.

But v as kind of an abstract entity is still basically the same thing,

regardless of your representation.

In kind of the same way that a certain object is the same object

regardless of what angle you see it from.

So hopefully the idea of representing a vector in

a different basis doesn't theme all that crazy.

However, what does it mean to represent a matrix in a new basis?

So let's we had A matrix A old =

( a11 old a12 old a21 old

a22 old) What does it mean to

change the basis of the matrix?

Well, unfortunately it's not quite as simply

representing each of the columns of A in the new basis.

But even if that would be more simple it wouldn't be very useful.

So what is a matrix?

Or what does it do?

What a matrix does, is to take one vector, so a matrix takes the vector v,

v old, and it maps it to a new vector.

So this new vector is A old times v old and that's all it does.

A matrix takes one vector and spits out a new vector.

And if it's a square matrix then it'll spit out the new vector in the same

vector space as the old vector.

So that's kind of nice.

And very often in neuroscience,

we will be working with square matrices, so that makes life easy.

Okay, so, what would it mean to write A new,

what would it mean to represent A in a new basis?

Well, we saw that v was kind of this abstract entity

that stayed the same in a lot of ways, regardless of which basis you were in.

So, if we change our representation of v old to v new,

so we do our change of bases formula and get the old represented in a new bases.

We want A new to have the same action on v

new as A old had on v old.

So we want to choose A new such that it does the same thing

to v new as A old did to v old.

We should find the representation of the matrix that preserves its action.

So A old, v old, that's just a vector, right?

So we can represent it

in our new representation by multiplying it by x, right?

So this is the new representation of A old v old.

So what do we want that to be equal to?

Well, we want the representation of A old, v old in the new basis

to be equal to A in the new basis multiplied by v in the new basis.

So all this says is that A new times v new is A old times v old but

represented in the new basis.

But what equation do we have for v new?

Well we know that v new is equal to x times v old.

So we can write A new times x

v old is equal to x times A old,

v old and then if we multiplied both

sides of this equation by x inverse,

we get x inverse times

A new times x times v old is

equal to A old times v old.