Let’s say I’ve got some

basis B, and it’s made up of k vectors. Let’s say it’s v1, v2,

all the way to vk. Let’s say I have some vector

a, and I know what a’s coordinates are with

respect to B. So this is the coordinates of

a with respect to B are c1, c2, and I’m going to have k

coordinates, because we have k basis vectors. Or if this describes a

subspace, this is a k-dimensional subspace. So I’m going to have k of

these guys right there. All this means, by our

definition of coordinates with respect to a basis, this

literally means that I can represent my vector a as a

linear combination of these guys, where these coordinates

are the weights. So a would be equal to c1 times

v1, plus c2 times v2, plus all the way, keep adding

them up, all the way to ck times vk. Now, another way to write

this is that– let me write it this way. If I had a matrix where the

column vectors were the basis vectors of B– so let me write

it just like that. So let me see I have some matrix

C that looks like this, where its column vectors are

just these basis vectors. So we have v1, v2, all

the way to vk. If we assume that all of these

are a member of Rn, then each of these are going to have n

entries, or it’s going to an n by k matrix. Each of these guys

have n entries. So we’re going to have n rows,

and we have k columns. So let’s imagine this

matrix right there. Another way to write this

expression right there, is to say that a is equal to the

vector c1, c2, all the way to ck, multiplied by this

matrix right there. This would be equal to a. This statement over here and

this expression over here are completely identical. If I take this matrix vector

product, what do I get? I get c1 times v1, plus c2 times

v2, plus c3 times v3, all the way to ck times

vk, is equal to a. We’ve seen this multiple

times in multiple different contexts. But what’s interesting

here is this expression is the same thing. And really I’m just applying new

words to things that we’ve seen probably 100

times by now. We can rewrite this

expression. This is C– and remember C is

just a matrix with our basis vectors as columns– C

is equal to this guy. This is just the coordinates

of a with respect to the basis B. So C times the vector that has

the coordinates of the vector a with respect to the basis B. That is going to

be equal to a. Now, why did I go through the

trouble of doing this? Because now you have a fairly

straightforward way of– if I were to give you this, if I were

to give you that right there, and say, hey, what is a,

if I wanted to write it in standard coordinates, or with

respect to the standard basis? Which is just kind of

the way we’ve been writing vectors all along? Then you just multiply it times

this matrix C, this matrix that has the basis

vectors as columns. The other way, if you have some

vector a that you know can be represented as a linear

combination of B, or it’s in the span of these basis vectors,

then you could solve for this guy right here to

figure out a’s coordinates with respect to B. So this little matrix right

here, what does it do? It helps us change bases. If you multiply it times this

guy, you’re going from the vector represented by

coordinates with respect to some basis, and you multiply

it times this guy, you’re going to get to the vector

just with standard coordinates. So we call this matrix right

here change of basis matrix, which sounds very fancy. But all it literally is is a

matrix with the basis vectors as columns. Let’s just apply this a little

bit to see if we can do anything vaguely constructive

with it. Let’s say that I have

some basis. let’s say B for basis. Let’s say I have two vectors. I’ll define the vectors

up here. Let’s say vector 1, let’s say

we’re dealing with R3. So vector 1 is 1, 2, 3. And let’s say that vector

2 is 1, 0, 1. And let’s say I’m going to

define some basis B as being the set of the vectors

v1 and v2. I’ll leave it to you to verify

that these are not linear combinations of each other,

so this is a valid basis. These aren’t in any way

linearly dependent. Now, let’s say that I know some

vector that’s in the span of these guys. All I know is how it happens

to be represented in coordinates with respect

to this basis. So let’s say I have

some vector a. And when I represent the

coordinates of a with respect to this basis, it’s equal

to 7, 7, minus 4. So how can we represent this

guy in its standard coordinates? What is a equal to? Well, you could just say a is

equal to 7 times v1, minus 4 times v2, and you’d be

completely correct. But let’s actually use this

change of basis matrix that I’ve introduced you

to in this video. So the change of basis matrix

here is going to be just a matrix with v1 and v2 as

its columns, 1, 2, 3, and then 1, 0, 1. And then if we multiply our

change of basis matrix times the vector representation with

respect to that basis, so times 7 minus 4, we’re going to

get the vector represented in standard coordinates. So what is this going

to be equal to? We have a 3 by 2 matrix,

times a 2 by 1. We’re going to get a 3 by 1

matrix, which makes complete sense because we’re

dealing in R3. a is going to be member of R3. So when we write it with

standard coordinates, we should have 3 coordinates

right there. Now when we represented a with

respect to the basis, we only had two coordinates, because a

was in the plane spanned by these two guys. Actually this is a good

excuse to draw this. So let me draw it in

three dimensions. Let’s say the span of v1

and v2 looks like this. Let’s say this is the 0

vector right there. So this right here is the

span of v1 and v2. Or another way, this is

the subspace that B is the basis for. So we know that a

is in this guy. So let’s say v1 looks like this,

and that v2– I’m not even looking at the numbers,

I’m just doing it fairly abstract– let’s say v2 looks

like this right here. Now, the fact that a can be

represented as a linear combination of v1 and v2, tells

us that a is also going to be in this plane in R3. In fact, it’s 7 times v1,

so it’s 1 v1, 1 v1’s, 3 v1’s, 4, 5, 6, 7. So it’s 7 in that direction, and

then it’s minus 4 in the v2 direction. So that’s 1 in the

v2 direction. This is minus 1 in the v2

direction, minus 2, minus 3, minus 4. Or we can do it here,

1, 2, 3, 4. So our vector a is going

to look like this. It’s going to sit

on the plane. So this is our vector a. It’s going to sit

on the plane. And when we represent it with

respect to this basis, when we represent these coordinates with

respect to our basis B, we say oh OK, it’s

7 of this guy. I’m just doing this

abstractly. Don’t pay attention to

the numbers just now. I just want you to understand

the idea. We said it’s 7 of this guy,

minus 4 of this guy. So it takes you back here. And you get this vector,

which is in this plane. So we only needed two

coordinates to specify it within this plane, because

this subspace was two-dimensional. But we’re dealing in R3. And if we just want the general

version of a in standard coordinates, we’ll have

to essentially get three coordinates. I want you to understand that

a is sitting on this plane. This plane just keeps going on

and on and on in all of these directions. a actually sits on that plane. It’s a linear combination of

that guy and that guy. But let’s figure out what

a looks like in standard coordinates. In standard coordinates, we get

the first term is going to be 1 times 7, plus

1 times minus 4. So that’s going to be 3. We get 2 times 7, plus

0 times minus 4. That is 14. You’re going to get 3 times

7, plus 1 times minus 4. So 3 times 7 is 21,

minus 4, is 17. So a is the vector 3, 14, 17. That is equal to a. Now let’s say we wanted

to go the other way. Let’s say we have some vector–

let me pick a letter I haven’t used recently– let’s

say I have some vector d, which is 8, minus 6, 2. And let’s say we know that d is

a member of the span of our basis vectors, the span of v1

and v2, which tells us that d can be represented as a linear

combination of these guys, or that d is in this subspace, or

that d can be represented as coordinates with respect

to the basis B. Remember, the basis B was just

equal to the set of v1 and v2. That’s all that basis B was. Now, we know that if we have

our change of basis matrix times the vector made up of

the coordinates, of d with respect to B– so let me write

that down, d with respect to B– is equal to d. We know that. We know if we have this guy’s

coordinates and we multiply it by the change of basis matrix,

we’ll just get the regular standard coordinate

representation of d. Now in this case, we have d. We’re given this. We of course know what the

change of basis matrix is. So if we wanted to represent d

in coordinates with respect to B, we’re going to have to

solve this equation. So let’s do that. So our change of basis matrix

is 1, 1, 2, 0, 3, 1. And we’re going to have to

multiply it times some coordinates. This thing right here, we can

represent it as– I’ll do it in yellow– we’re going to

need two coordinates. It’s going to be some

multiple of v1, plus some multiple of v2. So it’s c1, c2. We know it has to be two

coordinates because this matrix vector product is only

well-defined if this is a member of R2, because this

is a 3 by 2 matrix. We have two columns here,

so we have to have two entries here. Then that’s going to

be equal to d. So we have 8, minus 6, 2. So if we figure out what this

vector is, we’ve figured out what the representation, or

the coordinates of d with respect to B, are. So let’s solve this. So to solve this, we can just

set up an augmented matrix. That’s just our traditional

way of solving a linear equation. So we have 1, 1, 2, 0, 3, 1. We augment it with this

side right there. So we have 8, minus 6, and 2. And let’s keep my first

row the same. So I have 1, 1, augmented

it with 8. And let’s replace my second row

with the second row minus 2 times the first row. So I’m going to get 2 minus 2

times 1 is– actually, let me do it the other way. Let me replace my second row

with 2 times my first row, minus my second row. So two times 1, minus 2, is 0. 2 times 1, minus 0, is 2. 2 times 8 is 16,

minus 6, is 10. Now let’s replace the third

row with 3 times the first row, minus the third row. So 3 times 1, minus 3, is 0. 3 times 1, minus 1, is 2. And then 3 times 8 is 24, minus

2, is going to be 22. See it looks like I must have

made a mistake someplace, because I have these two would

lead to no solutions. Let me verify what I did, make

sure that I didn’t make any strange errors. So the second row, I replaced it

with 2 times the first row, minus the second row. So 2 times 1, minus 2, is 0. 2 times 1, minus 0, is 2. 2 times 8, minus minus 6– so

there’s my error– that’s equal to 22. That was my error. So these two things

are equivalent. I’ll do one step at a time. Let me replace my third row with

my third row, minus my second row, just get

it out of the way. So I”ll keep this 1,

1, 8, 0, 2, 22. And then the third row, I’m

going to replace it with my third row, minus

my second row. So it’s going to be 0, 0, 0. So that just gets zeroed out. Now let me divide my

second row by 2. So I get 1, 1, and 8. And then this one becomes

0, 1, and 11. Then of course the third row

is just a bunch of 0’s. Then let me keep my middle

row the same. So it’s 0, 1, and 11. Then let me replace my first

with my first row, minus my middle row. So 1 minus 0 is 1. 1 minus 1 is 0. 8 minus 11 is minus 3. And I’ll keep my last

row the same. So I put the left-hand side in

reduced row echelon form. So this right here

is essentially telling me my solution. So I could write it this way. I could write that 1, 0, 0, 1,

0, 0 times c1, c2 is equal to minus 3, 11, 0. Or another way of writing this,

is that 1 times c1, plus 0 times c2, or c1, is

equal to minus 3. Then we have 0 times c1, plus

1 times c2 is going to be equal to 11. So our solution to this equation

is minus 3, 11. Or another way of saying this,

is that if I wanted to write my vector d in coordinates with

respect to my basis B, it would be the coordinates minus

3, 11, which implies– let me write it this way– which

implies that d is equal to minus 3 times vector 1, plus

11, times vector 2. I’ll leave that for

you to verify. But just like that, using this

change of basis matrix, we can go back and forth. If you have this representation,

it’s very easy to take the product and

get the standard representation for d. If you have the standard

representation or the coordinates with respect

to the standard basis, it’s very easy. Well, it’s a little

more involved. But then you just solve

for your coordinates with respect to B.

I'm having my exam tomorrow, and this might save me from the devastating D-grade…

Thank you

Thank you so much for your videos! I'm a physics student and you have helped me understand so many principles of mathematics that I haven't been able to understand in lectures or textbooks! Thank you so very much!!

your video lectures rock man! thanks!

very nice explanation sir. Helped.

oh my god, it finally makes sense

I do know what it is about these videos but you seem to have a knack for making complex maths problems seem trivial. I have been trying to read about change of basis matrices in two textbooks and was terrified then I found this and went "ohhh that's all it is". Thanks.

Good job, you have done saved me again. My grade on tomorrows test will be 100% thanks to you.

In 13:56 he calculated wrong? 2*8 – (-6) is not 10, is 22!

please sal, can u make a vid about the DUAL SPACE and the QUOTENT SPACE!!!

@jolealdoneto he caught his error, thankfully about a minute later

I have learned more from you in the past 2 hours than I have from my professor in the last 2 months. You are awesome. Thank you so much!

Very talented 3D drawing (Y)

Very helpful but just a tip: I find it helps to write what row operations you are using beside the matrix before I move on. This would also make it a bit clearer, but overall great stuff thanks for the help!

I Really Like The Video Using a change of basis matrix to get us from one coordinate system to another From Your

You should come to carleton and teach the economics course load…just all the courses…this is better explained than almost any of my profs can manage. Great videos!

You sir, are a gentleman. Thank you for taking the time to create this.

great video, thanks for making it

I think its just a bit ambiguous. When he writes it he writes C="n x k matrix" and then adds on the other stuff.

He could have told us that he knew that [8; -6; 2] lie on the plane… One could think he just made that up – but he didn't. Try it yourself: make up a vector that "probably" isn't on the plane – you'll end up with an inconsistent augmented matrix, which lets you know there are no solutions, and therefore the "unlikely" vector isn't lying on that plane. Just wanted to point that out.

Gotta love being able to get the help I needed. Your discussion is clear and sound. Thanks!

wow. thank you!

Holy shit!!! I just learned change of basis in aprx18min…..I spent 1hr30min in my class and didnt learn crap. Thank You!

thanks sal, the 'abstract' visualization of the process you drew was what made the whole thing click in my brain! what I find most difficult about linear algebra is that I can't visualize it in my head easily, so drawings like that make the numbers much more meaningful to me!

you have saved my day good sir! you can have all my likes!!!!

<3 you

merci

yeeei 🙂

Yeah. You encapsulated the difference between a good teacher and a bad teacher in one comment. 😀

very clear. Than you!

What is the point of matirx if you do not solve equation in the matrix form?

tks for video!

listen to college class to learn this? I prefer to read books and watch this videos. Also this remembers me of preschool. If little kids saw this videos they would learn this stuff without knowing it was college stuff

what is the point if doing this?

So… Why can't my professor teach us like this… Everything he has been trying to teach us makes perfect sense now! Thank you Khan Academy for not being completely out of touch with reality! <3 Best lessons ever.

Sal, you did a mistake at 13.50!! 🙂

When finding the coordinates for D at 14 min, you make a mistake during row operations on the augmented matrix

this is just stupid, I was looking for numerical method and I land on that….can't believe I am watching that in 2016.

6:41 "change of bitches"

This is very helpful. I wouldn't have made it this far in Linear Algebra if it weren't for you Sal!

Thank you! Wonderful job!

What about when the basis is made up of 2×2 matrices. I struggle a lot with these:

B1 = (1 0) , (1 0 ), (1 0) B2 = (1 0) , (1 0) , (0 0)

(1 1) (1 -1) (-1 1) (0 1) (0 -1) (1 -1)

Asked to find the transition matrix P(B1 – B2). Please assist.

helpful

Another way to find c1, c2 – You could just multiply the two matrices and then solve a system of linear equations with 3 equations and 2 unknowns, which is must easier than using an augmented matrix.

very helpful and transparent lecture

do you know everything?

I LOVE UUUU

LMAO, so he's row reducing and I got 22 the first time and I'm beating myself up for being dumb enough to make mistakes, then a few seconds after, "looks like I made a mistake" … geez, my shaky heart, give me a break.

You are my hero!! Your explanation is very clear!

Mr.Sal, this is a great video, I can not express my gratitude. Thank you and keep making videos like this which are well explained in detailed.

Did he do row reduction incorrectly?

Newton Master Matrix thansk for you teaching class is Good Education for me.

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