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## Linear algebra

### Course: Linear algebra > Unit 3

Lesson 3: Change of basis- Coordinates with respect to a basis
- Change of basis matrix
- Invertible change of basis matrix
- Transformation matrix with respect to a basis
- Alternate basis transformation matrix example
- Alternate basis transformation matrix example part 2
- Changing coordinate systems to help find a transformation matrix

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# Invertible change of basis matrix

Using an invertible change of basis matrix to go between different coordinate systems. Created by Sal Khan.

## Want to join the conversation?

- Is the determinant defined for non-square matrices?(12 votes)
- The determinant is only defined for square matrices.(31 votes)

- How do we change from a basis other than the standard basis of R^n to another basis?(2 votes)
- You could multiply your coordinate in basis A by the matrix M that takes you from basis A to the standard basis, then by the matrix N that takes you from the standard basis to basis B

NM = another matrix, T, then becomes your transition matrix from basis A to B(8 votes)

- Can someone please give me a proof in which is shown that the change of basis matrix is always invertible?(2 votes)
- Proof (kind of informal)

Let B = {**b1**,... ,**bn**} be a basis for V (whatever vector space we are dealing with). Then the change of basis matrix Pb equals [**b1**...**bn**].

Note that Pb is a square nxn matrix. Well, by the definition of a basis, the column vectors of Pb are linearly independent. Therefore, by the Invertible Matrix Theorem, Pb is invertible.(3 votes)

- Hi, Let's say that B1, B2 are bases for vectorspace V. Is the transition matrix from B1 to B2 always invertible?

From what I understood transition matrices are always invertible, right?

Thnx(2 votes)- Yes the change of basis matrix is always invertible.(2 votes)

- At2:39, you mentioned that the matrix has to be square for it to be invertible. But if C is full column rank then it will have a left inverse.

We can just multiply the left inverse on the left hand side on both sides of the equation and we will get the vector with respect to the basis B, right? Or is there an issue here which I am missing?(2 votes)- Isn't that what he's showing at around5:34?(2 votes)

- Let's say that B and B' are bases for V; and P(B->B0), P(B'->B0) the change of basis matrix from B,B' to B0 respectively (where B0 is the standard basis).

If we were to change from coordenates with respecto to B to coordinates with respect to B' Can we use this matrix M?

M = P^(-1) (B'->B0) · P(B->B0)

this doesn't really make sense to me, because it's like going from B0 to B' and then from B to B0 again, but it's what is written in my school material.

Shouldn't it be rather like this:

M = P(B->B0) · P^(-1) (B'->B0)?

This way you start in B, go through B0, to go afterwards to B'.(2 votes)- It's about the order of operations. You textbook is correct. Let's go through the matrix-vector multiplication.

Let us start with the vector [x]_B. Then x = P(B->B0) [x]_B.

Then we find [x]_B' = P^(-1)(B'->B0) x.

Substitute the first into the second and we get

[x]_B' = P^(-1)(B'->B0) P(B->B0) [x]_B

Thus, M = P^(-1)(B'->B0) P(B->B0)

When reading a series of matrix multiplications, you generally need to read from right to left, not left to right, because the vector you are multiplying is on the far right.(1 vote)

- If V= R ^ n, What difference would there be between the basis being in R^n and not being in R^n? Is it just that when basis is in R^n, we now no longer need to check that the vector is in the span? or is there something more? Thanks!(2 votes)
- I think maybe you meant "spanning R^n", not "being in R^n".

What does it mean for a vector v to be in R^n?

(It means that v has n components. So, whenever the basis vectors have n components, they're in R^n, no matter how many vectors are in the basis. So basically they're always in R^n.)

When k=n (k = # basis vectors = the dimension) we have n basis vectors in R^n, which means that, as I think you meant to say, any vector in R^n is in the span of this basis (this can be proven). If k<n, then yes, not all R^n vectors are in the span of our basis.

If k<n, the transformation matrix is not onto R^n, so therefore it's not 1-1 and onto, and so not invertible, and we couldn't transform coordinates from one basis to another using matrices and their inverses.(1 vote)

- For those who watched Grants "essense of linear algebra" videos and think this new notation of
**A*x = b is weird, think of it this way : btw([x] = [x]sub(b) )**[x] = b could be interpreted as 'take the components of the vector your transforming and multiply with the corresponding basis vectors from A'. This is actually what you do with Ax = b, because x is some combination of the standard vectors (in this case[1 0]^t and [0 1]^T) with weights, b is the result

*A**in**your standard coordiate system when**A**transforms them in some way.

any question? ask and I'll answer(1 vote) - How do we find the "change in basis matrix" or C here in the video? Is it the alternative basis vectors expressed in our standard basis vectors i hat (1, 0) and j hat (0, 1)?(1 vote)
- Sal uses the conjugated matrix (1/det)(matrix*)

see the inverted matrix and determinant playlist(1 vote)

- What is the difference between as set of vectors being a basis "in" R^n and "for" R^n? If n=k is it "for" R^n and n>k then it is "in" R^n? Does this relate to the concept of one-to-one and onto (which I never fully understood)?(1 vote)
- {(1, 0, 0), (0, 1, 0)} = {i, j} is a basis in R^3 but not for R^3 (because it doesn't span R^3), and {(1, 0, 0), (0, 1, 0), (0, 0, 1)} = {i, j, k} is a basis both in R^3 and for R^3 (because it spans R^3). So yes, if k<n it's not "for".

It does relate to 1-1 and onto (It might be worth going back to those videos about inverses and studying there to get it all straight.). I don't yet have a complete explanation, but:

When k = n, the matrix formed from the basis vectors is a 1-1 and onto transformation, is square, and invertible.(1 vote)

## Video transcript

Like we've done in the last
several videos, let's assume that we have some set
of basis vectors B. And let's say our basis
is going to be v1 v2 all the way to vk. So this will span a subspace
of dimension k. And let's assume that each of
these guys are members of Rn. So v1 v2 all the way to vk. They're all members of Rn. Now in the last video, we saw
that we can define a change of basis matrix. And it's a fancy word, but all
it means is a matrix that has these basis vectors
as its columns. So v1, v2 all the way to
vk as its columns. So we're going to have k columns
and we're going to have n rows. Because each of these guys are
members of Rn, so they're going to have n entries. So we're going to have n rows. So it's going to be
an n by k matrix. And we saw in the last video
that if I have some vector a that is a member of Rn-- and
assuming that a is in the span of B-- I can represent a. I could say that a is equal to
the change of basis matrix times the coordinates of a with
respect to our basis. This is what we saw
in the last video. If I have the coordinates of
a with respect to B, I can multiply it by the change of
basis matrix and I'll get my vector a in standard
coordinates. Or if I have my vector a in
standard coordinates, then I can solve for my vector a in
coordinates with respect to B. We saw that in the last video. Now let's take a special case. Let's assume that
C is invertible. What does that mean? Or what does that
tell us about C? Well if C is invertible,
it's two things. It means that C is a square
matrix or has the same number rows and columns. And that its rows or columns--
You can pick either of them-- have to be linearly
independent. So linearly independent, let's
just pick columns. Now the second statement
is a bit redundant. We know that C has linearly
independent columns, because its columns are bases
for a subspace. So a basis, by definition, all
of the vectors have to be linearly independent. So we know this is
a bit redundant. But what's interesting is if we
know that C is invertible, C has to be square. And if all of these vectors are
members of Rn, then k has to be equal to n. So C is square means that k is
equal to n or that we have n basis vectors. Now if that's the case,
what is the span of B? Think about it. We have n linearly independent
vectors in Rn. So any time you have n linearly
independent vectors in Rn, those guys are
a basis for Rn. Because any basis that has n
entries-- and they're all linearly independent-- is going
to be a basis for Rn. So then B is a basis for Rn. So if we know that C is
invertible, we also know that you can get to any vector in Rn
by some linear combination of your basis vectors
right there. In the last video, we had to
make sure that this guy was in the span of these vectors. But now we don't have to make
sure, because if C is invertible, then the span of B
is going to be equal to Rn. Or another way you could say
it is if the span of B is equal to Rn. If we have n vectors here, if k
was equal to n, then we know that the span of B would
be equal to Rn. And so we'd have n vectors here,
n linearly independent columns here, and it would be
an n by n matrix with all of the columns linearly
independent. So then C would be invertible. So we could write
if and only if. And we could write
it the other way. If the span of B is Rn,
then C is invertible. And that's useful, because if
either of these things are true, then we can rewrite
the same equation. So let's say if we know this and
we're looking for that, we can just multiply
C times that. Let's say we know this and
we're looking for that. Before we had to do that
augmented matrix and solve for it, whatnot. But if we know C is invertible,
then one, we know that any vector here can
be represented in the span of our basis. So any vector here can be
represented as linear combinations of these guys. So you know that any vector can
be represented in these coordinates or with
coordinates with respect to our basis. We can multiply both sides of
this equation times C inverse. And what do you get
if you multiply? So it becomes C inverse C times
our coordinates of a with respect to B is equal
to C inverse times a. This is just the identity
matrix right there. Another way of writing this is
that the coordinates of a, with respect to our basis B,
which spans all of Rn, is equal to C inverse times
our vector a. Let's apply this a little bit. Let's apply this. Let's use this information, what
we've done in this video. Let's do some concrete
examples. So let's say I have
some basis. Let me define two vectors. I'll do it this way. So let's say I have v1 is equal
to the vector 1, 3. And let's say v2 is equal
to the vector 2, 1. And I have a basis that is equal
to the set of v1 and v2. Now I'll leave it for you to
verify that these guys are linearly independent. But if I have two linearly
independent vectors in R2, then B is a basis for R2. And if we write the change of
basis matrix, if we say C is equal to 1, 3, 2, 1, we know
that C is invertible. And actually to show that C
is invertible, we can just calculate its inverse. So what's the determinant
of C? The determinant of C is equal to
1 times 1 minus 2 times 3. So it's equal minus 5. That's the determinant of C. And so C inverse-- We figured
out a general formula for doing this for 2 by 2 matrices--
is equal to 1 over the determinant of C, so 1
over minus 5 times-- You switch these two guys, so you
switch the 1's, and you make these two guys negative. So minus 2, and then minus 3. And the very fact that this guy,
the determinant of C, was non-zero told us that
this was invertible. But anyway, this is C inverse. So let's say that I have
some vector a that is a member of R2. I'm just going to pick
some random numbers. Let's say that a is
equal to 7, 2. And I want to find out what the
coordinates of a are with respect to my basis B. Well we go to this situation. We know what a is, so we just
multiply a times C inverse to get this guy right here, to get
the coordinates of a with respect to B. So let me write that down. So what is C? So C is that. C inverse is that. So we could write the
coordinates of a with respect to B is equal to C inverse
times the standard coordinates of a. Or this is the same thing. Let me put the actual
numbers here. The coordinates of a with
respect to B are going to be equal to C inverse, which is
minus 1/5 times 1 minus 3 minus 2, 1, times
a, times 7, 2. And what is this equal to? This is equal to minus 1/5. And then we're going to get 1
times 7 plus minus 2 times 2. So it's minus 4. So 7 minus 4 is 3. And then we're going to get
minus 3 times 7, which is minus 21, plus 1 times 2. So minus 21 plus 2
is minus is 19. So the coordinates of a with
respect to the basis B are going to be equal to-- Let me
just multiply the negative 1/5-- you get minus 3/5. And then you get plus 19/5. So 19 over 5. Just like that. And let's verify that. This means that a is equal to
minus 3/5 times our first basis vector plus 19/5 times
our second basis vector. Let's verify that
that's the case. So let's see, minus 3/5 times
1, 3, plus 19/5 times 2, 1. Let's see what this is
going to be equal to. Let me write the two vectors. This is minus 3/5 times
3 is minus 9/5. And there we're going to
add it to this guy. So this guy is 2 times
19 is 38/5. Right? And then 19/5 times 1 is 19/5. And then if you add
these two vectors together, what do we get? We get minus 3/5 plus 38/5. That's 35/3. 35/5 is 7, minus
9/5 plus 19/5. That's 10/5 or 2. And there you have it. That was our original a. So we see that a can definitely
be represented as minus 3/5 times our first basis
vector plus 19/5 times our second basis vector. Now that was a case where we
had some vector a and we wanted to represent it in
coordinates with respect to B. What if we had the other way? What if we said that some vector
w's coordinates with respect to B are-- I'll do
something simple-- are 1, 1. Then what is w in standard
coordinates? Well there we can
just multiply. Remember w is just equal to
the change of basis matrix times w's coordinates with
respect to the basis B. So w is going to be equal to
the change of basis matrix, which is just 1, 3, 2, 1, times
the coordinates of w with respect to B times 1, 1. Which is equal to 1 times
1 plus 2 times 1 is 3. And then 3 times 1
plus 1 plus 1. So 3 times 1 is 3,
plus 1 is 4. So w is just equal to
the vector 3, 4. So there you see if our change
of basis matrix is invertible, which is really just another way
of saying that are basis spans Rn-- in this example it
was R2-- then you can easily go back and forth between
coordinate representations in our standard coordinates and
coordinate representations with respect to our basis. Right? This is with respect
to the basis. This is in standard
coordinates. And you can do that just simply
by either using this information or just saying,
oh, the coordinates with respect to the basis equal to
C inverse times a, or the inverse of our change of
basis matrix times a. Or saying our coordinates with
respect to the standard basis is just equal to the change
of basis matrix times the coordinates with respect
to the basis.