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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?
• The determinant is only defined for square matrices.
• How do we change from a basis other than the standard basis of R^n to another basis?
• 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
• Can someone please give me a proof in which is shown that the change of basis matrix is always invertible?
• 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.
• 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
• Yes the change of basis matrix is always invertible.
• At , 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?
• Isn't that what he's showing at around ?
• 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'.
• 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!
• 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) )
*A
[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 in your standard coordiate system when A transforms them in some way.
(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.