Invertible Change of Basis Matrix Using an invertible change of basis matrix to go between different coordinate systems
Invertible Change of Basis Matrix
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- 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.
- 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.
- 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.
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