# Lesson: Change of Coordinates and Linear Mappings

1 point

## Transcript

Let’s take a look at an example from the previous lecture. We let L from M(2, 2) to P2 be defined by L([a, b; c, d]) = (a + b) + (a + c)x + (a + d)(x-squared). In the previous lecture, we found the matrix for L with respect to two different pairs of bases. If B1 is the standard basis for M(2, 2) and C1 is the standard basis for P2, then we have that C1[L]B1 is the matrix [1, 1, 0, 0; 1, 0, 1, 0; 1, 0, 0, 1]; while if B2 is this set and C2 is this set, then we found that C2[L]B2 is the matrix [0, 1, 0, 0; 0, 0, 1, 0; 1, 1, 1, 2].

Since it is reasonably easy to find the coordinates for a matrix with respect to the standard bases, we can easily use our first matrix to compute L(x). But before we can use the second matrix to compute L(x), we would first need to find the B2-components of x. Now, if we only wanted to do that for one vector x, we would do so by setting up an equation and row reducing a matrix. However, the more likely situation is that we would want to compute L(x) for many, perhaps even all, matrices x in M(2, 2), and the faster way to do that would be to find the change of coordinates matrix from the standard basis B1 to B2.

We’ve already looked at how to do that, so the point now is that we can also use our change of coordinates matrix to find C2[L]B2 if we already have C1[L]B1. Let’s let R be the change of coordinates matrix from B1 to B2, and let Q be the change of coordinates matrix from C1 to C2. Well, then we have that the C2-coordinates of L(x) would be Q times the C1-coordinates of L(x). Well, that’s going to be Q times the C1[L]B1 times the B1-coordinates of x. We also have that the C2-coordinates of L(x) will equal C2[L]B2 times the B2-coordinates of x, and this can equal C2[L]B2 times R times the B1-coordinates of x.

So for every vector [x]B1 in Rn, which will, in fact, be every vector in Rn, we have that Q(C1[L]B1)([x]B1) = (C2[L]B2)R([x]B1). Now, it’s been a while, but if you recall Theorem 3.1.4, then you’ll see that this means that Q(C1[L]B1) must equal (C2[L]B2)R. Since R is a change of coordinates matrix, R is invertible, as R-inverse is the change of coordinates matrix from B2 to B1. And so we have that C2[L]B2 = Q(C1[L]B1)(R-inverse).

Now, I started this explanation in regards to our example, but I finished using only general notation, so, in fact, we’ve shown that this holds in general. To summarize, if we let L from V to W be a linear mapping, let B1 and B2 be bases for V, and C1 and C2 be bases for W, then if we let R be the change of coordinates matrix from B1 to B2, and Q be the change of coordinates matrix from C1 to C2, we have that the matrix C2[L]B2 = Q(C1[L]B1)(R-inverse).

Let’s go ahead and apply this result to our example. To do so, we will first need to find R-inverse and Q. Recalling that R-inverse is the change of coordinates matrix from B2 to B1, we know that R-inverse is the matrix whose columns are the B1-coordinates of our B2 matrices. Since B1 is the standard basis, so we can easily compute that R-inverse is the matrix whose columns are [1; 0; 0; 0], [1; 1; 0; 0], [1; 1; 1; 0], [1; 1; 1; 1]—or this matrix here. Now, Q is the change of coordinates matrix from C1 to C2. So that is to say, Q is the matrix whose columns are the C2-coordinates of our standard basis vectors. Luckily, these are still easy to find, and we see that Q is the matrix whose columns are [1; 0; 0], [-1; 1; 0], [0; -1; 1]—or this matrix seen here.

Now finding C2[L]B2 is a simple case of matrix multiplication. We will multiply Q(C1[L]B1)(R-inverse) to get this, which, of course, is the same as our previous result.

Note that, again, I have deviated from the text in order to provide the most general result. So let’s see how to take our result and make it the same as the result in the text. For starters, the text only looks at the case where L is a linear operator, so V and W are the same. It further makes the restriction that V is Rn, although that is rather extreme. Why would the text even bother with the notation [x]S when, of course, [x]S is the vector x? But there is value in assuming that one of our bases is the standard basis S. The other significant restriction is that the book sets B1 = C1 and B2 = C2. That is, not only does it have the domain and codomain equal to each other, but we are looking at the matrix with respect to the same basis for the domain and codomain.

So to get the result in the book, you have to set B1 = C1 = S (a standard basis) and B2 = C2, called B. Then we have that R is the change of coordinates matrix from S to B, and Q is also the change of coordinates matrix from S to B. As such, R = Q, and our result is now that the matrix [L]B = R([L]S)(R-inverse).

This is almost what the text writes, except that we need to replace R with P-inverse, and R-inverse with P, and that is because, of course, they are defined as inverses. While I found it more straightforward to use the change of coordinates matrix from S to B, the text based its result on the change of coordinates matrix from B to S.

So let’s do an example. Let L from R3 to R3 be defined by L([a; b; c]) = [2a; a + b; 4b + c], and we’ll let B be this set. To find the B-matrix for L, we will first find the standard matrix for L. So [L]S is the matrix whose columns are L of our first standard basis vector, L of our second standard basis vector, L of our third standard basis vector. So it is the matrix [2, 0, 0; 1, 1, 0; 0, 4, 1].

Next, we need to find P, which is the change of coordinates matrix from B to S. Well, that’s going to be the matrix whose columns are the S-coordinates of the B vectors. But of course, the S-coordinates are the standard coordinates, which is just the same as the vectors, so we quickly find that P equals the matrix [1, 2, -1; 0, 1, 1; 1, 1, 0].

Well, now we need to find P-inverse. To do that, we’ll use the matrix inverse algorithm. So we start with our matrix P, and augment it with the identity matrix. We then row reduce until we have the identity matrix on the left and P-inverse on the right. And so we see that P-inverse is this matrix.

At long last, we can compute [L]B by saying that [L]B = (P-inverse)([L]S)P, which is this matrix product, which we compute and find that it equals the matrix [0, 4, 7; 1, 1, -3; 0, 2, 3]. If you go back to the previous lecture, you will see that this is the same as the result we found there.