Lesson: Coordinates With Respect to an Orthonormal Basis

Question 2

1 point

Transcript

Now, I said that our goal was to create a class of basis that was not necessarily the standard basis but whose coordinates were still easy to calculate. As you may have guessed, an orthonormal basis is such a basis. Note first that since every orthonormal set is linearly independent, once we have a set of n vectors from Rn in our orthonormal set, we automatically know that it is a basis by the two-out-of-three rule. The following theorem shows how to find the coordinates with respect to an orthonormal basis.

If the set B of vectors {v1 through vn} is an orthonormal basis for Rn, then the ith coordinate of a vector x in Rn with respect to B is bi, and is equal to x dot vi. It follows that x can be written as x equals the sum of (x dot v1)v1 + through to (x dot vn)vn.

The proof of this theorem is similar to the proof that orthogonal sets are linearly independent, except that instead of looking at a linear combination that is equal to the 0 vector, we now look at a linear combination that is equal to x. To that end, since our set B is a basis for Rn, we know that we have scalars b1 through bn such that b1v1 + through to bnvn = x. Well, then for every i from 1 to n, we can take the dot product of vi with both sides of this equation. We will start by having (the sum of b1v1 through bnvn) dotted with vi equals x dot vi. Then we can distribute the dot product on the left to get that (b1v1 dot vi) + through to (bnvn dot vi) = x dot vi. Next, we use the associative property of the dot product to pull out our scalars, and we’ll see that b1(v1 dot vi) + through to bi(vi dot vi) + through to bn(vn dot vi) = x dot vi. But recalling that our set B is orthonormal, and therefore orthogonal, we know that our vj dot vi = 0 when j does not equal i. So, our sum becomes b1(0) + through to bi((the norm of vi)-squared) + through to bn(0) is still equal to x dot vi. Of course, all of our 0 terms go away, and remembering that B is orthonormal means that the norm of vi is 1, so our middle term simply becomes bi, and we have shown that bi = x dot vi, as desired.

So let’s look at an example. Suppose we want to find the coordinates of the vector x = [-4; -7] with respect to our orthonormal basis {[2/(root 13); -3/(root 13)], [6/(root 52); 4/(root 52)]}. Well, we’ll calculate them individually. Our first coordinate b1 will be [-4; -7] dot [2/(root 13); -3/(root 13)]. This equals -8/(root 13) + 21/(root 13), which is 13/(root 13). Meanwhile, our second coordinate b2 equals [-4; -7] dotted with our second vector [6/(root 52); 4/(root 52)]. This equals -24/(root 52) – 28/(root 52), which is -52/(root 52). And so we see that the B1-coordinates of x are [13/(root 13); -52/(root 52)].

To find the coordinates of the vector y = [1; 2; 3] with respect to the orthonormal basis B2 seen here, we again calculate them individually. b1 will equal [1; 2; 3] dotted with our first vector [3/(root 26); 1/(root 26); 4/(root 26)], which equals 3/(root 26) + 2/(root 26) + 12/(root 26), which is 17/(root 26). Our second coordinate equals the dot product of [1; 2; 3] and our second basis vector [-1/(root 3); -1/(root 3); 1/(root 3)], and this equals -1/(root 3) – 2/(root 3) + 3/(root 3), which equals 0. And our third coordinate equals [1; 2; 3] dotted with our third basis vector [5/(root 78); -7/(root 78); -2/(root 78)]. This equals 5/(root 78) – 14/(root 78) – 6/(root 78), which is -15/(root 78). And so we see that the B2-coordinates of y are [17/(root 26); 0; -15/(root 78)].

Looking at these strange coordinates, you may be wondering if orthonormal bases are really worth the trouble. But in forcing the vectors in our basis to have length 1, it turns out that coordinates now preserve length. That is to say that the norm of x equals the norm of x with respect to B. In fact, we see that coordinates with respect to an orthonormal basis B will also preserve dot products.

Theorem 7.1.a: Let x and y be vectors in Rn, let B, the set {v1 through vn}, be an orthonormal basis for Rn, and let the B-coordinates of x be [x1 through xn] and the B-coordinates of y be [y1 through yn]. Then we have two properties.

  1. x dot y = [x]B dot [y]B, which is to say that the dot product of x and y is the same as the dot product of the B-coordinates of x and the B-coordinates of y. And,
  2. That the norm of x equals the norm of [x]B. That is to say, that x and the B-coordinates of x have the same length.

So how do we prove this? We’ll first note that x dot y equals the dot product of (x1v1 + through to xnvn) with (y1v1 + through to ynvn). First, we’ll distribute our x dot products, and make this equal to (x1v1 dotted (y1v1 through ynvn)) + through to (xnvn dotted with (y1v1 through ynvn)). Now we distribute all of our y dot products to get this is equal to (x1v1 dotted with y1v1) + through to (x1v1 dotted with ynvn) + all the way through to (xnvn dotted with y1v1) + through to (xnvn dotted with ynvn). So basically, we’re looking at every pair of dot products xivi with yjvj. Again, remembering that our xi’s and yi’s are scalars, we can pull them out. We see that this equals x1y1(the dot product v1 dot v1) + through to x1yn(the dot product v1 vn), now we’ll plus x2y1(the dot product of v2 and v1) + through to x2yn(the dot product of v2 and vn), and again, we continue with all of our pairs, eventually ending up with xny1(vn dot v1) + through to xnyn(vn dot vn).

At this point, we finally use the orthogonality of B, and we’ll note that any pairing vi dot vj where i does not equal j will go to 0. So all that’s left are the pairings (v1 dot v1), (v2 dot v2), through (vn dot vn). And again, since B was orthonormal, those dot products all equal 1. So we have that x dot y = x1y1 + x2y2 + through to xnyn. Now, this may not seem surprising, but remember that our xi’s and yi’s are the B-coordinates, not the standard coordinates, for x and y. And so we have, in fact, proved our result, that our dot product of x dot y equals the dot product of the B-coordinates and y-coordinates of x and y.

Now, our property 2, that the norm of x will equal the norm of [x]B follows from this first property since, of course, (the norm of x)-squared = x dot x, which we now know to equal [x]B dot [x]B, which equals x1-squared + through to xn-squared. And so we must have that the norm of x equals the square root of (x1-squared + through to xn-squared), which, by definition, is the norm of the B-coordinates of x.

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