Lesson: Orthogonality in \( \mathbb{C}^n \)

Question 2

1 point

Transcript

Now that we have the inner product correctly defined, we can define orthogonality for complex vector spaces just as we did for real vector spaces.

Let V be an inner product space over C, with inner product denoted with two brackets. then two vectors v and w in V are said to be orthogonal if the product of <v with w> equals 0. The set of vectors {v1 through vn} in V is said to be orthogonal if the product of <vj with vk> equals 0 for all j not equal to k. If we additionally have that the product of <vj with vj> equals 1 for all our j from 1 to n, then the set is said to be orthonormal.

For example, the vectors u1 = [1; i; 2 + i] and u2 = [1 + i; 1 – i; 0] are orthogonal, since the product of <u1 with u2> equals u1 dotted with u2-conjugate, which equals the vector [1; i; 2 + i] dot [1 – i; 1 + i; 0], which equals 1 – i + i + i-squared + 0, which equals 0. Now the vectors v1 = [1 + i; 1 – i; -2i] and v2 = [i; 1; i] are not orthogonal, since the inner product of <v1 with v2> equals v1 dotted with v2-conjugate, which equals [1 + i; 1 – i; -2i] dotted with [-i; 1; -i], which equals –i – i-squared + 1 – i + 2(i-squared), which is -2i, which does not equal 0.

Now, as I almost word-for-word copied the definition from Chapter 7, there should be no surprises here. We can also copy the definition for projection, as no changes are needed. Let’s write it out.

Let B, the set of vectors {v1 through vk}, be an orthogonal basis for a subspace S of Cn. Then the projection of z in Cn onto our subspace S is given by proj(S)(z) = ((the product of <z with v1>)/(the product of <v1 with v1>))v1 + through to ((the product of <z with vk>)/(the product of <vk with vk>))vk.

While this is the same definition we’ve always used, it’s important to remember that our inner product is no longer symmetric, so we must make sure we take our inner product in the correct order. If we were to use the product of <v1 with z> instead of the product of <z with v1>, we’ll end up with the wrong answer.

Let’s work through an example. Let’s let u1 be the vector [1; i; 2 + i], and u2 be the vector [1 + i; 1 – i; 0], as I’ve used in the previous example. Then we already know the set B = {u1, u2} is an orthogonal basis for the subspace S equal to the span of {u1, u2}. If we let our vector z be the vector [1 + i; 1 + 2i; 1 + 3i], and we can look at the projection of z onto S. Well, from our definition, we see that proj(S)(z) = ((the product of <z with u1>)/(the product of <u1 with u1>))u1 + ((the product of <z with u2>)/(the product of <u2 with u2>))u2.

First, let’s compute all the inner products. We’ll see that the product of z with u1 equals z dotted with u1-conjugate, so that will equal [1 + i; 1 + 2i; 1 + 3i] dotted with [1; -i; 2 – i]. When we multiply all this out, we’ll see that it equals 8 + 5i. Now remember, when we look at the product of <u1 with u1> we know that all of our plus/minus i terms will cancel out; we’re simply needing to square all of our real and imaginary terms, so we know that this will equal 1-squared + 1-squared + 2-squared + 1-squared, so that equals 7. We’ve got to do all the work again, when we calculate the product of <z with u2>, which again, will be z dotted with u2-conjugate, so we’re looking at our vector z, [1 + i; 1 + 2i; 1 + 3i] dotted with [1 – i; 1 + i; 0]. So we can multiply all of this out, and we’ll see that it equals 1 + 3i. And then we get back to the easy case, we’re looking at the product of <u2 with u2>, so this’ll simply be the square of all of our real and imaginary parts, so we’ll get a 1-squared + 1-squared + 1-squared + (-1)-squared, and this equals 4.

Plugging in all of these inner products, we’ll see that the proj(S)(z) will equal (((8 + 5i)/7) times our vector [1; i; 2 + i]) + (((1 + 3i)/4) times our vector [1 + i; 1 – i; 0]). I’m going to go ahead and pull out a 1/28 to get rid of my fractions. This will leave me having to multiply the first part by 4 and the second part by 7. But now, if I go ahead and multiply everything out, I get this one huge vector, and once I do my calculations, I’ll see that this equals (1/28)[18 + 48i; 8 + 46i; 44 + 72i], which, if I wanted, I could write as [9/14 + (12/7)i; 2/7 + (23/14)i; 11/7 + (18/7)i]. Either one is just fine.

Now, since the definition of perp(S)(z) and the Gram-Schmidt procedure are all based on the definition of proj(S), these also work the same as they did in Rn—again, making sure that we pay attention to the order when we compute any inner products.

So let’s now look at an example of this. Let’s let v1 be the vector [1 + i; 1 – i; -2i] and v2 be the vector [i; 1; i], again as in a previous example. In that example, we saw that v1 and v2 are not orthogonal, so the set {v1, v2} does not form an orthogonal basis for S, the span of {v1, v2}. Of course, the set does form a spanning set, so we can use it in the Gram-Schmidt procedure to find an orthogonal basis for S.

We start by setting w1 = v1, and our S1 equal to the span of {w1}. Well, then w2 will be perp(S1)(v2), which will equal v2 – proj(S1)(v2), which will equal v2 – ((the product of <v2 with w1>)/(the product of <w1 with w1>))w1. So let’s compute the inner products that we need. First, we want to compute the product of <v2 with w1>. Well, of course, this is just the product of <v2 with v1>, and if we want to make use of our earlier work, we can think about this as the conjugate of (the product of <v1 with v2>), which, again, pulling from our previous example, we know to be the conjugate of (-2i), which is 2i. To compute the product of <w1 with w1>, we use our lovely shortcut to see that it equals 1-squared + 1-squared + 1-squared + (-1)-squared + (-2)-squared, which equals 8.

So we have that w2 would equal our vector v2, which is [i; 1; i], minus ((2i/8) times our vector w1, which is also our vector v1, [1 + i; 1 – i; -2i]). Thankfully, 2i/8 isn’t too difficult to deal with, so we’ll simply pull that in, and we’ll see that our vector w2 = [1/4 + (3/4)i; 3/4 – (1/4)i; -1/2 + i]. However, I’m still not a big fan of fractions, and as before, we can go and pull out the scalar 1/4 from our w2 and still have an orthogonal basis vector. And so, by the Gram-Schmidt procedure, we have that our set B = {[1 + i; 1 – i; -2i] and [1 + 3i; 3 – i; -2 + 4i]} is an orthogonal basis for S.

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