Lesson: Properties of Complex Inner Product Spaces

Question 2

1 point

Transcript

When doing the practice problems assigned for the previous lecture, you hopefully noticed that the standard inner product for complex numbers is not symmetrical—that is, that the product of <z with w> does not equal the product of <w with z>. So, right away we know that our definition of an inner product will have to be different than the one we used for the real numbers. But hopefully, you also noticed that the product of <z with w> equals the conjugate of the product of <w with z>, so this is yet another case where we need to introduce conjugation to extend a result from the reals to the complex numbers. Bilinearity also needs some adjustment in the complex numbers.

So let’s take a look at the definition of a generic inner product on Cn. Let V be a vector space over C. A complex inner product on V is a function from (V cross V) to C such that

  1. Property 1: For all z in V, we have that the product of <z with z> is a non-negative real number, and that the product of <z with z> equals 0 if and only if z is the 0 vector.
  2. Property 2: For all vectors w and z in V, the product of <z with w> equals the conjugate of the product of <w with z>. And,
  3. Property 3: For all vectors u, v, w, and z, and all complex numbers alpha, we have the following four properties:
    1. Property (1), that the product of <(v + z) with w> will equal (the product of <v with w>) + (the product of <z with w>).
    2. (2), that the product of <z with (w + u)> will equal (the product of <z with w>) + (the product of <z with u>).
    3. (3) The product of <((alpha)z) with w> will equal alpha times (the product of <z with w>). And lastly,
    4. (4) The product of <z with ((alpha)w)> will equal (the conjugate of alpha) times (the product of <z with w>).

Now, our property 1 is still the same as we saw in Rn, and is still referred to as being positive definite. Property 2 is known as the Hermitian property of the inner product instead of the symmetric property. Because the complex inner product is not symmetric, we cannot find a simple counterpart to bilinearity, but we can combine the statements of property 3 into one statement as follows. Let u, v, w, and z be vectors, and let alpha, beta, gamma, and delta be complex numbers, and let’s expand out the product of <((alpha)u + (beta)v) with ((gamma)w + (delta)z)>. So one use of part (1) will get us to (the product of <((alpha)u) with ((gamma)w + (delta)z)>) + (the product of <((beta)v) with ((gamma)w + (delta)z)>). Then we can use our part (2) twice, on each of these, to find that this equals (the product of <((alpha)u) with ((gamma)w)>) + (the product of <((alpha)u) with ((delta)z)>) + (the product of <((beta)v) with ((gamma)w)>) + (the product of <((beta)v) with ((delta)z)>). Now we’ll use part (3) four times to get that this is (alpha times (the product of <u with ((gamma)w)>) + (alpha times (the product of <u with ((delta)z)>) + (beta times (the product of <v with ((gamma)w)>) + (beta times (the product of <v with ((delta)z)>). Lastly, we’ll use part (4) again four times to see that this expands out to ((alpha)(gamma-conjugate)(the product of <u with w>)) + ((alpha)(delta-conjugate)(the product of <u with z>)) + ((beta)(gamma-conjugate)(the product of <v with w>)) + ((beta)(delta-conjugate)(the product of <v with z>)). So we see that the inner product is almost bilinear; we simply need to remember to take the conjugate of any scalar we pull out of the right side of the inner product.

We won’t spend much time on non-standard inner product spaces, but we should at least verify that the standard inner product we defined is, in fact, an inner product. So we’ll do this as an example.

Let’s start by looking at property 1. Let’s let z be any vector in Cn. Then the product of z with itself is the sum from j = 1 to n of (zj(zj-conjugate)), which is the sum from j = 1 to n of ((the modulus of zj)-squared). Now, since this is the sum of non-negative real numbers, it must be a non-negative real number. Moreover, the only way to have that this product of z with itself, the sum from j = 1 to n of ((the modulus of zj)-squared), is equal to 0 is to have (the modulus of zj)-squared equal to 0 for all our j from 1 to n, and since the only complex number with a modulus of 0 is 0, we see that the inner product of <z with z> equals 0 if and only if z is the 0 vector.

Now let’s look at property 2. Let’s let z and w be complex vectors. Then we can use properties of the conjugate to see that the conjugate of the product of <w with z> will be the conjugate of the sum from j = 1 to n of (wj(zj-conjugate)). Now, since our conjugation distributes across sums, we can rewrite this as the sum from j = 1 to n of ((wj-conjugate)((zj-conjugate)-conjugate)). Of course, (zj-conjugate)-conjugate simply is zj, and complex multiplication is commutative, so we can switch the order to see that this is equal to the sum from j = 1 to n of (zj(wj-conjugate)), which is the definition of the product of <z with w>.

Now we’ll move on to various properties in property 3. Let’s first look at property 3(1), so we’ll let v, w, and z be complex vectors. Then the product of <(v + z) with w> could be the sum from j = 1 to n of ((vj + zj)(wj-conjugate)). Well, wj-conjugate is just some complex number, so we can still distribute it, and we’ll get the sum of j = 1 to n of (vj(wj-conjugate) + zj(wj-conjugate)). And now, of course, we can simply break this sum since complex addition is commutative and associative, we can split this into the two sums, (the sum from j = 1 to n of our (vj(wj-conjugate)) terms) + (the sum from j = 1 to n of our (zj(wj-conjugate)) terms). So this will equal (the inner product of <v and w>) + (the inner product of <z and w>).

Next up is property 3(2), so we’ll let u, w, and z be complex vectors. Then the product of <z with (w + u)> will equal the sum from j = 1 to n of (zj times the conjugate of (wj + uj)). Again remembering that our complex conjugation can be distributed over addition, this becomes the sum from j = 1 to n of (zj(wj-conjugate + uj-conjugate)). We can distribute our zj, and then divide up our sum to see that this equals (the product of <z with w>) + (the product of <z with u>), as desired.

For property 3(3), we’ll let w and z be complex vectors, and alpha be a complex number. Well, then the product of <((alpha)z) with w> will equal the sum from j = 1 to n of (((alpha)zj)(wj-conjugate)). Well, we could easily just pull this alpha out of our sum, and see that this equals alpha times (the product of <z with w>).

Lastly, we’ll look at property 3(4), which is where things get a bit interesting. Let w and z be complex vectors, and alpha be a complex number. Then the product of <z with ((alpha)w)> will equal the sum from j = 1 to n of (zj(((alpha)wj)-conjugate)). Now again, keeping in mind that alpha and wj are simply complex numbers, and we can distribute that conjugation over the product, so this becomes the sum from j = 1 to n of (zj((alpha-conjugate)(wj-conjugate))). And then we can just go ahead and pull our alpha-conjugate out from the sum, and we see that we do, in fact, end up with alpha-conjugate times (the product of <z with w>).

There are two additional properties that hold of the complex inner product: the Cauchy-Schwarz Inequality and the Triangle Inequality. We’ll state them in the following theorem.

Let V be a complex inner product space. Then for all w and z in our vector space, we have that the Cauchy-Schwarz Inequality holds, that the absolute value of (the product of <z with w>) is less than or equal to the product of (the length of z) times (the length of w). And we also have the Triangle Inequality, that the length of (z + w) is less than or equal to (the length of z) + (the length of w).

For once, we cannot simply copy our proof from the proof we used in the reals. I’ll prove the Triangle Inequality here. First, we want to note that the Triangle Inequality would be equivalent to the statement that the square of the length of (z + w) is less than or equal to the square of the sum of (the length of z) + (the length of w). Or equivalently, that (the length of (z + w))-squared – (the square of (the length of z) + (the length of w)) will be less than or equal to 0. To see if this is, in fact, the case, let’s expand the left side. So first of all, we can distribute out this ((length of z) + (length of w))-quantity-squared to become (the length of z)-squared + 2(the length of z)(the length of w) + (the length of w)-squared. And then we can start putting things into inner product notation. So our (length of (z + w))-quantity-squared is the product of <(z + w) with (z + w)>, and then we’re going to subtract out our (product of <z with z>), subtract our (product of <w, w>), and we’re still left with also a - 2(the length of z)(length of w). We can use our properties 3 of the inner product to split up our product of <(z + w) with (z + w)> into a (product of <z with z>) + (a product of <z with w>) + (a product of <w with z>) + (a product of <w with w>), and then cancel things out so that we’re left with (the product of <z with w>) + (the product of <w with z>) – 2(the length of z)(the length of w). Well, if you remember that the product of <w with z> is the conjugate of the product of <z with w>, then we get this, and if you remember that in general, for a complex number, a number plus its conjugate will equal 2 times the real part of the number. So, somehow we’ve changed this into needing to show that 2 times the real part of (the product of <z with w>) – 2(the length of z)(the length of w) is less than or equal to 0, and of course, divide out our 2, and we’ll see that this is the same as showing that the real part of (the product of <z with w>) is less than or equal to (the length of z)(the length of w).

At this point, we’re going to make use of the Cauchy-Schwarz Inequality by first showing that the real part of <z with w> is less than or equal to the absolute value of (the product of <z with w>). To see this, we first note that (the absolute value of (the product of <z with w>))-squared will equal (the real part of (the product of <z with w>))-squared + (the imaginary part of (the product of <z with w>))-squared. Again, I’m simply thinking this as a complex number, and the absolute value squared of a complex number is (its real part)-squared + (its imaginary part)-squared. Now, since (the imaginary part)-squared must be greater than or equal to 0, we see that (our real part)-squared must be less than or equal to the square of (the product of <z with w>). And thus, we have that the absolute value of the real part of (the product of <z with w>) must be less than or equal to the absolute value of the product of <z with w>. But since, of course, the real part of (the product of <z with w>) is less than or equal to its absolute value, we have managed to show that the real part of (the product of <z with w>) is less than or equal to the absolute value of (the product of <z with w>). And our Cauchy-Schwarz Inequality tells us that the absolute value of (the product of <z with w>) is less than or equal to (the length of z)(the length of w), so we see that the real part of (the product of <z with w>) is less than or equal to (the length of z)(the length of w), which completes our proof of the Triangle Inequality.

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