Lesson: The Standard Inner Product for \( \mathbb{C}^n \)

Question 2

1 point


When we first studied Rn, we introduced a dot product for vectors. We will extend this definition to the complex numbers as well. So let z = [z1 through zn] and w = [w1 through wn] be vectors in Cn. Then the dot product of z and w is z dot w, which equals z1w1 + through to znwn.

For example, the dot product of [2 – i; 1 + i; 3] and [-i; 5 – i; 3 + i] will equal (2 – i)(-i) + (1 + i)(5 – i) + 3(3 + i), which, if we perform the calculation, we see equals 14 + 5i.

In Rn, we ended up doing quite a lot with the dot product, but unfortunately, much of this does not translate well into the complex numbers—the exception being the definition of matrix multiplication, where we still have that the jkth entry of the matrix product AB is the dot product of the jth row of A with the kth column of B, even when A and B have complex entries.

But if we tried to use the dot product to define length, we would find that z dot z = z1-squared + through to zn-squared. At just a glance, this may seem like a fine start to defining length, but consider that z dot z is a complex number. So, what would the square root of (z dot z) be? Sure, we get two square roots, just like in the real numbers, but in the real numbers, we know that one of these is a positive and one is a negative, and we take the symbol “square root” to mean the positive square root. But the complex numbers do not have a notion of positive, negative, or even greater than or less than. So, trying to decide which complex number should be the length of a vector is an impossible task. Instead, we want our notion of length to be a real number, and as such, the dot product cannot be used for this purpose.

Instead, we will seek inspiration from C itself, which we can think of as C1. For in this space, we already have the notion of length: the modulus of a complex number. Recalling that (the modulus of z)-squared = z(z-conjugate) gives us the inspiration to define the following.

In Cn, the standard inner product, which will be notated with brackets, is defined by the product of <z and w> equals z dot (w-conjugate), so this will equal z1(w1-conjugate) + through to zn(wn-conjugate) for all w and z in Cn.

So note that if z and w contained only real entries, then (the conjugate of wj) = wj, and this inner product is the same as the dot product. So while it may not have seemed so at first, this is the correct way to extend the definition of the dot product to the complex numbers, and we can use this to define the length of complex vectors.

Let z be a vector in Cn. Then we define the length of z by the square root of (the product of <z with z>), which equals the square root of (z dotted with z-conjugate).

Let’s look at an example. The product of <[1 + 3i; -1 + 2i] and [3 + 4i; 2 – i]> will equal (1 + 3i)(3 – 4i) + (-1 + 2i)(2 + i). Multiplying this out and combining, we’ll see that this equals 11 + 8i. So again, our product has turned out to be a complex number.

To calculate the length of the vector [1 + 3i; -1 + 2i], we will first calculate that the square of the length equals the product of [1 + 3i; -1 + 2i] with itself. To calculate this product, we’ll see that it equals (1 + 3i)(1 – 3i) + (-1 + 2i)(-1 – 2i). Again, you multiply it all out, and recombine to find that this equals 15. So since the square of the length equals 15, it means that the length of our vector [1 + 3i; -1 + 2i] equals the square root of 15.

If we take a closer look at our example, we see that all the i terms cancelled out when we took the inner product. This was, of course, the point of this definition, but it means that we can simplify our calculation of the length. For if z = [z1 through zn], which we can write as [(x1 + y1i) through (xn + yni)], where our xj and yj are real numbers, then we’ll see that the product of z with itself will be the sum from j = 1 to n of ((xj + yji)(xj – yji)), but this is simply the sum from j = 1 to n of (xj-squared + yj-squared). So to find the length of z, we sum the squares of the real parts and the squares of the imaginary parts, and then take the square root.

For example, if I wanted to find the length of [3 + 4i; 2 – i], then I’ll take the square root of the sum of the squares of each real and imaginary part I see. So I’ll take (3-squared + 4-squared + 2-squared + (-1)-squared), which turns out to be the square root of 30. That’s a bit easier than actually multiplying out all those complex values.

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