 One of the main activities of mathematics is generalizing concepts. In this case, the dot product can be generalized to the notion of an inner product. Suppose I have some vector space with u, v, and w vectors in our space. We're going to define an inner product to be a function that assigns each pair of vectors, u and v, to a scalar where the inner product satisfies the following properties. First, the inner product of any vector with itself is going to be zero, if and only if the vector is itself the zero vector. For all other cases, the inner product of a vector with itself is going to be greater than zero. Next, if we alter the order in which we take the vector, then the inner product of the vectors u and v is going to be the conjugate of the inner product of the vectors v and u. Next, we have a restricted form of additivity. The inner product of u with the sum of two vectors is the same as the inner product of u with the first vector, plus the inner product of u with the second vector. Note that this only applies to the sum of the vectors in the second position. And of course, we want to deal with scalar multiplication. So now for any scalar c, the inner product of u with a scalar multiple of the second vector should be the product of the scalar with the inner product of the two vectors. And if we have our inner product together with our vector space, we can call the vector space an inner product space. We'll make a couple of quick observations on this definition. First of all, if our scalars are drawn from the set of real numbers, the inner product of u and v being the conjugate of the inner product of v and u gives us commutativity. Don't believe me, you should prove it. Likewise, our requirement that the inner product of u with a scalar multiple of v must be equal to the scalar c times the inner product of u and v does not commit us to anything about the inner product of a scalar multiple of u with the vector v. And likewise, our additivity says nothing about what happens when we add two vectors in the first component. So let's start off by proving that the dot product is itself an inner product. So the first things we want to check are that the inner product of a vector with itself is zero if and only if the vector itself is zero and that the inner product of a vector with itself is greater than or equal to zero for all vectors. It's usually convenient to check these two at the same time. So we'll form the dot product of a vector with itself which is going to be the sum of the squares of the vector components and since the vector components are real numbers, this sum is going to be greater than or equal to zero which confirms the second requirement. And it's worth noting the only way that we can have this sum equal to zero is if all of the v i's are themselves also equal to zero which is only going to be true for the zero vector and that confirms the first requirement. The next requirement we want to check is that the inner product of two vectors is the conjugate of the inner product of the two vectors taken in the opposite order. So we'll take two vectors in Rn and form their dot product. And here's where the magic of proof begins. Well, like all magic tricks, it's not really a trick once you understand how it works. For proofs, one important thing to keep in mind is the last thing that you write is what you've actually proven. And what this means is that a good way to build a proof is to start at the end with what you want to prove and work your way back to the beginning. In this case, because we want the dot product of u and v to be the conjugate of the dot product of v and u, write down that conjugate at the very end of our proof and it's important to recognize that this is at the end all of the proof should be above this line. So now we find the conjugate of the dot product which is going to be the conjugate of the component-wise product and sum but because u and v have components in the real numbers the conjugate of the sum is just the sum itself and finally it's worth noting that the second line has the same products as the first line just that the products aren't done in the reverse order but because these components are in the real numbers and multiplication is commutative our second line does in fact follow from the first. Now one caution about when constructing proofs in the reverse order like this you should always make sure that they read correctly forward. So let's go through our proof line by line. So we form the dot product of u and v as the sum of the component-wise products multiplication is commutative so we can reverse the order of the products because these components are in the real numbers the conjugate is the same as the value itself and that conjugate is in fact the conjugate of the dot product of v and u and so this checks our third requirement. Since the first three properties held then the rest need to be checked in a similar fashion. We'll leave that up to you. Remember trust but verify. One useful thing is that once we know that we have an inner product space then given even a single inner product we can find a large number of inner products including all linear combinations of the vectors in the inner product. Let's see how that might work. So suppose I have an inner product space and I know that the inner product of a and b is 5 and I want to find another inner product say of 3a and b. Well since we know that v is an inner product space then we know that the inner product of c, a and b is equal to the inner product of c, a and b. The only thing we know about inner product and scalar multiples is where we're taking a scalar multiple of the second vector. So we have to try something else. It'll be useful to have the definition of inner product space. Let's see if there's anything here that can help us. Well we don't seem to have the zero vector so the first requirement doesn't seem to be useful and the second requirement gives us an inequality and we don't really care about an inequality. We don't really care about what something is equal to. So how about this third requirement? That the inner product of two vectors is the conjugate of the inner product taken in the reverse order. So let's see if we can do anything with that. Now you might wonder, do we actually need this? And the answer is, who knows? But it is a fact and facts are often useful. The worst thing that will happen if we make use of this is a little bit of paper, or even in this case a little bit of screen space, and paper is cheap. So let's try it and see where it takes us. So we know the inner product of B and A is the conjugate of the inner product of A and B and that's going to be the conjugate of five which is just five because five is itself a real number. Now remember we want to say something about the inner product of vector 3A with something. So let's pull our definition of inner product back in and see if there's something useful here that can help. And in our definition we see that the inner product of a vector with a scalar multiple of a second vector is defined. So this would at least allow us to say something about an inner product involving a vector 3A. So the inner product of B and 3A is going to be three times the inner product of B and A. And we just found that inner product is equal to five. So we can replace it in our equation and find that the inner product of B and 3A is equal to fifteen. Since we know that V is an inner product space we also know that the inner product of two vectors is the conjugate of the inner product taken in the reverse order. And so if I reverse the order of the vectors and find the inner product of 3A and B it's going to be the conjugate of the inner product of B and 3A. And I know that's going to be the conjugate of fifteen which is just going to be fifteen.