 The dot product is a very specific instance of what's known as an inner product. The inner product emerges from a mathematician's inclination to generalize a concept. There are several useful features of the dot product that we want to preserve when we generalize it, and so we define an inner product as follows. Let V be a vector space where U, V, and W are all vectors in V. Suppose this inner product, U, V, assigns each pair of vectors U and V to a scalar and satisfies the following properties. Again, a quick observation, there are only so many symbols, and here we're recycling these symbols that we use to indicate vectors, but this time what we mean by this set of symbols is the inner product of the vectors U and V. So what does the inner product do? First of all, we want the inner product of any vector with itself to be zero if and only if the vector is itself the zero vector. Next, we want the inner product of any vector with itself to be greater than or equal to zero for all vectors. These first two requirements are usually bundled as a single requirement. Next, we want the inner product of U and V to be the conjugate of the inner product of V and U. Fourth, we want the inner product of U with the vector sum V plus W to be the sum of the inner product of U and V with the inner product of U and W. Finally, for any scalar C, the inner product of U and the scalar multiple C, V is going to be C times the inner product of U and V. If all of these things hold, then the function U, V is an inner product and our vector space V is an inner product space. So we claimed that the inner product is a generalization of the dot product, but let's make sure that the dot product is actually an inner product. So we'll check our first two requirements. If I have some vector V, then the dot product of V with itself is going to be V1 squared plus V2 squared and so on, and that is always going to be greater than or equal to zero. Moreover, the only way this can possibly ever be zero is for all of the VIs to be zero. And that's only true for the zero vector. So that satisfies our first two requirements. Next, we want to prove that the inner product of U and V is going to be the conjugate of the inner product of V and U. And one way we can do this is we can start with the inner product of U and V and see if we can build a bridge to the conjugate of the inner product of V and U. So we can write down the dot product of U and V and we'll work a step back from the end and write down the conjugate of the dot product of V and U. Since these are vectors in Rn, all of our components are going to be real numbers and so the conjugate of a sum of products of real numbers is just going to be the same sum of the products of real numbers. And again, since the components are real numbers, I know that the product of two real numbers is commutative. So V1, U1 is the same as U1, V1, U2, V2 is the same as V2, U2, and so on. And so I have a bridge from the inner product of U and V to the conjugate of the inner product of V and U. The next thing I want to show is that the inner product of U with the sum V plus W is the same as the inner product of U and V plus the inner product of U and W. So again, we'll set down our starting point, we'll set down our ending point, and try to work towards the middle. So I know the vector U has components U1 through Un and the vector V plus W has components V1 plus W1, V2 plus W2, and so on to Vn plus Wn. And when I form the dot product, I will find the sum of the component-wise products. And we'll expand that out a little bit. On the other hand, let's take a step backwards from our ending point. This inner product of U and V, since we're working with the dot product, is going to be the sum of the component-wise products of U and V. And likewise, the inner product of U and W will be the sum of the component-wise products of U and W. And we see that if we rearrange the terms in our sum, we'll get the previous line. And that connects the starting point to the ending point, and we can conclude that the dot product satisfies the fourth of the requirements necessary to be an inner product. And finally, we'll check that last requirement that the inner product of U with a scalar multiple of the second factor is the scalar multiple of the inner product of U and V. So I'll set down my starting point, set down my ending point, and see if I can work towards the middle. So the dot product of U with a scalar multiple of V is going to be C U1V1 plus C U2V2 and so on. Meanwhile, C times the inner product of U and V, that's our dot product, is going to be C times U1V1 plus U2V2 and so on. And if we expand, we'll see the two are the same. So the dot product satisfies all of the requirements for being an inner product, and that means our vectors in Rn with the dot product form an inner product space. So who cares? Well, one of the advantages to inner product spaces is that we can determine quite a bit of information from a very small amount. For example, suppose V is an inner product space, and I know one inner product, namely the inner product of the vector A and the vector B. Given this single inner product, can I find other inner products? Since V is an inner product space, we know the inner product of C U and V is equal to... Well, actually, we don't know what happens when the first vector is multiplied by a scalar. So all that we know is the inner product of C U and V is the inner product of C U and V. And because of this, we have to try something else. So because V is an inner product space, we know that the inner product of U and V is the conjugate of the inner product of V and U. And so that tells me that the inner product of B and A is the conjugate of the inner product of A and B. Well I know the inner product of A and B, it's five, and the conjugate of five is just five, and so that means I know the inner product of B and A. And the natural question to ask is, is this relevant? And the important answer is, who knows? But it's a fact, and facts are often useful. Looking at our question, we want to know something about the inner product of 3A. And now that A is the second component of our inner product, we can actually find this using our last rule. So I can find the inner product of B with 3A is 3 times the inner product of B with A. And so that tells me that that inner product is 3 times 5 is equal to 15. And we now have the right vectors B and 3A, but they're in the wrong places. But one of the properties of an inner product is that if we switch the order, we'll get the conjugate. And so this means that the inner product of 3A and B is the conjugate of the inner product of B and 3A. And we already know what that is. And that gives us the inner product of 3A and B.