 In the previous video, I talked a lot on why it was so important to find the inner product the way it was. The inner product for real vectors is the dot product. The inner product for the complex vectors is the Hermitian product. And I emphasize it had to do with that positive definite condition that when you take the inner product of a vector with itself, this cannot equal zero unless of course, you know, V was the zero vector itself. That's the only exception to that rule, the positive definite condition. Let me show you why we want the positive definite condition. With an inner product, we can define something called the length of the vector or more commonly in linear algebra, it's called the norm of the vector. In which case, it's going to be denoted as you take your vector and you write in this case like a double absolute value. Some people actually use absolute value symbols. You see that in practice. That's perfectly fine. For us here, we're going to use double absolute values to describe the norm, the length of the vector. And it really is meant to describe how long a vector is. Remember, we kind of first thought of vectors as these arrows in space and like the length of that vector, that is what we're trying to describe right now as its norm. And we use, we can define the norm using this inner product. So in fact, the norm of a vector is defined to be the square root of V dot V, which when V is a real vector, that's just the usual dot product. When V is a complex vector, it's going to be the Hermitian product. And so if V is just a column vector in FN, where FN is either RN or CN, then the dot product with itself is to produce this sum V1 squared plus V2 squared plus V3 squared all the way down to VN squared, where the entries of V are V1, V2 up to VN, right? And then you put that inside the square root. Great. That gives us the length of the vector, the norm of the vector. If a vector's norm happens to be one, we call this a unit vector. The vector of length one, unit vectors will be very important for us. Also, a nice thing to mention here, and we use this identity all the time, is if you take a vector dot itself, this is equal to the square of its norm. That's just taking this equation, squaring both sides. Now what important property you should know about norms is the following, that if you take the norm of a scalar product, this is equal to the scalar times the norm. Although you do have to take the absolute value of that when it's a real number. If c were, of course, were a complex number, when you talk about the norm here of c, the absolute value, well, let's pretend this is some complex number, right? So it looks like a plus bi. In that situation, the norm, the scalar there is going to be the square root of a squared plus b squared. Or in other words, this is just equal to a plus bi times a minus bi. So it's the product of the complex number with its conjugate. This is the exact, this is a generalization of the absolute value. It's sometimes called the modulus of the complex number or the norm of the complex number. And so when you take absolute values of complex numbers, you're talking about this modulus, this norm of the complex number, be aware of that. And so this property right here is really one of the critical observations when it comes to the norm. In fact, in general linear algebra, any vector space which has a function that basically behaves like this, it has the positive definite condition. It has this multiplicative property with scalar multiplication. You also need the triangle inequality. Then that's what we call a normed vector space. We can construct norm vector spaces using inner products. If a vector space has an inner product, we call it an inner product space and that leads to this norm vector space, which we're going to see these for our vector spaces here. So let's just do an example in R4. So if you have the vector 1, 0, 2, negative 2, then the length of the vector we calculate its norm, this is going to look like v dot v, v dot v inside the square root. And so you just take the product of the elements with itself. So you're going to get 1 times 1 plus 0 times 0 plus 2 times 2 plus negative 2 times negative 2, which in case you're going to get a sum of squares, you're going to get 1 plus 0 plus 4 plus 4. And so that's usually how you do these calculations. So if you want to take the product, you're just going to sum up the squares of the entries there inside of the square root. That adds up to be the square root of 9. And so the length of this vector turns out to be 3. Oftentimes when you compute the length of the vector, you're going to get some irrational number like the square root of 7 or whatever. I want to show you though that by the multiplicative property we saw just a moment ago, this one right here, it should be true that if I take 1 third times the vector v right here, by the multiplicative property, this should equal the absolute value of 1 third times the norm of v, which the absolute value of 1 third is 1 third and the norm of v turned out to be 3. This should equal 1. And therefore we would have that the vector 1 third v, this is going to be a unit vector. It has length 1. And I would encourage you to check this on your own here. If we take, you know, if you just add a third to all of these things, you're going to see that in fact its length will be 1. You're going to get 1 ninth plus 0 plus 4 ninths plus 4 ninths all inside the square root that adds up to be 1. Take the square root, you get 1. All right, I told you to do it yourself, but I just did it. There you go. And this is something we can actually do in general, right? So given any nonzero vector, so if v, if v is a nonzero vector, then it's so called normalization. It's normalization, which would look like 1 over the norm of v times v. So the norm of a vector is a scalar. 1 over that scalar would also be a scalar. So this is just a scalar multiple of v. It's normalization. This is going to be the unit vector, the unit vector. So it has a length 1. It's the unit vector in the so-called same direction, the same direction as v itself. Now direction is a very geometric idea here. What does direction mean? Well, geometrically, or at least from the linear algebra perspective, what we're trying to say is if you have some vector v, right, that vector will naturally span a one-dimensional subspace, which we call a line. And any other vector which spans that same line, we would say, is in the same direction. And so if we had our vector, let's say the original vector v was really long like this, the unit, it's normalization, which is a unit vector. We're just saying, oh, we're going to take a smaller version of that, but now it's length is 1. And so that's the normalization. We're going to see later in this chapter the importance of normalizing vectors. But I want to be before I conclude this video, I want to do a norm calculation using a complex vector. Things are a little bit different, but not by much, right? When it comes to computing the norm, this is going to be the square root of u dot u. Now the most common mistake when people compute inner products with complex vectors is they forget that the inner product is not always the dot product, probably because they focus on real vectors so often. The Hermitian products will be used for complex vectors. So the Hermitian product would look like 1 plus i bar times 1 plus i. And then you add to that i bar times i. That's just the number i, it's not a vector. And then you're going to get 3 minus i bar times 3 minus i. And this sits inside of a square root. For which case then you're going to take conjugates of these things. You're going to get 1 minus i times 1 plus i. You're going to get negative i times i plus, then you're going to get 3 plus i times 3 minus i. This all sits inside of the square root. Now you have to foil this stuff out. It seems like it's painful, but one thing to remember with complex numbers is if ever you have a complex number a plus bi and you multiply by its conjugate a minus bi, when you foil that thing out, you're going to get an a squared, you're going to get a negative a bi, you're going to get a positive a bi, and then you're going to get a negative b squared i squared. Well, things to remember is that the negative a bi will cancel with the positive bi, and then i squared itself is equal to negative 1, which will cancel with this negative. So when you multiply a complex number by its conjugate, you always get a squared plus b squared, the sum of the squares of its real part and imaginary part. And so as such, if I was computing the norm of this vector, I would skip all of this stuff right here, and I would jump when I calculate the norm to the following, I'm going to get the square root of 1 squared plus 1 squared. So looking at the sum of squares of the first part, then you're going to get a 1 squared from the i right there. And you're not going to have any i, so that's an important thing there. And then, so like this was our first part, the second part. And if you don't like the second part there, let me write it like this, 0 squared plus 1 squared, if you insist upon it. And then for the last one, you're going to get a 3 squared plus a 1 squared from this part right here. The signs don't matter. This is all going to be positive when we're done. 3 squared plus 1 squared. And so then we compute this thing. So what we have, we have a lot of ones, right? So we have 1 plus 1 plus 1, that's a 3 plus 1, that's a 4. And then 3 squared is 9. So we end up with the square root of 13 when we're done with this thing. And that would then be the length, the norm of this complex vector. It will be a real number. There won't be any i's. They will all cancel out through the process, because of all these conjugations. Again, that's why we use the conjugate transpose when we define the Hermitian product. Because by using conjugates, this will guarantee that the final result is going to get a positive real number.