 One of the primary tasks of mathematics is to try and generalize things that we do. So, for example, many times what we do is we use a symbol to indicate that we're going to do something to a particular object. And these symbols are actually referred to as operators, and the way to interpret them is they are instructions that we're going to do something to what follows. So, for example, when I write this particular symbol, I read this as the square root operator, and what it tells us is we want to find the principal square root of whatever we put inside. So, if I write this 4, this says find the principal square root of 4, and I can calculate that to be 2. Likewise, I have things like the differential operator. This says you're going to differentiate with respect to x whatever follows it. So, I can write down d dx x cubed, and I know I'm going to do something to x cubed, and so, in this case, that gives me 3x squared. And we can actually extend this to even simpler operations. So, for example, I might write 5 plus, and that's an operator if I choose to interpret it as adding whatever follows to the value 5. So, when I write this before 3, this says take 3 and add it to 5, and that gets me 8. Well, there's many, many different types of operators, but those that are particularly nice are known as linear operator, and what makes this a very nice operator is that it is something that acts on a vector, which is kind of our now generic term for something, and it's a linear operator if we preserve scalar multiplication and vector addition. And what that means in particular, if I apply the operator to a scalar multiple of a vector, what I get is the scalar multiple of the linear operator applied to the vector. And if I have the linear operator applied to the sum of two vectors, it's going to give me the sum of the linear operators applied to the vectors individually. So, let's take an example. Let's take an actual vector, u equals u1, u2, u3, and I'll define a particular operator. I'll give it a name capital D, and if I apply that to the vector v, it's going to be u, this given vector dotted with whatever vector I have. And we want to prove or disprove that D is a linear operator. So, we want to determine whether D meets the requirements, so we'll check those requirements. So, first thing, it does our operator D preserve scalar multiplication. If I apply D to c times a vector, do I get c times D applied to the vector? So, let's see if we can do this. We'll build a bridge. We'll go from the left-hand side to the right-hand side. So, I have some vector v, D applied to cv. What I'd like to determine is whether it's going to be c times Dv. And here's my starting point. Here's where I'd like to get to, and if I can build a bridge from this side to this side, then I have proven that we've preserved scalar multiplication. Conversely, if I can't build the bridge, if there's something that happens in here that prevents me from getting here, then I know that these two are not equal, and I don't have a linear operator. Well, let's go ahead and fill in the gaps. And again, we have our definition c times vector v. Well, that's a scalar multiple of this vector v, so I know what that's going to be. It's going to be cv1, cv2, cv3. And my definition D applied to any vector v is going to be the dot product of that vector with u. So, I'll find the dot product of this vector with this vector, and that's going to be this. Let's take a step back. Let's meet in the middle. So, vector v is going to be this v1, v2, v3. C applied to D. Well, so D applied to this vector. D applied to a vector is u dot v. Here's my vector. Here's u. So, I'll find that dot product. And c is going to be multiplied by that whole thing. And these two lines I can get from this line to this line by factors A. And so what that means is I do actually have a bridge that goes from my initial line here to my final line there. And so my operator D does in fact preserve scalar multiplication. Now, the other thing we have to show in order to claim that D is a linear operator is we have to show it also preserves a vector addition. So, D applied to the sum of two vectors is the same as the sum of D applied to the vectors individually. So, I need two vectors. So, let v equal v1, v2, v3. W equals w1 through 3. And let's go ahead and do the same sort of bridge building. Here's the starting point D of v plus w. Here's my ending point D of v plus D of w. And I want to go from here to here and see if I can build that bridge. So, this is going to be D. Well, v plus w, that's a vector addition. So, I can add the component wise. I can add the vectors component wise. And my definition says if I apply D to a vector, it's the dot product of that vector with my predefined vector u. So, there's my dot product. And let's see. Yeah, let's take a step backwards. So, here I have D applied to v, D applied to w. So, I'm going to dot v with u. I'm going to dot w with u. And so, that's going to be those two dot products. And again, the question is, will these two bridges meet in the middle? And if they do, we're in good shape. And let's see. Well, that takes a little bit of, let's go ahead and expand that out. I'll expand each of these products out. And let's see. There's u1, v1. That's u1, w1 is over here. u2, v2 is here. u2, w2, u3, v3, u3, w3. So, just rearranging these terms will give me this set of terms. And so, I now have the bridge from here to there. It connects. And so, my starting point and my ending point are connected. And my operator D satisfies all the requirements for a linear operator. I have the preservation of vector addition and the preservation of scalar multiplication. So, I have everything. And again, it's nice to have a little tag line that summarizes all the work that we've done. So, we might say D satisfies all the required properties for a linear operator.