 So in the previous video, we considered the following linear transformation. T is a map from R3 to R2. That follows the following rule. T of x1, x2, x3 maps to x1 plus 2x2, x3 minus 3x2. And the question I want to ask is, how does one compute the kernel of T? What is the kernel of this linear transformation? Well, if you're computing the kernel, remember what the definition of the kernel is all about, the kernel is trying to find out those vectors, you have to solve the equation T of x equals the zero vector. Let's unwrap that a little bit. T of x is given by this formula right here. You have x1 plus 2x2 and x3 minus 3x2. This is equal to the zero vector. This is the zero vector in R2. And as such, this is just the vector with two zero entries. So this is the equation, this is a vector equation that we're trying to deal with. But it turns out if you think about it, it's like, okay, what does it mean for two vectors to be equal? Two vectors are equal if their first entries are equal. And if their second entries are equal, that's what equality of vectors means. And so therefore, looking at the first entries, we have x1 plus 2x2, this has to equal zero. And x3 minus 3x2, this has to likewise equal zero. And that's what it would mean for those two vectors to equal each other. But it's like, hmm, x1 plus 2x2 equals zero, x3 minus 3x2 equals zero. If I draw a little curly sign in front of that, this then looks like a system of linear equations. This is a linear system. Trying to solve that vector equation comes down to solving a linear system, which will come down to computing the kernel of this linear transformation. Well, how do we go about doing this? Well, if you want to solve the linear system, there's plenty of options. You can use the technique of elimination, the technique of substitution. These, of course, are all options. I'm going to try to solve this one using substitution. That was always one of my favorite ones when I was working on these things. And so let's say, for example, the first equation, let's solve it for x1. If you solve for x1, you see that x1 is going to equal negative 2x2. All right, you could substitute that into the second equation. You'll notice the second equation also has an x2 in it. If you solve for x3 in that equation, you'll see that x3 equals 3x2. And that's about the best you can do, right? You can't really get, you can't solve it really better than that. And that's because in this system, you can see that x2 is going to be an independent variable. It's independent. And on the other hand, you're going to see that x1 and x3 are going to be dependent variables. Whatever you choose for x2, which there's no restriction on when we can choose for x2, it could be any real number, whatever you choose for x2, we can, x1 and x2 will be determined by that. So as an example, right, if we chose, if we chose x2 to equal 0, that would imply that x1 and x3 are likewise equal to 0. That is, our vector x is the 0 vector. This hopefully should come as no surprise because we've seen before that for any linear transformation, the 0 vector always maps to the 0 vector. So one thing, the one thing I want you to be aware of is that when it comes to a kernel, when it comes to a kernel, every kernel for every linear transformation under the sun will contain the 0 vector. The 0 vector is always part of the kernel. But are there other vectors in there? And it turns out there are, right? If we take x2 to equal 1, then you would get that x1 is equal to negative 2, and you get that x3 is equal to 3. And so your vector x, we're saying here, is the vector negative 2, 1 and 3. Like so. And let's verify this. Let's verify this. If you take t of negative 2, 1 and 3, you're going to end up with the following. Take the first component, n times 2, add to that 2 times the second component, and then take 3, the third component, minus 3 times the second component. And what do we get there? We end up with a negative 2 plus 2. That's a 0. You get 3 minus 3, which is a 0. This is the 0 vector. So this shows us that in this situation, negative 2, 1, 3 is likewise inside of the kernel of t. So what else is inside the kernel? Well, since we have this free variable of x2, we can just choose x2 to be whatever we want, and we're going to see the following. The kernel, the kernel of this map, is going to be the set of all vectors of the following form. x2 could be whatever, whatever you want. x1 will just be negative 2 times that number you chose, and then x3 will just be 3 times that number you chose, where t could be an arbitrary real number. And so the kernel is going to be all vectors of this form right here. The two examples we saw already is when t was 0 and t was 1. But notice that every component here has a multiple of t in it. If we factor that t out, we would see that the kernel is just t times negative 2, 1, and 3. And so the kernel of this vector, sorry, the kernel of this linear transformation is just the scalar multiple of a specific vector, the vector we got when we plugged in t equals 1. We're going to see in the future that this type of phenomenon happens all the time with kernels, that the kernel will just be a linear combination of a certain set of vectors, which we will find out is pretty important. And we'll learn about those more in the future. For the meanwhile, though, I just want to show us how one could compute, I should say, the set of vectors that map to 0, a.k.a. the kernel of the transformation.