Understanding Subspaces in R4

Okay, here we have a true-false question about subspaces. And it says the set V, which is defined in this weird-looking way, which we'll talk about that, it says that set is a subspace of R4, true or false. Okay, let's talk about what is a subspace. First of all, the first term I want to define is a subset. And all that is, it's just any collection of vectors, right? So you could call it a subset, or you could just call it a set. There's nothing special about that. It's just a collection of vectors. But once you start talking about subspaces, Now you have to satisfy these special conditions. So a subspace is just a special subset. So it's a special collection of vectors. And it satisfies the following three conditions. The first one is it has to contain the zero vector, right? So that's pretty straightforward. The second one, if you add any two vectors in the set together, the sum is another vector that's still in the set. So you've probably heard this be referred to as being closed under addition. And all it means is you take two vectors, add them together, and you're staying in the set. right the sum is still in the set then the third one is called being being closed under scalar multiplication and all that means is if you multiply any vector in the set by any scalar then that scaled vector is still in the set so by multiplying a vector by some scalar you're not leaving the set you could say okay those are the three conditions so if we're given a set defined in this way how can we actually determine if it's a subspace or not so to do that um First, let's talk about how it's defined. So the set V is defined such that if you're looking at a vector, and then you can take here, it says x minus 4z equals 0. So that means you're looking at this vector, and you see that the first entry minus 4 times the third entry equals 0. So if that condition is met, then the vector that you're looking at is in the set V. So to solve this question, I'm going to define two vectors, a and b, such that they're in the set V. So let's say the A vector is some vector with components A1 through A4, and I'm going to define A to be in the set V, meaning its first entry minus four times its third entry equals zero. So we're going to say A is an element of the set V, meaning A1 minus four A3 is equal to zero. Now I'm going to define another vector B. and say this is also in the set V. So it'll have components b1 through b4. And I'll say b1 minus 4 times b3 equals 0, because that's the condition that vectors have to meet to be in the set. And I'm defining these vectors to be in the set. So we know these two equations are true. And now we just want to see, if we add them together, is the sum still in the set V? So right now, we're checking the second condition. But really quick, it's pretty easy to check this first condition. Does the set V contain the zero vector? Well, if X through W, all these components are equal to zero, does it satisfy this condition? Zero minus four times zero equals zero. Yes. So the zero vector satisfies this condition, which means the zero vector's in the set V. Okay, so for this set that's defining the problem, it meets the first condition. But to solve, to see if it meets the second and third condition, you have to have these example vectors up here. So let's see, if we add the vectors a and b, which are each in the set, do we get another vector that's still in the set? So let's see, what is the vector a plus b? Well, you just add the components. So you get a1 plus b1 is the first entry, a2 plus b2, a3 plus b3, and a4 plus b4. So here's the vector a plus b, and we need to see is the sum still in the set? And if so, then it would be closed under addition because we've written these vectors generally, right? So it would be true for all cases. So let's see. We want to see if a1 plus b1, right, the first entry, minus 4 times the third entry. We want to see if that equals 0 because if that's true, then this vector a plus b is in the set by definition. So minus 4 times the third entry, which is a3 plus b3. We want to see. does that equal zero? So we don't know, we have to find out. And to do this, we're gonna use these two equations that we know are true. So we can rearrange this left-hand side of the equation, and we can substitute in for a1 minus 4a3 and b1 minus 4b3 to be zero, and you'll see that this is actually true. So if we rearrange this equation, we say a1 minus 4a3, I'm doing this, minus four times a3, plus... b1 minus 4 times b3. You see I distributed the 4 and then I rearranged the terms. And this is equal to 0. We have to find out. But we know that this part here, a1 minus 4a3 is 0 because we defined an a to b in the set. So it has to meet that condition. So this is 0. And we know b1 minus 4b3 is also 0. And yes, 0 plus 0 does equal 0. So the sum a plus b is also in the set. That means that the set is closed under addition. So we know that this subset of vectors satisfies the second condition. So now we just have to see, does it satisfy the third one? To do that, we'll look again at just, now we just need to look at one vector a. Let's say it's the same a1 through a4. And we're defining a to be in the set, which means its first entry minus 4 times the third entry equals 0. That's how the set is defined. And now we just want to see, is it closed under scalar multiplication? So if we multiply by some scalar, C times A, for all values of C, will this still be in the set? So let's see. C times A is what? Well, you just distribute the scalar through, so we get CA1, CA2, CA3, and CA4. And we want to see, is this vector still in the set? To determine that, we just have to see, does the first entry minus four times the third entry equal zero? So we get CA1 minus four CA3 equals zero? Question mark? And then this is pretty straightforward. If you just factor out the C, you get C times A1 minus four A3 equals zero. Maybe we have to determine that. But look, we have over here that A1 minus four a three equals zero. So this part here is zero. And so no matter what we pick for our scalar multiple, we get c times zero equals zero, always, right? So we say that yes, this set is closed under scalar multiplication. And since this subset of vectors v satisfies these three conditions, we say that v is a subspace of R four. Being a subspace of R four just means that all that all the vectors in the subset are vectors in R4, which we can see is clearly true because it defines the set so that all vectors are in R4, okay?

First of all, the first term I want to define is a subset. And all that is, it's just any collection of vectors, right? So you could call it a subset, or you could just call it a set. There's nothing special about that. It's just a collection of vectors.

But once you start talking about subspaces, Now you have to satisfy these special conditions. So a subspace is just a special subset. So it's a special collection of vectors.

And it satisfies the following three conditions. The first one is it has to contain the zero vector, right? So that's pretty straightforward.

The second one, if you add any two vectors in the set together, the sum is another vector that's still in the set. So you've probably heard this be referred to as being closed under addition. And all it means is you take two vectors, add them together, and you're staying in the set. right the sum is still in the set then the third one is called being being closed under scalar multiplication and all that means is if you multiply any vector in the set by any scalar then that scaled vector is still in the set so by multiplying a vector by some scalar you're not leaving the set you could say okay those are the three conditions so if we're given a set defined in this way how can we actually determine if it's a subspace or not so to do that um First, let's talk about how it's defined. So the set V is defined such that if you're looking at a vector, and then you can take here, it says x minus 4z equals 0. So that means you're looking at this vector, and you see that the first entry minus 4 times the third entry equals 0. So if that condition is met, then the vector that you're looking at is in the set V. So to solve this question, I'm going to define two vectors, a and b, such that they're in the set V. So let's say the A vector is some vector with components A1 through A4, and I'm going to define A to be in the set V, meaning its first entry minus four times its third entry equals zero.

So we're going to say A is an element of the set V, meaning A1 minus four A3 is equal to zero. Now I'm going to define another vector B. and say this is also in the set V. So it'll have components b1 through b4. And I'll say b1 minus 4 times b3 equals 0, because that's the condition that vectors have to meet to be in the set.

And I'm defining these vectors to be in the set. So we know these two equations are true. And now we just want to see, if we add them together, is the sum still in the set V? So right now, we're checking the second condition.

But really quick, it's pretty easy to check this first condition. Does the set V contain the zero vector? Well, if X through W, all these components are equal to zero, does it satisfy this condition? Zero minus four times zero equals zero. Yes.

So the zero vector satisfies this condition, which means the zero vector's in the set V. Okay, so for this set that's defining the problem, it meets the first condition. But to solve, to see if it meets the second and third condition, you have to have these example vectors up here. So let's see, if we add the vectors a and b, which are each in the set, do we get another vector that's still in the set?

So let's see, what is the vector a plus b? Well, you just add the components. So you get a1 plus b1 is the first entry, a2 plus b2, a3 plus b3, and a4 plus b4. So here's the vector a plus b, and we need to see is the sum still in the set?

And if so, then it would be closed under addition because we've written these vectors generally, right? So it would be true for all cases. So let's see. We want to see if a1 plus b1, right, the first entry, minus 4 times the third entry.

We want to see if that equals 0 because if that's true, then this vector a plus b is in the set by definition. So minus 4 times the third entry, which is a3 plus b3. We want to see. does that equal zero? So we don't know, we have to find out.

And to do this, we're gonna use these two equations that we know are true. So we can rearrange this left-hand side of the equation, and we can substitute in for a1 minus 4a3 and b1 minus 4b3 to be zero, and you'll see that this is actually true. So if we rearrange this equation, we say a1 minus 4a3, I'm doing this, minus four times a3, plus... b1 minus 4 times b3. You see I distributed the 4 and then I rearranged the terms.

And this is equal to 0. We have to find out. But we know that this part here, a1 minus 4a3 is 0 because we defined an a to b in the set. So it has to meet that condition. So this is 0. And we know b1 minus 4b3 is also 0. And yes, 0 plus 0 does equal 0. So the sum a plus b is also in the set.

That means that the set is closed under addition. So we know that this subset of vectors satisfies the second condition. So now we just have to see, does it satisfy the third one?

To do that, we'll look again at just, now we just need to look at one vector a. Let's say it's the same a1 through a4. And we're defining a to be in the set, which means its first entry minus 4 times the third entry equals 0. That's how the set is defined.

And now we just want to see, is it closed under scalar multiplication? So if we multiply by some scalar, C times A, for all values of C, will this still be in the set? So let's see.

C times A is what? Well, you just distribute the scalar through, so we get CA1, CA2, CA3, and CA4. And we want to see, is this vector still in the set? To determine that, we just have to see, does the first entry minus four times the third entry equal zero? So we get CA1 minus four CA3 equals zero?

Question mark? And then this is pretty straightforward. If you just factor out the C, you get C times A1 minus four A3 equals zero.

Maybe we have to determine that. But look, we have over here that A1 minus four a three equals zero. So this part here is zero.

And so no matter what we pick for our scalar multiple, we get c times zero equals zero, always, right? So we say that yes, this set is closed under scalar multiplication. And since this subset of vectors v satisfies these three conditions, we say that v is a subspace of R four.

Being a subspace of R four just means that all that all the vectors in the subset are vectors in R4, which we can see is clearly true because it defines the set so that all vectors are in R4, okay?

Transcript for:Understanding Subspaces in R4

Transcript for:
Understanding Subspaces in R4