Span and linear independence example | Vectors and spaces | Linear Algebra | Khan Academy

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I want to unite everything we have learned for linear independence and dependence, for a linear envelope of a set of vectors in a rather complex task, because if you understand what this task is about, that means you understand what we’re doing, and that’s the key to an understanding of linear algebra, these two concepts. The first question I will ask is about a set of vectors s, which are three-dimensional vectors, have three components, and the question is whether the linear shell of s is equal to R3. It seems possible. If each of these vectors adds new information, then it seems likely that I can describe each vector in R3 with these three vectors, with some combination of these three vectors. The second question I will ask is are these vectors linearly independent? Maybe we can answer both questions at the same time. Let’s deal with the first question first. Is the linear shell equal to R3? For their linear shell to be equal to R3, this means some linear combination of these three vectors to be able to give each vector in R3. I will make a linear combination of the three vectors. We can have c1 on the first vector, [1; –1; 2], plus another arbitrary constant c2, some number, according to the second vector, [2; 1; 3] plus some third scaling constant on the third vector [–1; 0; 2]. I have to be able, using random constants, to get some combination of these vectors that gives each vector in R3. I will represent each vector in R3 by vectors with coordinates a, b and c, as a, b and c are arbitrary real numbers. So if you give me some numbers a, b and c, I will be able to give you a formula to tell you what your c3, c2 and c1 are, which means that this combination has a linear shell equal to R3, because if you give me a vector, I can always tell you how to get this vector through these three vectors. Let’s see if I can do that. From our definition of scalar vector multiplication, we know that c1 on the first vector, I can rework it if I want. I usually skip this step, but now I want everything to be clear. So c1 on this … I can present it as 1 on c1 … each component multiplied by c1. Similarly, c2 on this vector is equal to each component on c2. And c3 on this vector is equal to each component on c3. I want to show you that everything we do it simply follows from the definition of scalar vector multiplication, which is what we just did, or collecting vectors, which we will now do. Adding vectors tells us that this component plus this component, plus this component must be equal to this component. I will write it down. We get c1 + 2c2 –c3 is equal to a. We will do the same with the next line. –C1 + c2 + 0 over c3 must be equal to b. We get -c1 + c2 + 0 over c3 – we don’t even need it to write it – is equal to b. Finally, let’s do the same with the third row. 2c1 + 3c2 + 2c3 is equal to c. Now let’s see how we can find the different constants. I will use elimination. Maybe you know this method. I think we’ve done it in some earlier videos of linear algebra, before I started doing it formally. We will negotiate again after a few videos, but I think you understand how it’s solved that way. I will first eliminate these first two members, and then I will eliminate this article, and finally I will find the different constants. If I want to eliminate this article first, then I can to add this equation and this equation. It would be even better to replace this equation with the sum of these two equations. Let’s do it. I’m just going to add these two equations and I will replace this with this sum. So -c1 + c1, this is equal to 0. I can ignore it. Then c2 + 2c2 is equal to 3c2. Then 0 plus -c3 is equal to -c3. -C3 is equal to … I replace this with the sum of these two, then b + a. This is equal to b + a. I will now rewrite the first equation above. The first equation does not change at all. So I get c1 + 2c2 – c3 = a. In the last equation, I want to eliminate this term. I take this equation and subtract it 2 times the first equation. It’s like adding this equation to the above equation multiplied by -2. Since we’re almost done using this, we can actually do it to write it down, just multiply this equation by -2. This becomes -2c1 -4c2 + 2c3 is equal to -2a. If we just multiply each of these members … I have to be careful, I don’t want to make a mistake inadvertently. –2 over c1, minus 4 over c2, plus 2 over c3, and then minus 2 over a. Now we can add these two. What happens? 2c1 minus 2c1 is equal to 0. No need to write it. 3c2 minus 4c2 is equal to -c2. Then we have our 2s3 plus 2s3, which is equal to 4s3, and this is equal to (c – 2a). I just replaced that with this minus 2 on this, and I got that. Now I will not change the above equation again. I won’t do anything with it, I’ll just copy it on the right. I get c1 + 2c2 – c3 = 0. I will keep the second equation, I get 3c2 – c3 = b + a. I’ll move a little. And then in this last equation I want to eliminate … My goal is to eliminate this article here. I want to multiply the lower equation 3 and then add it to the mean equation, to eliminate this article here. If I multiply this lower equation by 3 … I don’t want to create chaos, so this happens -3 + 3, which are destroyed. This becomes 12 minus 1. 12c3 minus c3 is obtained, which is equal to 11c3. And that’s happening … oh, I’m sorry, I already did. When I collect 3 on this plus this, they are destroyed. Then when I multiplied 3 by that, I got 12s3 minus c3, so this is 11c3. I multiplied this by 3 plus this, so I got 3s minus 6a … just multiply this by 3 … plus this: plus b + a. How can I process this? I want to clarify something. This c is different from those c1, c2 and c3 here. I think you understand that. But now I see I’ve used the letter twice, I don’t want to get confused. This has no index and is a different constant of all these constants here. Let’s see if we can simplify this. We have a and -6a, which we can add. Let’s get rid of this a, we get -5a. If we divide the two sides of this equation by 11, what will we get? We get c3 is equal to 1/11 (3c – 5a). If you give me the values \u200b\u200bof a and c, I will be able to immediately to say how much is c3. How much is c2? c2 is equal to … I will simplify this equation here. I’ll do it here. If I add c3 to both sides of the equation, I get 3c2 is equal to b + a + c3. If I divide both sides by 3, I get c2 is equal to at 1/3 (b + a + c3). I will leave it like that for now. How much is equal to c1? I can just rework the above equation if I subtract 2c2 and adding c3 to both sides, I get that c1 is equal to –2c2 + c3. What did I just show you? You can give me any vector in R3 you want to find. You can give me any real number for a, any real number for b and any real number for s. And if you give me those numbers, I say I can always give you some combination of these three vectors, whose sum is equal to this vector. And in fact, I’ve already found how much to multiply each of these vectors to add them to this third vector. If you give me the numbers a, b and c, I just have to substitute those a and with here. Oh, I’m sorry. Here I forgot b. There are also b. It was suspicious that I didn’t need b. There are also b. This here is 3c – 5a + b. Let me write it down. Here in the brackets there is b. But I think he understood the general idea. You give me your a, b and c, can be any real number. There is no division here, so I don’t have to worry about dividing by 0. So this is a linear combination of random real numbers, so I can get another real number. You give me your a, b and c, and I will give you c3. You give me a, b and c, I get c3. This will be just another real number. And then I can take that with your previous a and b and I can give you c2. We have already managed to find c2 and c3, and then just use a, and we’ll find how much is c1. I hope you see that no matter what the numbers a, b and c are, i can find c1, c2 or c3. There is no reason for any a, b or formula not to work. We don’t have a division, so even with 0 it works. I can definitely say that the set of these three vectors has a linear shell equal to R3. Now I will ask you another question. I already asked it. Are these vectors linearly independent? We said that in order to be linearly independent, the only solution for c1 in the first vector, [1; –1; 2], must be c1 = c2 = c3 = 0, plus c2 on the second vector [2; 1; 3], plus c3 on the third vector [–1; 0; 2]. If we have linear independence, that means the only solution of the equation … i.e. I want to find some set of combinations of these vectors that gives the zero vector, and I did that in the previous video. If they are linearly dependent, then there must be some non-zero solution. One of these constants, at least one of these constants, must be nonzero for this solution. You can always make them zeros, no matter what, but if they do are linearly dependent, then one of these coefficients will be nonzero. If they are linearly independent, then all these coefficients … the only solution to this equation will be c1, c2, c3, all three constants equal to zero. c1, c2 and c3 must be zero. Linear independence implies that this implies linear independence. This is exactly the same as what we did here, but in this case i just choose my a, b and c to be zero. This is a, this is b, and this is c, right? I can choose any vector from R3 for my a, b and c. Now I choose the zero vector. Let’s see how many are our c1, c2 and c3. My a is equal to b, is equal to c, is equal to zero. I choose them to be equal to the zero vector. What linear combination of these three vectors is equal to the zero vector? Well, if a, b and c are equal to 0, this term is equal to 0, it’s zero, it’s zero. We get 1/11 by 0, minus 0, plus 0. It’s just 0. So c3 is equal to 0. If c3 is equal to 0, we already know that a is equal to 0 and b is equal to 0. c2 is 1/3 over 0, so it is also equal to 0. Now how much is c1? It is equal to c3, which is 0, c2 is also zero, so 2 over 0 is 0. So c1 will just be equal to a. But I just said that a is equal to 0. So the only solution to this equation here, the only linear combination of these three vectors, from which the zero vector is obtained is when you multiply all three vectors by zero. I just showed you that c1, c2 and c3 must be zero. And since they are all zeros, we know it is a set of linearly independent vectors. Or that none of these vectors can be represented as a combination of the other two. This is interesting. We have exactly three vectors whose linear envelope is R3, and they are linearly independent. Linearly independent for me means that there are no redundant vectors, nothing that can to be constructed by the other vectors, and we have exactly three vectors whose linear shell is R3. So, basically, I didn’t prove it to you, but I could, if you have exactly three vectors and they have a linear shell R3, they must be linearly independent. If they are not linearly independent, then one of these will be redundant. Let’s say that this vector is redundant. I always choose the third, but let’s say this vector is redundant, which means that the linear shell will be equal to the linear envelope of these two vectors, right? Because if this vector is redundant, it just could be part of the linear envelope of these two vectors. And the linear envelope of two vectors can never be R3. Or vice versa, if you have three linearly independent vectors, three stacked threes, and they are independent, then you can always say that their linear shell is R3. I didn’t prove it to you, but I hope you got the idea, that each of them brings a new dimension, right? One points this way. They are not completely orthogonal to each other, but they give us enough focus that we can to add a new direction to the situation. I hope this is somewhat useful to you, and see you in the next video.

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