But we have this first equation right here, that c1, this first equation that says c1 plus 0 is equal to x1, so c1 is equal to x1. Because we're just scaling them up. Is this an honest mistake or is it just a property of unit vectors having no fixed dimension? So this is some weight on a, and then we can add up arbitrary multiples of b.
So it's really just scaling. You get 3c2 is equal to x2 minus 2x1. Sal was setting up the elimination step. I'm going to assume the origin must remain static for this reason. So let's say I have a couple of vectors, v1, v2, and it goes all the way to vn. Write each combination of vectors as a single vector. →AB+→BC - Home Work Help. So this is a set of vectors because I can pick my ci's to be any member of the real numbers, and that's true for i-- so I should write for i to be anywhere between 1 and n. All I'm saying is that look, I can multiply each of these vectors by any value, any arbitrary value, real value, and then I can add them up. So it's just c times a, all of those vectors.
My a vector was right like that. Now you might say, hey Sal, why are you even introducing this idea of a linear combination? So if you add 3a to minus 2b, we get to this vector. Denote the rows of by, and. So this vector is 3a, and then we added to that 2b, right? Let me remember that. C2 is equal to 1/3 times x2. One term you are going to hear a lot of in these videos, and in linear algebra in general, is the idea of a linear combination. Let me draw it in a better color. Write each combination of vectors as a single vector. (a) ab + bc. Now, if I can show you that I can always find c1's and c2's given any x1's and x2's, then I've proven that I can get to any point in R2 using just these two vectors. He may have chosen elimination because that is how we work with matrices. This is j. j is that. And that's why I was like, wait, this is looking strange.
If you have n vectors, but just one of them is a linear combination of the others, then you have n - 1 linearly independent vectors, and thus you can represent R(n - 1). My a vector looked like that. We get a 0 here, plus 0 is equal to minus 2x1. So that's 3a, 3 times a will look like that. So c1 is equal to x1. What does that even mean?
Over here, I just kept putting different numbers for the weights, I guess we could call them, for c1 and c2 in this combination of a and b, right? But this is just one combination, one linear combination of a and b. Well, I know that c1 is equal to x1, so that's equal to 2, and c2 is equal to 1/3 times 2 minus 2. Around13:50when Sal gives a generalized mathematical definition of "span" he defines "i" as having to be greater than one and less than "n". What combinations of a and b can be there? So b is the vector minus 2, minus 2. This happens when the matrix row-reduces to the identity matrix. Linear combinations and span (video. And in our notation, i, the unit vector i that you learned in physics class, would be the vector 1, 0. So it equals all of R2. So you scale them by c1, c2, all the way to cn, where everything from c1 to cn are all a member of the real numbers.
That would be the 0 vector, but this is a completely valid linear combination. So you call one of them x1 and one x2, which could equal 10 and 5 respectively. Is this because "i" is indicating the instances of the variable "c" or is there something in the definition I'm missing? Since we've learned in earlier lessons that vectors can have any origin, this seems to imply that all combinations of vector A and/or vector B would represent R^2 in a 2D real coordinate space just by moving the origin around. So 2 minus 2 times x1, so minus 2 times 2. If nothing is telling you otherwise, it's safe to assume that a vector is in it's standard position; and for the purposes of spaces and. This is minus 2b, all the way, in standard form, standard position, minus 2b. That's going to be a future video. Oh, it's way up there. We haven't even defined what it means to multiply a vector, and there's actually several ways to do it. Write each combination of vectors as a single vector.co.jp. Why do you have to add that little linear prefix there? This was looking suspicious. I made a slight error here, and this was good that I actually tried it out with real numbers. So that one just gets us there.
Since you can add A to both sides of another equation, you can also add A1 to one side and A2 to the other side - because A1=A2. My text also says that there is only one situation where the span would not be infinite. Since L1=R1, we can substitute R1 for L1 on the right hand side: L2 + L1 = R2 + R1. Let's call that value A. So it could be 0 times a plus-- well, it could be 0 times a plus 0 times b, which, of course, would be what? Let's figure it out. Write each combination of vectors as a single vector icons. Sal just draws an arrow to it, and I have no idea how to refer to it mathematically speaking. What is the linear combination of a and b? Define two matrices and as follows: Let and be two scalars. So I had to take a moment of pause. In order to answer this question, note that a linear combination of, and with coefficients, and has the following form: Now, is a linear combination of, and if and only if we can find, and such that which is equivalent to But we know that two vectors are equal if and only if their corresponding elements are all equal to each other. And now the set of all of the combinations, scaled-up combinations I can get, that's the span of these vectors. It would look something like-- let me make sure I'm doing this-- it would look something like this.
Or divide both sides by 3, you get c2 is equal to 1/3 x2 minus x1. This is for this particular a and b, not for the a and b-- for this blue a and this yellow b, the span here is just this line. Another way to explain it - consider two equations: L1 = R1. Let me write it down here. Note that all the matrices involved in a linear combination need to have the same dimension (otherwise matrix addition would not be possible). This lecture is about linear combinations of vectors and matrices. I wrote it right here. So in which situation would the span not be infinite? Is it because the number of vectors doesn't have to be the same as the size of the space? Let me do it in a different color. 2 times my vector a 1, 2, minus 2/3 times my vector b 0, 3, should equal 2, 2. The first equation is already solved for C_1 so it would be very easy to use substitution. So let's say that my combination, I say c1 times a plus c2 times b has to be equal to my vector x.
These form a basis for R2. The next thing he does is add the two equations and the C_1 variable is eliminated allowing us to solve for C_2. At17:38, Sal "adds" the equations for x1 and x2 together. Let me write it out. And then we also know that 2 times c2-- sorry. No, that looks like a mistake, he must of been thinking that each square was of unit one and not the unit 2 marker as stated on the scale. If we want a point here, we just take a little smaller a, and then we can add all the b's that fill up all of that line.
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