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Studying for a test? Prepare with these 5 lessons on Alternate coordinate systems (bases).
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I've got this 2 by 2 matrix A here, and I've got this other member of R2, this vector b. Let's figure out all of the interesting things that we can figure out about this matrix and this vector. So the first thing of interest, and this is all going to essentially help us visualize what we learned in the last video, is the null space of A. To figure out the null space of A, we know that the null space of A is equal to the null space of the reduced row echelon form of A. So let's just find out the reduced row echelon form of A. Let's say we leave the first row the same, we get 3 minus 2. And let's replace the second row with the second row, minus 2 times the first, so 6, minus 2 times 3 is 0. And minus 4, minus 2 times minus 1 is 0. And then let's replace the first row with the first row divided by 3. So then that becomes 1 minus 2/3, and the second row is still 0. So that's the reduced row echelon form of A. And we want to find its null space. So we want to find all of the vectors that we can multiply it by. So this is the vector x1, x2, that is equal to the 0 vector in R2. So the second row gives us no information, 0 times x1, plus 0 times x2 is equal to 0. No information there. So our only constraint is the first row. 1 times x1-- let me write it here-- so 1 times x1, plus-- let me write it this way-- minus 2/3 times x2 is equal to that 0. Is equal to that 0 right there. Or we could write that x1 is equal to 2/3 x2. So if we wanted to write the null space of A-- actually, before I write that, just to simplify things and just to show you that x2 isn't some special number, let's just say that x2 is equal to t, where t is some real number. And then we would have x1 is equal to 2/3 times t. So the null space of our matrix, the null space of A, is going to be equal to the set of all x1, x2's, that are equal to some real number t times the vector. x2 is equal to t times 1, and x1 is equal to 2/3 t. So 2/3, 1. Just like that. That is our null space. All of the multiples of the vector 2/3, 1. And actually just to make this a little bit simpler, we can pick c, let's pick t, sorry, let's pick t to be equal to 3c. So then what do you have? If this is equal to 3c, if you multiply the 3 times these, we could rewrite this as being equal to the x1's, x2's that are equal to the scalar c, where that is some other real number, times 3 times 2/3 is 2, 3 times 1 is 3. And the whole reason why I did that is just to write the null space with a slightly simpler basis vector. The one that didn't have any fractions in it. Or we could also write that the null space is equal to the span. The span of the vector 2, 3. All of these statements are equivalent. So we figured out the null space. Maybe another interesting thing, especially if we're going to make this relate to what we did in the last video, is find a solution set to the equation Ax is equal to b. To do that, we just set up an augmented matrix. So we set up an augmented matrix, 3 minus 2, 6 minus 4, and we augment it with b, 9, 18. And then we put the left-hand side in reduced row echelon form, and we get-- let's keep the first row the same-- we get 3 minus 2, and then 9. Let's replace the second row with the second row minus 2 times the first. So we get 6 minus 2 times 3, is 0, minus 4 minus 2 times minus 2, that's 0 as well. Minus 4 plus 4. 18 minus 2 times 9, that's 18 minus 18, that's also 0. Almost there. Now let's just replace the first row with the first row divided by 3. So the second row is going to stay the same. And the first row is going to be 1 minus 2/3, and then we get a 3 there. So if we wanted to rewrite this is as an equation, we would write that the matrix 1 minus 2/3, 0, 0, times the vector that's in our solution set, x1, x2, is equal to the vector 3, 0. Or another way to say it, the second row gives us no real constraint, so we don't have to pay attention to it. The first row tells us that x1 minus 2/3, times x2 is equal to 3. Or, that x1 is equal to 3, plus 2/3 x2. Let's do the same exercise here. If we say that x2 is equal to t, then x1 is equal to 3 plus 2/3t. Or we could say that the solution set to Ax is equal to b is equal to the set of all x1, x2's that are equal to-- let's see, x1 is equal to 3 plus 2/3, so plus t times 2/3. Let's write it like that. So plus right there. And x2 is just equal to t, so let's just go 1 times t plus, well let's just say 0. So that's our solution set right there. And you might immediately recognize that it's some particular solution plus some scaled up version of our null space right there. That's our solution set. I could do that same substitution that we did over here. I could say that let's replace t with 3c, and so we could rewrite this as being equal to-- let's rewrite it neatly-- x1, x2 is equal to the vector-- I'm going to run out of space-- 3, 0, plus c, times-- let me scroll over a little bit-- plus c, times the vector, 3 times 2/3 is 2, 3 times 1 is 3. So it's equal to the vector 3, 0 plus some member of our null space, right? The null space was the span of 2, 3, or all the scalar multiples of our null space. And this is where c is any real number. So that's our solution set. That's our null space. Now, one other thing that might be of interest, just because it's the orthogonal complement to the null space, and this relates to what we were doing in the last video, is what the row space of A is. So the row space of A is just the column space of A transpose. This is the row space. This is the row space of A. And what is this equal to? This is equal to the span of the row vectors of A. So we have 3 minus 2, and we have 6 minus 4. But this guy is just 2 times that vector right there, so we can just ignore it. So it's just the span of the vector 3 minus 2. Now let's graph all of these. So we have the row space, we have the solution set and we have the null space. Let's see if we can graph it. We can make a reasonable graph of this thing right here. So I have my vertical axis right there. I have my horizontal axis right there. Now what does the null space look like? It's all the multiples of 2, 3. So if I go out 1, 2 and then I go up 1, 2, 3, the vector 2, 3 looks like this in the standard position. And the null space is all of the multiples of this. So, all the multiples of that. So if you do all the multiples of that, you get a line that looks something like this. If you do all the multiples of that vector, you get a bunch of vectors that point to every point on this green line right there. So that right there is the null space of A. Now what is the solution set? The solution set is the vector 3, 0. So it is the vector 3, 0, so 1, 2, 3. So it is the vector 3 , 0, plus members of the null space. So if you take any of these -- let's say if you add 2, 3-- 2, 3 would look like this. But you're adding any multiple of 2, 3. So when you add all the different multiples of 2, 3, you get this line right here. It's shifted, essentially, by that vector 3, 0. So this right here is the solution set. So solution to Ax is equal to b. That's the solution set right there. Now, what's our row space? The row space is all of the multiples of the vector 3 minus 2. So what does 3 minus 2 look like? The vector 3 minus 2, we go 1, 2, 3, and then we go down 2. So you go 1, 2, and it's going to look like this. That's going to look something like this, and then if you were to graph it and you can see, well, it's going to look something like-- I want to do it relatively neatly-- so it's going to look something like this. Let me do it in another color actually. It's going to look like that right-- I can't do it at all straight. Oh, this is horrible. Something about the bottom right hand side of my-- that's a pretty decent attempt. So that right there is my row space. Because if you have 3 minus 2 you get roughly to that point, and then you want all of the multiples of that vector, right? 3 minus 2, if you multiplied it by minus 1, you would get the vector minus 3, 2, so it would look like this. It would be minus 3, 2. It would look like this, 1, 2, 3 and then up 2, it would be like that, so all of the scalar multiples of that as well. That is our column space,. And notice-- I'm sorry that is our row space-- and notice that our row space is orthogonal to our null space. Now this is a nice visual representation of everything we can do essentially with this matrix right here, which is our matrix A. But in the last video, we had a very interesting result. We had a very interesting result. We said that if we have some b-- let me do it in a new color-- we found out that if we had some b that is a member of the column space of A, then the solution with the smallest length, or we can say the smallest, or the shortest solution to Ax is equal to b is a unique member of the row space of A. Now let's see, this is our big takeaway from the last video. And it might have seemed a little hard to visualize before. But now that we have graphed it all out, I think we can visualize it. This blue line right here is a solution set. The row space right here is this line that's perpendicular to the solution set. But notice, one of the vectors on it is both pointing to something, pointing to a position on the solution set, and it is in my row space. This vector right here, that I'm doing in a bold green, That vector right there, we could call that vector r, because that vector is both a member of my row space, right? It points-- if you do it as a position vector, it points to a point on the line that represents our row space. All of the members of our row space point to points on this line. But at the same time, it also points to a point right there that is a member of our solution set. And notice, it is the only vector in our row space that points to a point that is a member of our solution set. And if you just look at it from a geometric point of view, all of the other solutions point to all of the other points on that line. So, that's a solution right there, this is a solution right there, that is a solution right there. All of the vectors that point to things on this solution line right there, there's all solutions. But the very shortest one is this green one. This green one is essentially orthogonal to it because it's a member of the row space. And this essentially has the same slope as our null space. It's orthogonal, so it's kind of the shortest path distance to getting to our solution set. And for this exact example, we can actually figure out what this shortest vector r is. We've got the vector 3, 0, over here, right? The vector 3, 0 is this vector right here. So if you take 3, 0 minus your vector-- so let me say r. Let me write this way. Our vector r, the special shortest solution, it's going to be a member of our column space. Our column space is a span of 3 minus 2. So it's going to be equal to some scalar multiple of 3 minus 2. Now, we know that 3, 0 points to another solution on our solution set. But we know that if we take the difference of these two vectors-- if I take 3, 0 and from that I subtract the vector r, what do I get? I get this vector right there. Let me do that in another color. I get this pink vector right there. That pink vector, it's not in a standard position, but that pink vector is going to be a member of our null space. That pink vector is going to be a member of our null space. So if I take 3, 0. That, minus my r, so minus c, times 3 minus 2, right? That's going to be this pink vector right there. So if I were to take the dot product, so this vector right here-- let me make this clear; that vector is this pink vector right there-- that's in my null space. I could shift it over. It's not in standard position there, but [? I'll have to ?] do it in the standard position, it points to something on my null space line. So that's in my null space. So if I were to dot it with any member of my row space, it's going to be equal to 0. The row space is the orthogonal complement of the null space. So let me take the dot product of that, with some member of my row space. Might as well just take the basis vector for my row space. That's a member of my row space. So if I dot it with 3-- sorry, I don't want you to write that as a fraction-- 3 minus 2. If I dot it with 3 minus 2, that will be equal to 0. So let's see if we can solve for c. So we get, let's see if we do this, we get this inside part, this pink vector right there is going to be 3 minus 3c, right? And then you're going to have 0 minus minus 2c, so that's just 2c. So this part right here simplifies to that. I just perform the scalar multiplication and then the subtraction. And then if we dot that with my basis vector for my row space, 3 minus 2, this should be equal to 0. So what do we get? We get 3 times 3, which is 9, minus-- let me see, 3 times 3-- minus 3, or minus 3c times minus 2-- let me write it this way-- plus-- let me do it this way. This is probably the easiest way to do it. Sorry, I was actually doing it wrong. This is the first entry, so you have 3 minus 3c, times 3, right? 3 minus 3c-- the first entry times the first entry here-- plus 2c, times minus 2, is equal to 0. I was performing that dot product in a strange way. I think I got a little caught off guard by these two terms. But anyway, the first term times the first term, plus the second term times the second term. That's what our dot product is. This isn't the matrix. This is just a first and a second term right there. So then what do we have here? We have 3 times 3 is 9, minus 9c, minus 4c, is equal to 0. This is 9 minus 13c is equal to 0. Or we get 13, let's see, 9 is equal to 13c, or c is equal to 9 over 13. So just like that we've solved for our special vector r. We said, hey look, if you take this 3, 0 vector and the difference between that and our vector r, you're going to get some vector, right here, that is in our null space. And if you take the dot product of that with some member in our row space, you're going to get 0. And the dot product is that entry times that entry, plus that entry times that entry. And so you get c is equal to 9 over 13. So our special vector r, the unique shortest solution to the equation Ax is equal to 0, is 9 over 13 times the vector, times our basis vector 3 minus 2, or we could write that as 27 over 13, right? And then we have 3 times-- and then 2 times 9 is minus 18 over 13. And that right there is our shortest vector in our row space that satisfies this equation. Let me write it better. It's the unique member of our row space that also happens to be the shortest solution to Ax is equal to b. So this is an example of what we talked about in the last video, and hopefully you see visually why this is so.