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# Visualizing a column space as a plane in R3

## Video transcript

In the last video, I started with this matrix right here, and right from the get go, we said the span of this matrix is just the span of the column vectors of it, and I just wrote it right there. But we weren't clear whether this was linearly independent, and if it's not linearly independent, it won't be a sufficient basis. And then we go off and we figure out the null space of A. We find out that the null space of A contains more than just the zero vector. It's just the span of these two vectors here, which means that these columns are not linearly independent. We saw that several videos ago. And we used that information that they're not linearly independent to try to make them linearly independent by getting rid of the redundant ones. So we were able to get rid of this guy and this guy, because these were essentially the columns associated with the free variables. And we were able to do it using this little technique down here, where we set one of these equal to 0, the other one equal to negative 1 and then solved for the pivot variables. And then we set the other one equal to 0 and the other one equal to negative 1 and solved for the pivot variables. And you could imagine that this is a generalizable process. If you have a bunch of free variables, you can set all of them but one to equal 0, and then that one that you didn't set to 0, you set it to equal negative 1. And you can express it as a sum of the pivot variables, where the pivot variables are a function of the free variables. In general, this would be a quick way of doing it. We had to do it more slowly over here. If I have some matrix A, and I want to figure out the basis for the column space, the column space is just the span of these, but if I wanted a linearly independent basis, I need to figure out some set of these that are linearly independent. What I can do is put this guy into reduced row echelon form. When I put them in reduced row echelon form, and I did that over here, this is the reduced row echelon form of A, I can look at the variables that are associated with the pivot entries. So this is x1. Let me scroll up a little bit. This is associated with x1, right? When you multiply this times x1, you get this column times x1, this column times x2, this column times x3, this column times x4 like that. When you look at just regular A, when you look at just your matrix A, it's the same thing. If you were to write Ax equal to 0, this column would be associated with x1, this column would be associated with x2, x3 and x4 like that. What you can do is you put it in reduced row echelon form. You say which columns have my pivot entries or are associated with pivot variables? You say, OK, x1 and x2 are associated with pivot variables, or they are the pivot variables, and they're associated with these first two columns, and so those first two columns would be a basis for the column space. How did I get this? Am I just making up things on the fly? Well, no! It all comes from the reality that you can always construct a situation where the vectors associated with the free variables you can write them as linear combinations of the vectors associated with the pivot variables, and we did a special case of that last time. But a very quick and dirty way of doing it, and I don't know if it's actually dirty, is just take your matrix, put it in it reduced row echelon form, and you say, look, this column and this column are associated with my free variables. Therefore, this column and this column must be associated with my free variables. The solution sets are the same to Ax equal to 0 or the reduced row echelon form of Ax is equal to 0. So they're the same. So if this column and this column are associated with free variables, so are this column and this column, which means that they can be expressed just by judiciously selecting your values of your free variables as linear combinations of the columns associated with the pivot variables, with the pivot entries, which are that column and that column. So this column and this column would be a basis for A, and we see that. We see that all the way down here. 1, 2, 3 and 1, 1, 4, We did a lot of work and we got all the way there, and we said this is a basis for the column span of A. Now, doing all of that work, let's see if we can actually visualize what the column space of A looks like. I have a strange feeling I might have said column span a couple of times, but the column space, what does it look like? Well, there's a couple of ways to think about what it looks like. One way is we can say, look, this span this is 2-- that's a member of R3. That's a vector in R3 and this is a vector in R3. Let me draw my x, z and-- well, normally it's drawn-- this is normally y, x, and z-axes in R3, if I'm want to represent them as three-dimensional space. Then the vector 1, 2, 3 might look like this where it's one, one, two, one, two, three, so we go out one down here, then up three. So the vector will look like that in its standard form. That's that one right there. And the vector 1, 1, 4 would be one, one and go up four, so it might look something like this. It's hard to actually draw them very well in three dimensions, but you get the idea. But the column space is the span of these, so all of the linear combinations of these two guys. So all of the linear combinations of these two guys are going to create a plane that contains these two vectors. If you just sum these guys up in multiple combinations, you can get any vector up there. If you just add them up, you'll get that vector right there. If you add this guy plus 2 times that, you'll get some vector up here. So if you view them as position vectors, they'll essentially form a plane in R3. But let's see if we can get the equation for that plane. Well, how can we do that? Well, we know that we can figure out the equation of a plane based on the fact that a normal vector dotted with any-- let me write my normal vector like this. The normal vector dotted with any position vector specifying a position on the plane. So let me call that x minus any point on the plane or any vector position on the plane. So I could do that minus the vector 1, 2, 3 has to be equal to 0. And we can use this information to figure out the equation for this plane. But what is a normal vector? How can we find a normal vector to this plane? So this would be a vector. Let me see if I can draw this in a way that doesn't confuse the issue. If the plane is like that, the normal vector would come out like that. So how can I create a normal vector? Well, we learned that you take the cross product of any two vectors in R3, and the cross product I've only defined so far in R3, and I will get a vector that's normal to both of those vectors. So let's take the cross product. This is a nice way of thinking about it, because it's really integrating everything that we've covered so far. So let me define my normal vector to be equal to 1, 2, 3 cross 1, 1, 4. And what does this equal? So my first term, I ignore that. I get 2 times 4 minus 3 times 1. 2 times 4 is 8. 2 times 4 minus 3 times 1. 8 minus 3. Then for my second row, I have 1 times 4, and my temptation is do 1 times 4 minus 3 times 1. But you reverse it. You do 3 times 1, so it's 3, minus 1 times 4. We've done that multiple times. You might want to review the cross product video if that seems strange. You ignore the middle row, and you normally do 1 times 4 minus 3 times 1, but the middle row you switch. We're only defining it for R3, so instead, we do 3 times 1 minus 1 times 4. And then finally for the last row, we ignore it, and we say 1 times 1, which is 1, minus 2 times 1, which is minus 2. And this is equal to the vector 5, minus 1, minus 1, which by definition of the cross product, and I've shown this to you multiple times, is normal to both of these vectors, and so it'll be normal to any linear combination of these two vectors. So now that we have our normal vector, we can define the traditional equation for the plane. So we now know that our normal vector 5, minus 1, minus 1, that I got by taking the cross product of our basis vectors dot any vector in our plane. So let me just write any vector. Let me just write it x, y, z. So x, y, z since that's how I defined my axes up here. This was the x-axis. x, y and z. x, y, z minus-- I just picked one of these. I could have picked either of them-- minus 1, 2, 3 has got to be equal to 0. So what's this? This is going to be equal to-- let me write it a little smaller, a little neater-- 5, minus 1, minus 1 dot-- what's this guy going to be? x minus 1, y minus 2, and z minus 3 has got to be equal to 0. And what's the dot product? It's 5 times x minus 1 plus-- or maybe I should say minus 1, so it's plus minus 1 times y minus 2 plus minus 1 times z minus 3 is equal to 0. That's just the definition of our dot product. If I simplify this, I get 5x minus 5 minus y plus 2 minus z plus 3 is equal to 0. You have 2 plus 3 is 5 minus 5, so those all cancel out. Those will equal 0. And we get 5x minus y minus z is equal to 0. This plane in R3 is the column space of A. So we've now shown you that it's truly a plane in A. And actually, it makes sense that this plane intersects the origin. If you set x, y, and z equal to 0, it satisfies this equation. And that makes sense, because we said that a column space of a matrix has to be a valid subspace, and a valid subspace has to contain the zero vector. And in R3 that's the coordinate 0, 0, 0. Now, what I want to do now is see if we can get at the same answer going a completely different way, or approaching it in a completely different way. Let me get my original A, which I've forgotten. I've marked it up a good bit, but let me just copy and paste it. So that's my original A right there. Let me copy it. Let me paste it. Nope. That's not what I wanted to do. Let's see, so my original A-- I copied and pasted the wrong thing. Let me do it a little-- don't want to waste your time. Edit, copy, edit, paste. There you go and let me scroll this down and get to a point that's relatively clean. Bring my A down. I've used a lot of space. So here you go. This is my original A right there. And what I want to do is see if I can get this result completely different. I got this result by figuring out the basis of the column span, finding a normal vector by taking the cross product of our two basis vectors, and then using the dot product of the normal vector with the difference-- this vector right here, where you take any vector on our plane minus one of our basis vectors, that's to find some vector in the plane. This is some vector in the plane. So any vector in the plane dotted with my normal vector is going to be equal to 0. And actually, I should probably make a side note here, that the only reason I was able to say that the normal vector is a cross product of my two basis vectors, is because I knew that these two basis vectors, not only do they specify some point on the plane-- so let's say that this guy right here is this blue vector. Not only does he specify some point on the plane right there, but the vector lies completely on the plane. How did I know that? Because I knew from the get go that the 0, 0 vector is in my span, right? I knew that if I draw this guy in just standard position, the point 0, 0, 0, is in my span, and I know its end point is in the span, so this whole vector has to be in my plane, and likewise, this whole vector has to be in my plane. So if I take the cross product, anything normal to these guys or any combination of these guys is going be normal to the plane, and we got this result right here. But let me take this right here and use our other definition of column span. Our other definition, or it's really an equivalent definition, is all of the valid solutions to Ax where x is a member of Rn. Or another way we could think of it is, we could view it as all of the valid b's where Ax is equal to b, and x is a member of Rn. These are equivalent statements. I'm just defining b here to be Ax, so these are equivalent statements. But let's run with this a little bit. So let's say that I define my b, so b is going to be a vector in R3, right? We already have an intuition like that. When I take Ax, I get a b that's equal to x, y, z. And I want to figure out what x, y's and z's can I get valid solutions for? So if I take my vector A and then let me multiply it times-- well, actually, the best way to do it, I think we're used to it right now. If I'm solving the equation Ax is equal to b, I can essentially just create the augmented matrix, where I have the matrix A and I can augment it with b, and put this in reduced row echelon form, and that'll essentially represent the solution set. So let me do that. So if I just augment this matrix right here with b, so I write x, y, z. So this is A augmented with b. This is A, this is b. let me put this in reduced row echelon form and find the solution set. And these are x, y, and z's that define a valid b. So what do I get? The first thing I want to do, and we've done this exercise before, let's keep my first row the same. 1, 1, 1, 1, and I get an x. And let's replace our second row with the second row minus the first row. Actually let me do it this way. Let me replace the second row with 2 times the first row minus the second row. So 2 times the first row minus the second row, we're going to get a 2x minus y up there. And then 2 times 1 minus 2 is 0. 2 times 1 minus 1 is 1. 2 times 1 minus 4 is minus 2. 2 times 1 minus 3 is minus 1. Fair enough. And now let me replace my third row with the third row minus 3 times the first row. So we're going to do the third row minus-- no, let me do it this way. It's the third row minus 3 times the first row. So I'm just doing the b column first, because I can remember what I did. The third row minus 3 times the first row. 3 minus 3 times 1 is 0. 4 minus 3 times 1 is 1. 1 minus 3 times 1 is minus 2. And then 2 minus 3 times 1 is minus 1. Now, I could go all the way to reduced row echelon form, but something interesting is already happening. So let me from the get go try to zero out this third row. And the best way to zero out this third row is to just replace the third row. So the first row-- well, I won't even write the first row. The second row is 0, 1, minus 2, minus 1, and 2x minus y. I'm not even going to worry about the first row right now. But let's replace the third row, just in our attempt to go into reduced row echelon form. Let's replace it with the second row minus the third row. So you get 2x minus y minus z plus 3x. I just took this minus this. So minus z plus 3x. So 0 minus 0 is 0. 1 minus 1 is 0. Minus 2 minus minus 2 is 0, and that's also 0. So we're only going to have a valid solution to Ax equals b if this guy right here is equal to 0. What happens if he's not equal to zero? Then we're going to have a bunch of zeroes equaling some number, which tells us that there's no solution. So if I pick a b where this guy does not equal zero, then I'll have no solution. If this guy equals 5, if I pick x, y, and z's such as that this expression is equal to 5, then Ax equal to b will have no solution, because it will have 0 is equal to 5. So this has to equal 0. So 2x minus y minus z plus 3x must be equal to 0 in order for b to be valid, in order for b to be a member of the column space of A, in order for it to be a valid vector that Ax can become, or the product A times x can become for some x. So what does this equal to? If we add the 2x plus the 3x, I get 5x minus y minus z is equal to 0, which is the exact same outcome we got when we figured out the basis vectors. We said oh, you know what? The basis vectors, they have to be in the column space themselves by definition. So let me find a normal vector to them both by taking the cross product. I did that, and I said the cross product times any valid vector in our space minus one of the basis vectors has to be equal to zero, and then I got this equation. Or we could have done it the other way. We could've actually literally solved this equation setting our b equal to this. We said what b's will give us a valid solution? And our only valid solution will be obtained when this guy has to be equal to zero, because the rest of his row became zero. And when we set that equal to zero, we got the exact same equation. So, hopefully, you found this to be mildly satisfying, because we were able to tackle the same problem from two different directions and get the same result, and it kind of shows you the beauty of linear algebra, how it all starts to fit together.