- Introduction to the inverse of a function
- Proof: Invertibility implies a unique solution to f(x)=y
- Surjective (onto) and injective (one-to-one) functions
- Relating invertibility to being onto and one-to-one
- Determining whether a transformation is onto
- Exploring the solution set of Ax = b
- Matrix condition for one-to-one transformation
- Simplifying conditions for invertibility
- Showing that inverses are linear
Exploring the solution set of Ax=b (non homogeneous equations). Created by Sal Khan.
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- What is a homogeneous equation vs. a non-homogeneous equation?(4 votes)
- The equations are referred to as homogeneous equations if they equal zero.. So for the matrix equation Ax=b, this is a set of homogeneous equations if b = 0(3 votes)
- Is that always the case Sal, for a non-homogenous linear equation? or are there exceptions?(4 votes)
- When Sal tries to define the actual vectors in the domain that are mapped to the co-domain, using the b1 as a variable, isn't it corresponding to the inverse of the transformation? And we know that this transformation is NOT invertible!(2 votes)
- I think the distinction is that to be the inverse of a transformation you need to map to a unique x in the domain. Instead, here we are mapping to an infinite number of x's in the domain that fall on the (5, 0) + x2(3, 1) line for that particular b (5,-5).
Also in this case if you took the following A^(-1)Ax you should get x back if A in invertible, but in this case you would not get x back (if this were even possible) because you would instead get a line.(1 vote)
- So I came across 2 questions that said,
write in the form of ax=b
does that basically mean:
2 0 3 x = 4
-1 4 -2 y = 1
1 -3 5 z = 2
and then it asks what is the condition that allows matrix A to be inverted?
What does that mean?
Thank you.(1 vote)
⎡ 2 0 3⎤⎡x⎤ ⎡4⎤
⎢-1 4 -2⎥⎢y⎥ = ⎢1⎥
⎣ 1 -3 5⎦⎣z⎦ ⎣2⎦
This is the Ax = b form. The brackets are important, indicating which set is A, x, and b respectively. A is the 3x3 matrix containing the 9 numbers.
By the definition of invertibility, A is considered invertible if there exists a matrix
B, such that
AB = BA = Iwhere
Iis the identity matrix (in this case the 3x3 identity). All this means for our problem is that If we can solve for x, y, and z (all three, not just some of them), then the matrix is invertible. The easiest way to solve for x, y, and z is to augment the matrix and put it into some row echelon form (Sal prefers reduced row echelon form):
⎡ 2 0 3 | 4⎤
⎢-1 4 -2 | 1⎥(this is the augmented matrix)
⎣ 1 -3 5 | 2⎦
(skipped some legwork with the magic of code; put into rref)
⎡1 0 0 | 0.92⎤
⎢0 1 0 | 0.84⎥
⎣0 0 1 | 0.72⎦
You'll notice that the A portion of our augmented matrix now looks like the identity matrix with 1's down the diagonal and 0's everywhere else. This alone proves that there is a matrix
AB = BA = I, so the matrix is invertible. However, putting the matrix in this form has the added bonus of solving for x, y, and z for us (which is the right column from top to bottom).(3 votes)
- so is this example one-to-one? I would think it is since it's null space is zero but from the geometric interpretation it does not look like it's one-to-one. I'm confused(1 vote)
- In this example, the null space isn't the zero vector and it's not one-to-one.
Given the equation:
T(x) = A x = b
All possible values of b (given all values of x and a specific matrix for A) is your image (image is what we're finding in this video). If b is an Rm vector, then the image will always be a subspace of Rm. If we change the equation to:
T(x) = A x = 0
Then all possible values of x (we have to solve for x) is known as the nullspace of A, or as the kernel of T. If x is an Rn vector, then the nullspace of A will always be a subspace of Rn.(2 votes)
- At about8:00the vector x1,x2 is set equal to the vector b1,0 plus x2 times the vector 3,1. Does this imply the x1,x2 is equal to the augmented matrix specified earlier in the lecture?(1 vote)
- The augmented matrix is used to "solve" for x1 and x2, it is not "equal" to x1 and x2. In the example, this gives x1 = b1 + 3x2, while x2 = x2 (free variable). Then the vector x is written in terms of its components.
x = x1,x2 = (b1 + 3x2),x2 = (b1 + 3x2),(0 + 1x2) = b1,0 + 3x2,1x2 = b1,0 + x2(3,1) in your notation.(2 votes)
- At3:06and5:43I'm really confused on what exactly is the matrix and what is the vector and by not knowing that it gets me confused throughout the whole video.Can someone help me understand.(1 vote)
- If a non-homogeneous linear system Ax=b has the solution set x_p+s(u_1)+t(u_2), where x_p is a particular solution vector of Ax=b, and s(u_1)+t(u_2) is an arbitrary vector in the solution space of Ax=0, then the dimension of the null space will be 2, but will the solution set be a vector space as well and have a dimension?(1 vote)
- Around2:30Sal talks about how to find the solutions when a row of your matrix is 0. But in the last video, "Determining whether a transformation is onto," Sal said that case would imply that there are no solutions. The case I mean is when you have a zeroed out row in your matrix but there's some expression of your b variables in the augmented part. What am I not getting?(1 vote)
Let's say I have some linear transformation T from R2 to R2. So that is R2, and then this is R2. T just maps from any member of R2 to another member of R2, just like that. And I'm going to define T. It's a linear transformation. I'm going to define it. When I take the transformation of some member of R2, it's equivalent to multiplying it by this matrix, by the matrix 1 minus 3, minus 1, 3. So just to kind of understand this transformation a little bit more, let's think about all of the values that it can take on in our codomain. So if we say that 1 minus 3, minus 1, 3, times any vector in our domain-- so x1, x2-- it's going to be equal to some other vector in our codomain. Let me call that vector b. B is a member of R2. So it's going to be equal to b1, b2. All I'm doing here is I'm essentially-- if I call this right here, I call that matrix A-- I'm just trying to find out what are all the possible values for Ax is equal to b? I'm trying to find out all of the possible b's in this case. So if I were to solve this equation for any particular b, what I would do is I would put this in reduced row echelon form. Well, first I would make an augmented matrix of it. So I would put a 1, a minus 3, a minus 1, and a 3. And I would augment it with the member of our codomain we're trying to be equal to. So b1, b2. And then, I would put it in reduced row echelon form. So how can we do that? How can we put it in reduced row echelon form? I'll keep my first row the same. 1 minus 3, and then b1. And then my second row, I'll replace it with my second row plus my first row. So minus 1 plus 1 is 0. 3 plus minus 3 is 0. And then b2 plus b1 is-- well I can just write that as b1 plus b2. So this is only going to have solutions. I have an interesting situation here. You've seen it before. I have a row of 0's. And the only way that this is going to actually have solutions is if this thing right here is also going to be equal to 0. So the only members b that are members of Rm that have solutions are the ones where if you add up their two terms-- so let's say b is equal to b1 and b2-- are the ones where the two terms added up have to be to 0. b1 plus b2 have to be equal to 0. Or another way you could write it is that b2 has to be equal to minus b1. So if we were to actually draw our codomain-- let's do it. We always stay in the abstract, but sometimes it's useful to draw an actual example. Let's say that our codomain is R2. Let me draw some axes right here. Let's say that this is my b1-axis, and this is my b2-axis. I could have called that x and y, but I called that b1 and b2. What are all of the members of my codomain that have a solution, that have a mapping? Well, b2 has to be equal to minus b1. So it's going to look like this. It's just going to be a line with a slope of negative 1. This is all the b's that have a solution. Because if you're not on this line. If you're a member of your codomain-- this is the codomain right here, R2. R2 is also our domain, but let me make it very clear that this is the codomain that I've drawn. This is what we're mapping into. It's very clear that if we're not on this line, if you pick somebody whose two terms don't add up to equalling 0, or aren't the negative of each other, if you pick someone over here in our codomain, and then you try to solve the equation over here, 0 is going to equal some non-zero number here and you're not going to have a solution. And we touched on this in the last video. And so in this case we could say that this right here is the image of our transformation. Or even another way of thinking about it. This is our codomain. Let me draw a domain. So if you take any member of R2, and it's always mapping onto something onto that line, clearly each point on that line is going to be mapped to by more than one vector. So we're not dealing with an onto transformation. We saw that in the last video. In order for something to be onto, when you put it in reduced row echelon form, you cannot have all 0's in one of the rows. Or another way to say it in reduced row echelon form, every row has to have a pivot entry. Well, let's focus on the b's that actually do have a solution. So let's focus on these b's, so that when you take a b1 plus b2, it actually is going to be equal to 0. So we could have the b, I don't know, that could be the b5 minus 5. Obviously the b0, 0 is going to work. You could have 1, negative 1. Maybe that's right there. Let's focus on those for a second and see how many members of our domain map into them. So if we take this right here and then we apply this equation up here, we only have one constraint. We're assuming that this is going to be equal to 0. So let's assume that we're dealing with the b's in our image. Let's assume that we're dealing with something where we can get a solution. That b1 plus b2 is equal to 0. Then what is our constraints? What will map to our vector b that we're dealing with? If we just take this topic equation right here, we have 1 times x1-- let me switch colors-- minus 3 times x2 is equal to b1. And then this row will give us no constraints, because it's just going to be a bunch of 0's. So this is the only constraint for a member of our domain that will map to some particular b now that we are picking, some particular b that satisfies this constraint. So we could write this solution set-- let me rewrite it as x1 is equal to b1, plus 3x2. Or if we wanted to write the entire solution set, it would look like this. x1, x2 is equal to the vector b1, 0, plus-- x1 is b1, plus 3, times x2, so plus x2 times 3. And x2 is just going to be equal to x2. That's a free variable. So x2 is equal to 0 plus x2, times 1. So what we're saying here is this transformation just maps to this line here for all of the vectors in our codomain, where their two entries add up to each other. Let's assume that we actually have one of those vectors. And so first of all, this is definitely not an onto transformation. But let's assume that we are dealing with one of those guys. So if we are picking a particular one of those guys for a particular b-- let me write this down-- that has a solution to Ax is equal to b, the solution set will be equal to this thing right here. It'll be equal to x1, x2 is equal to the first entry of your b. b1, 0 plus x2, times the vector 3,1. And if you think about this, if you pick a particular b-- so let's say we pick our-- let me draw this out, because I think it's nice to visualize it all. Actually, maybe I'll draw it like this. I don't want to draw our blobs anymore. So let me draw my axes. So my axes look like this. We know that the image of our transformation is the line with a negative 1 slope, because the two entries have to be equal to the negative of each other. Let's pick a particular b that has a solution. So let's say we pick that b right there. In order for it to have a solution, its entries have to be the negative of each other. So let's say that's the entry 5, minus 5. That is our b. So what we just showed is that the solution set, if we want to say, hey, what in our domain maps to this guy? So let's think about what in our domain maps to this right here. Let's think about what in our domain maps to this point right here, to this particular b. So that's going to be all of the x's that satisfy Ax is equal to 5, minus 5. So x1, x2 is going to be equal to b1, it's going to be equal to 5, 0 plus x2, plus any scale or multiple of the vector 3, 1. So our solution set is going to be, you take the vector 5, 0-- so maybe the vector 5, 0, specifies this position right here-- and then you're going to add to it multiples of the vector 3, 1. The vector 3, 1 looks like this, 1, 2, 3 and you go up 1. The vector 3, 1 is going to look like this. So if you add multiples of this-- multiples of that could stretch out like that or could go negative like that-- to this vector 5, 0, you're essentially going to-- let me see if I can draw this neatly-- you're going to end up with a solution set that looks like that right there. So if you pick a particular b right there that has a solution, we just said that everything on this line will map to that point in our solution set. And in fact, if you picked another point, let's say you picked the point minus 5, 5. Then the solution set that maps to that would actually be-- this first term would be the minus 5. It would be here. And all of these guys would map to that. Well, this all interesting. I mean, we've been doing a lot of abstract things. And I think it might be satisfying that you're actually seeing something more concrete in this example. But I'm doing all of this for a reason. Because I want to understand what the solution set is to a general non-homogeneous equation like this. And to understand it a little bit better, let's imagine what is the solution set if I were to pick this guy, the 0 vector? Then what's going to be the solution set? So if we say that Ax is equal to 0, then our solution set is going to the vector 0, 0, plus x2, times 3, 1. So it's going to be what? This is just the 0 vector, so it's going to be here. It's going to be here, and it's going to just be multiples of 3, 1. It's going to look something like that. What's this? What is the solution set to the equation of Ax equal to 0. This is the null space. This, by definition, is the null space of A. So notice-- and this is the big takeaway from this video-- that for any solution-- we're picking b's that actually do have solutions, because we're picking them on this line. We're picking them in the image of our codomain. The solution set to any Ax is equal to some b where b does have a solution, it's essentially equal to a shifted version of the null set, or the null space. This right here is the null space. That right there is the null space for any real number x2. Any scale or multiple of 3, 1 is the null space. I just showed it right there. It's going to be that. And so all of these other solution sets are just some particular vector, some x particular, plus the null space. Obviously, this vector by itself would also be a solution to Ax is equal to b, because you can just set x2 to be equal to 0. So in general-- and I haven't proven this to you rigorously, but hopefully you kind of get the intuition behind it. The solution-- and I'll do this in the next video, just because I realize I'm running long on time. Assuming Axb has a solution-- in the example we just did, we can assume it has, if we pick one of these points here. So assuming it has a solution, if we pick a point off of our image, then we're not going to have a solution. But assuming Axb has a solution, the solution set is going to be equal to some particular vector-- so you could just think of it as a set with just one vector right there-- with, or combined with, or the union of that set, with your null space of this matrix right here. I haven't proven this to you yet, but hopefully you get the intuition of why this is true. We just solved it for particular cases that do have solutions. We say, hey, it's going to take this form. And I just showed you that this is the form of the null space. And the reason why we're doing that is because we've been talking about invertibility. And in order to be invertible you have to be onto, and 1 to 1. And for something to be 1 to 1, you have to have at most one solution that maps to a particular vector. You might have none. But you might have, at most, one. So in order to have, at most, one solution-- and the solution set is always going to be equal to this, so you're always going to have this solution-- so in order to have, at most, one solution, your null space can't have anything in it. Or it can just have the zero vector. So that means that your null space of A has to be trivial, or to be empty, or to just have the zero vector. I'll do this a little bit more rigorously in the next video, but I think when you do it sometimes with the rigor, you don't necessarily get the intuition. But this is going to be a very interesting takeway, and I think you already understand it's leading up to conditions for invertibility.