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Showing that inverses are linear

Showing that inverse transformations are also linear. Created by Sal Khan.

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  • leaf green style avatar for user Fares
    : "best way to get rid of the T is to multiply by T^{-1} on both side of the equality"

    No ! Use the fact that T is invertible and therefore bijective : if T(x) = T(y) then x = y
    (16 votes)
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    • male robot donald style avatar for user Jeremy
      Sal hasn't used the word "bijective" yet in this series of videos, so he can't rely on that property to solve his problem. Therefore, based on the bed of knowledge that we've built so far taking the inverse transformation of both sides might actually, "be the best way to get rid of T." There may be simpler ways that Sal hasn't yet covered, such as you mention, but since they haven't yet been covered they aren't in our toolkit yet.
      (16 votes)
  • hopper cool style avatar for user Cole Wyeth
    If the first condition for linearity is satisfied, won't the second condition be satisfied too? If we know that the transformation of two added vectors is the same as their transformations added, then multiplying one vector by c could be seen as adding another vector to it also, right?
    (10 votes)
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  • orange juice squid orange style avatar for user FishHead
    So what's the point of finding the inverse transformation? Is it simply to find the perpendicular line to our equation, or rather the equation of the original equation flipped over the y=x axis?
    (6 votes)
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  • male robot donald style avatar for user Jeremy
    At about Sal says that we know T is a linear transformation (and it has to be linear to represent it as a matrix), and the whole video is based on finding out whether T inverse is also linear.

    My question is, are there any non-linear transformations that are invertible? I have the sense that the transformation from x to y such that y=x^3 should be invertible, but does anyone know for sure? Or does invertibility imply linearity?
    (3 votes)
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  • blobby green style avatar for user Ji Wang
    At ," I just change the associativity of this" . I think we cannot use associativity before we proof that T-inverse is linear transformation . Look forward to answers. Many thanks
    (3 votes)
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    • leaf blue style avatar for user Matthew Daly
      That really wasn't associativity per se, more like the definition of the composition operation. But it is true that any mappings follow the associative rule whether they have any special properties or not. It's really beyond the scope of linear algebra, but if you know any abstract algebra, it follows from the fact that the set of all mappings from a set to itself is a monoid under the operation of commutation. Hope this helps!
      (2 votes)
  • leaf green style avatar for user pickyourfavouritememory
    To add on to Fares comment: don't we only need the fact that T is injective to show that T(a) = T(b) implies a = b?
    (2 votes)
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  • aqualine ultimate style avatar for user liar3524
    at does associativity only apply to linear transformations?
    (2 votes)
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  • blobby green style avatar for user ricksanchez
    I'm not sure if this is a valid question,
    But at , he expresses the sum of the vectors a and b, as the sum of the Identity Transformations of each individual vectors.
    But we don't know for sure that a and b are members of the domain of T-1 right?
    We assumed at the start that a+b is a member of the domain of T-1, so we used the identity transformation of it.
    How did he do the same for each a and b?
    (1 vote)
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  • leaf green style avatar for user Rajat Kathuria
    At 17.50, Sal "calls" the matrix [in matrix-vector product representing linear transformation of T^(-1)] as A^(-1). But, later relates it to A as its inverse. How did we, in the first place, know that linear transformation T^(-1)(x) will be represented by A^(-1) if transformation T(x) is being represented as Ax
    (1 vote)
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    • blobby green style avatar for user InnocentRealist
      Sal assumes that in general T(x) (for some linear transformation T from R^m to R^n) = Ax (for some mxn matrix A of fixed real number coefficients). If you accept this as true, then:

      For T:X->Y (X, Y in R^n), if T is invertible, call the inverse of T "Ti", Ti:Y->X. He shows that Ti is also a linear transformation, which means that it's also a matrix vector multiplication (Ti(y) = By, for some nxn matrix B). So let's just call B "Ai" (A inverse). Now T(x) = Ax and Ti(y) = (Ai)*y.
      (1 vote)
  • leafers tree style avatar for user Daniel Kent
    Regarding syntax, at , T^-1(b), b, should have a vector arrow above it, and at , Ax, should also have a vector arrow above the x?
    (1 vote)
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Video transcript

I've got a transformation T. When you apply the transformation T to some x in your domain, it is equivalent to multiplying that x in your domain, or that vector. by the matrix A. And let's say we know the linear transformation T can be-- that's transformation matrix when you put it in reduced row echelon form-- it is equal to an n by n identity matrix, or the n by n identity matrix. Well this alone tells a lot of things. First of all, if when you put this in reduced row echelon form, you get a square identity matrix. That tells us that the original matrix had to be n by n. And it also tells us that T is a mapping from Rn to Rn. And we saw in the last video, all of these are conditions, especially this one right here, for T to be invertible. So If we know that this is true-- T is a linear transformation, it's a reduced row echelon form of the transformation matrix of the identity matrix right there-- we know that T is invertible. Let's remind ourselves what it even means to be invertible. To be invertible means that there exists some-- we used the word function before, but now we're talking about transformations, they're really the same thing. So let's say there exists some transformation-- let's call it T-inverse like that, or T to the minus 1-- such that the composition of T-inverse with T is equal to the identity transformation on your domain, and the composition of T with T-inverse is equal to the identity transformation on your codomain. Just like that. And just to remind you what this looks like, let's draw our domains and codomains. Our domain is Rn, and our codomain is also Rn. So if you take some vector in your domain, apply the transformation T-- you're going to go into your codomain, so that is T. And then if you apply the T-inverse after that, you're going to go back to that original x. So this says, look, you apply T and then you apply T-inverse. You're just going to get back to where you started. It's equivalent to the identity transformation. Just like that. This is saying if you start here in your codomain, you apply this inverse transformation first, then you apply your transformation, you're going to go back to the same point in your codomain. So it's equivalent to the identity transformation in your codomain. It just happens to be in this case that the domain and the codomain are the same set, Rn. Now we know what a transformation-- what it means to be invertible. we know what the conditions are for invertibility. So this begs the next question. We know this guy is a linear transformation, in fact that's one of the conditions to be able to represent it as a matrix. Or any transformation that can be represented as a matrix vector product is a linear transformation. So this guy's a linear transformation. But the question is, is T-inverse a linear transformation? Let's review what the two conditions are that we need to have to be a linear transformation. So we know T is a linear transformation. So we know that if you apply the transformation T to two vectors-- let's say x and y-- if we apply to the sum of those two vectors, it is equal to the transformation of the first vector plus the transformation of the second vector. That's one of the conditions, or one thing that we know is true for all linear transformations. And the second thing we know is true for all linear transformations is, if we take the transformation of some scaled version of a vector in our domain, it is equal to the scaling factor times the transformation of the vector itself. These are both conditions for linear transformations. So let's see if we can prove that both of these conditions hold for T-inverse, or this guy right here. So to do this, let's do this little exercise right here. Let's take the composition of T with T-inverse of two vectors, a plus b. Remember, T-inverse is a mapping from your codomain to your domain, although they're both going to be Rn in this case. But T-inverse maps from this set to that set. Let's write it up here. T-inverse is a mapping from your codomain to your domain. Although it looks identical, just like that. OK, so what is this going to be equal to? Well we just said, by definition of your inverse transformation, this is going to be equal to the identity transformation on your codomain. So assuming these guys are members of your codomain, in this case Rn, this is just going to be equal to a plus b. This thing, the composition of T with its inverse, by definition is just the identity transformation on your codomain. So this is just whatever I put in here. If I put in an x here, this would be an x. If I put in an apple here, this would be an apple. It's going to be the identity transformation. Now what is this equal to? Well I could use the same argument to say that this right here is equal to the identity transformation applied to a. And I'm not writing the identity transformation, I'm writing this. But we know that this is equivalent to the identity transformation. So we could say that is equivalent to the composition of T with the inverse applied to a, and we could say that this is the equivalent to the identity transformation, which we know is the same thing as T, the composition of t with T-inverse applied to b. So we can rewrite this thing right here as being equal to the sum of these two things. In fact we don't even have to rewrite it. We can just write it's equal to-- this transformation is equal to this. And maybe an easier way for you to process it is, we could write this as T of the T-inverse of a plus b, is equal to T of the T-inverse of a plus T of the T-inverse of b. And this should-- I don't know which one your brain processes easier, but either of these, when you when you take the composition of T with T-inverse, you're going to be left with an a plus b. You take the composition of T with T-inverse, you're left with an a. You take the composition of T with T-inverse, you're just left with a b there. So in either case you get a plus b-- when you evaluate either side of this expression you'll get the vector a plus the vector b. Now what can we do? We know that T itself is a linear transformation. And since T is a linear transformation, we know that T applied to the sum of two vectors is equal to T applied to each of those vectors and summed up. Or we can take it the other way. T applied to two separate vectors-- so we call this one vector right here, and this vector right here. So in this case I have a T applied to one vector, and I'm summing it to a T applied to another vector. So it's this right here, which we know is equal to T applied to the sum of those two vectors. So this is T applied to the vector T-inverse of a-- let me write it here-- plus T-inverse of b. It might look a little convoluted, but all I'm saying is, this looks just like this. If you say that x is equal to T-inverse of a, and if you say that y is equal to T-inverse of b. So this looks just like that. It's going to be equal to the transformation T applied to the sum of those two vectors. So it's going to equal the transformation T applied to the inverse of a plus T-inverse of b. I just use the fact that T is linear to get here. Now what can I do? Let me let me simplify everything that I've written right here. So I now have-- let me rewrite this. This thing up here, which is the same thing as this. T, the composition of T, with T-inverse, applied to a plus b is equal to the composition-- or actually not the composition, just T-- applied to two vectors, T-inverse of a plus T-inverse of vector b. That's what we've gotten so far. Now we're very close to proving that this condition is true for T-inverse, if we can just get rid of these T's. Well the best way to get rid of those T's is to take the composition with T-inverse on both sides. Or take the T-inverse transformation of both sides of this equation. So let's do that. So let's take T-inverse of this side, T-inverse of that side, should be equal to T-inverse of this side. Because these two things are the same thing. So if you put the same thing into a function, you should get the same value on both sides. So what is this thing on the left-hand side? What is this? This is the composition-- let me write it this way-- this is the composition of T-inverse with T, that part, applied to this thing right here. I'm just changing the associativity of this-- applied to T-inverse of the vector a plus the vector b. That's what this left hand side is. This part right here, T-inverse of T of this, these first two steps I'm just writing as a composition of T-inverse with T applied to this right here. That right there is the same thing as that right there. So that was another way to write that. And so that is going to be equal to the composition of T-inverse with T-- I'll write that in the same color-- a composition of T-inverse with T. That's this part right here, which is very similar to that part right there-- of this stuff right here, of T-inverse of a plus T-inverse of the vector b. Now by definition of what T-inverse is, what is this? This is the identity transformation on our domain. This is the identity transformation on Rn. This is also the identity transformation on Rn. So if you apply the identity transformation to anything, you're just going to get anything. So this is going to be equal to-- I'll do it on both sides of the equation-- this whole expression on the left-hand side is going to simplify to the T-inverse of the vectors a plus the vector b. And the right-hand side is just going to simplify to this thing. Is equal to-- because this is just the identity transformation-- so it's just equal this one, T-inverse of the vector a plus T-inverse of the vector b. And just like that, T-inverse has met its first condition for being a linear transformation. Now let's see if we can do the second condition. Let's do the same type of thing. Let's take the composition of T with T-inverse, let's take the composition of that on some vector, let's call it ca. Just like that. Well we know what this is equal to, this is equal to the identity transformation on Rn. So this is just going to be equal to ca. Now what is a equal to? What is this thing right there-- I'll write it on the side right here, let me do it in an appropriate color. Or we could say that a, the vector a is equal to the transformation T with the composition of T with T-inverse applied to the vector a. Because this is just the identity transformation. So we can rewrite this expression here as being equal to c times the composition of T with T-inverse applied to my vector a. And maybe it might be nice to rewrite it in this form instead of this composition form. So this left expression we can just write as T of the T-inverse of c times the vector a-- all I did is rewrite this left-hand side this way-- is equal to this green thing right here. Well I'll rewrite similarly. This is equal to c times the transformation T applied to the transformation T-inverse applied to a. This is by definition what composition means. Now T is a linear transformation. Which means that if you take c time T times some vector, that is equivalent to T times c times T applied to c times that vector. This is one of the conditions in your transformation. So this is always going to be the case with T. So if this is some vector that T is applying to, this is some scalar. So this thing, because we know that T is a linear transformation, we can rewrite as being equal to T applied to the scalar c times T-inverse applied to a. And now what can we do? Well let's apply the T-inverse transformation to both sides of this. Let me rewrite it. On this side we get T of T-inverse of ca is equal to T of c times T-inverse times a. That's what we have so far. But wouldn't it be nice if we could get rid of these outer T's? And the best way to do that is to take the T-inverse transformation of both sides. So let's do that. T-inverse-- let's take that of both sides of this equation, T-inverse of both sides. And another way that this could be written. This is equal into the composition of T-inverse with T applied to T-inverse applied to c times our vector a. This right here, I just decided to keep writing it in this form, and I took these two guys out and I wrote them as a composition. And this on the right-hand side, you can do something very similar. You could say that this is equal to the composition of T-inverse with T times-- or not times, let me be very careful. Taking this composition, this transformation, and then taking that transformation on c times the inverse transformation applied to a. Let me be very clear what I did here. This thing right here is this thing right here. This thing right here is this thing right here. And I just rewrote this composition this way. And the reason why I did this is because we know this is just the identity transformation on Rn, and this is just the identity transformation on Rn. So the identity transformation applied to anything is just that anything. So this equation simplifies to the in T-inverse applied to c times some vector a, is equal to this thing, c times T-inverse times some vector a. And just like that, we've met our second condition for being a linear transformation. The first condition was met up here. So now we know. And in both cases, we use the fact that T was a linear transformation to get to the result for T-inverse. So now we know that if T is a linear transformation, and T is invertible, then T-inverse is also a linear transformation. Which might seem like a little nice thing to know, but that's actually a big thing to know. Because now we know that T-inverse can be represented as a matrix vector product. That means that T-inverse applied to some vector x could be represented as the product of some matrix times x. And what we're going to do is, we're going to call that matrix the matrix a-inverse. And I haven't defined as well how do you construct this a-inverse matrix, but we know that it exists. We know this exists now, because T is a linear transformation. And we can take it even a step further. We know by the definition of invertibility that the composition of T-inverse with T is equal to the identity transformation on Rn. Well what is a composition? We know that T, if we take-- let me put it this way. We know that T of x is equal to Ax. So if we write T-inverse, the composition of T-inverse with T applied to some vector x is going to be equal to first, A being applied to x is going to be equal to Ax, this a right here, Ax. And then you're going to apply A inverse-x, you're going to apply this right here. And we got that this is the equivalent to-- when you take the composition, it's equivalent to, or your resulting transformation matrix of two composition transformations is equal to this matrix matrix product. We got that a long time ago. In fact that was the motivation for how a matrix matrix product was defined. But what's interesting here is, this composition is equal to that, but it's also equal to the identity transformation on Rn applied to that vector x, which is equal to the identity matrix applied to x. Right? That is the n by n matrix, so when you multiply by anything, you get that anything again. So we get a very interesting result. A-inverse times A has to be equal to the identity matrix. A-inverse, or the matrix transformation for T-inverse, when you multiply that with the matrix transformation for T, you're going to get the identity matrix. And the argument actually holds both ways. So we know this is true, but the other definition of an inverse, or invertibility, told us that the composition of T with T-inverse is equal to the identity transformation in our codomain, which is also Rn. iRn. So by the exact same argument, we know that when you go the other way, if you apply T-inverse first and then you apply T-- so that's equivalent of saying apply T-inverse first, and then you apply T to some x vector, that's equivalent to multiplying that x vector by the identity matrix, the n by n identity matrix. Or you could say, you could switch the order. A times A-inverse is also equal to the identity matrix. Which is neat, because we learned that matrix matrix products, when you switch the order they don't normally always equal each other. But in the case of an invertible matrix and its inverse, order doesn't matter. You can take A-inverse times A and get the identity matrix, or you could take A times A-inverse and get the identity matrix. Now we've gotten this far, the next step is to actually figure out how do you construct. We know that this thing exists, we know that the inverse is a linear transformation, that this matrix exists. We see this nice property, that when you multiply it times the transformation matrix you get the identity matrix. The next step is to actually figure out how to figure this guy out.