- 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
Introduction to the inverse of a function. Created by Sal Khan.
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- Why do we need to satisfy two conditions for the inverse function ? i.e. g(f(x)) = Ix AND f(g(x)) = Iy? Is the one not automatically implied if you state the other? Could somebody give me an example of a function that does not satistfy both conditions?(10 votes)
- When g is the inverse of f, for everything in the range of f .. but g also maps some other things too.. and these happen not to be in the domain of f. So then g acts as an inverse to f, but f is not an inverse to g. This can be referred to as a partial inverse, and an actual inverse might be found by restricting the domain of g to the range of f. A concrete example is f is 'principle square root' and g is 'square'.. Doing f first (i.e. take a square root) then doing g (i.e. squaring that number) always gets back to the starting number for anything f can cope with. However, if starting with g, then g(-1) =1 but now f(1) is not -1.(17 votes)
- Can you tell me about difference between liner and inverse functions ?(4 votes)
- A linear function is simply any function that has a straight line as it's graph. That is functions with one x and no degree. A linear function has to do with a single function and tells you what type it is, nothing more.
Inverse functions, on the other hand, are a relationship between two different functions. They can be linear or not. The inverse of a function basically "undoes" the original.
As a simple example, look at f(x) = 2x and g(x) = x/2. To see what I mean, pick a number, (we'll pick 9) and put it in f.
f(9) = 2(9) = 18. Now put this answer in g.
g(18) = 18/2 = 9. You are back to your original number.
Of course, this is a very simple example, they can get much more complicated than this. To find an inverse of a function, you switch the x and y in the original problem and then solve for y to put in function form.(6 votes)
- I don't understand why at12:14Sal says that g dot f maps to the identity matrix of f. Why shouldn't it be the identity matrix of y? If g maps from Y-->X and f from X-->Y shouldn't the composition of these functions be a mapping from Y-->Y?(5 votes)
- gof = g(f) maps from the domain of f (= X) to the codomain of f (= Y), and "g" maps from Y back to X (first you do f, then g - "right to left").(2 votes)
- Hello. I found the inverse of y=x^3, which is y=x^1/3. The problem is, I don't know if it is a function or not. Is it a function? Thanks!(3 votes)
- Indeed, if we look at the cube root function, we see that there is only one value for each x. It is not always the case that a function is invertible, but x cubed is. If we look at x squared, the inverse is (plus or minus) the square root function. This means there are actually two values for every x, therefore the inverse of x squared is not invertible for its entire domain. If we restrict the domain of x squared to values greater or equal to zero and take the inverse (think absolute value), then we will get a function!(5 votes)
- I'm not entirely sure if what he just explained eliminates the possibility that I'll explain below.
Let X and Y be sets.
Let a be an element in X
Let b and c be elements in Y
Let the function g be a mapping from Y to X and when applied to the element b or c in Y it will resolve to the element a in X.
Let the function f be a mapping from X to Y such that the element a in X maps to b in Y.
This creates a situation where element c in Y maps to a in X which maps to b in Y. The identity element of Y applied to c returns c and the identity element of Y applied to b returns b.
My questions are as follows:
Thus it would seem to me that the function f is invertible, and it's inverse is the function g?
Does this mean that the composition of f with g is not the identity element from Y?
What about the composition of g with f? Is that the identity element from X?
Maybe another way of putting my question is: Is g the inverse of f? Is f the inverse of g?
I'm quite confused. Have I just constructed a system that is impossible or at least isn't allowable in linear algebra?(3 votes)
- One of the conditions for a function to be invertible is that is one-to-one, meaning that each element in the domain maps to one (and only one) element in the co-domain. From what you present,
fappears for fulfil this, but
gdoesn't, since multiple values in it's domain map to the same value in it's co-domain. Therefore
fmight be invertible, but
gis clearly not invertible, and for that it cannot be the inverse of any function.(4 votes)
- I'm sorry I don't understand. What is an identity?
Also why is a Linear Algebra video in the normal Algebra section?(6 votes)
- Can someone help me with this question please?
Let T and L be linear transformations. T: R3 to R2. L: R2 to R3.
a. Prove that the composition LT cannot be invertible.
b. Give an example to show that TL can be invertible.(4 votes)
- Is y=x^2-2 the inverse of y= sqroot (x+2)?
Secondly, how can you show algebraically where these two graphs intersect?(2 votes)
- If they intersect, then at the point of intersection, they'd have to have the same y, right? So: what if you set those two equations for y to be equal to one another?(5 votes)
- i dont see how the proof at the end proves the uniqueness(3 votes)
- He supposed there were two inverses g and h, and derived that they have to be the same. This is precisely how one proves uniqueness.
The definition of a thing satisfying a property being unique is whenever two things satisfy that property they have to be equal.(3 votes)
Let's say we have some function, f, and it's a mapping from the set X to Y. So if I were to draw the set X right there, that's my set X. If I were to draw the set Y, just like that. We know, and I've done this several videos ago, that a function just associates any member of our set X-- so I have some member of my set X there-- if I apply the function to it, or if we're dealing with vectors, we can imagine instead of using the word function, we would use the word transformation. But it's the same thing. We would associate with this element, or this member of X, a member of Y. So that's why we call it a mapping. When I apply this function-- I'll do it in a different color-- this little member of X is associated with this member of Y. If this is right here, this is a capital X. Let's say we call this a, and let's call that b. We would say that the function where a is a member of X and b is a member of Y, we would say that f of a is equal to b. This is all a review of everything that we've learned already about functions. Now I'm going to define a couple of interesting functions. The first one-- I guess it's really just one function, I said it's a couple-- but I'll call it the identity function. This is a function. I'll just call it a big capital I. This identity function operates on some set. So let's say this is the identity function on set X, and it's a mapping from X to X. What's interesting about the identity function is that if you give it some a that is a member of X-- so lets say you give it that a-- the identity function applied to that member of X, the identity function of a, is going to be equal to a. So it literally just maps things back to itself. So the identity function, if I were draw it on this diagram right here, would look like this. It would look like-- we pick a nice suitable color-- it would look like this. It would just kind of be a circle. It just points back at the point that you started off with. It associates all points with themselves. That's the identity function on X, especially as it applies to the point a. If you apply it to some other point in X, it would just refer back to itself. That's the identity function on X. You could also have an identity function on Y. So let's say that b is a member of Y. So I drew b right there. Then the Y identity function-- so this would be that identity function on Y applied to b-- would just refer back to itself. So that would be equal to b. This is the identity function on Y. So you might say, hey Sal, these are kind of silly functions. But we'll use them. They're actually at least a useful notation to use as we progress through our explorations of linear algebra. But I'm going to make a new definition. I'm going to say that a function-- let me pick a nice color, pink-- I'm going to say that a function, let me say f since we already established it right over here. I'm going to say that f is invertible, introducing some new terminology. f is invertible if and only if the following is true. I could either write it with these two-way arrows like that, or I could write it as iff with two f's. That means that if this is true, then this is true, and only if this is true. So this implies that, and that implies this. So f is invertible-- I'm kind of making a definition right here-- if and only if there exists a function, I'll call it nothing just yet. I'll call it something in a second. I'll write it as this f with this negative 1 superscript on it. So f is invertible if and only if there exists a function f inverse-- well I guess I just called it something. Let me do that in purple. Remember f is just a mapping from X to Y. So this function, f inverse, is going to be a mapping from Y to X. So I'm saying that f is invertible if there exists a function, f inverse, that's a mapping from Y to X such that if I take the composition of f inverse with f, this is equal to the identity function over X. So let's think about what's happening. This is just part of it actually. Let me just complete the whole definition. This is true, this has to be true, and the composition of f with the inverse function has to be equal to the identity function over Y. So let's just think about what's this saying. There's some function-- I'll call it right now, this called the inverse of f-- and it's a mapping from Y to X. Let me draw it up here. So f is a mapping from X to Y. We showed that. This is the mapping of f right there. It goes in that direction. We're saying there has be some other function, f inverse, that's a mapping from Y to X. So let's write it here. So f inverse is a mapping from Y to X. f inverse, if you give me some value in set Y, I go to set X. So this guy's domain is this guy's codomain, and this guy's codomain is this guy's domain. Fair enough. But let's see what it's saying. It's saying that the composition of f inverse with f, has to be equal to the identity matrix. So essentially it's saying if I apply f to some value in X-- right, if you think about what's this composition doing-- this guy's going from X to Y. This guy goes from Y to X. So let's think about what's happening here. f is going from X to Y. Then f inverse is going from Y to X. So this composition is going to be a mapping from X to X, which the identity function needs to do. It needs to go from X to X. They're saying this equals the identity function. So that means when you apply f on some value in our domain, so you go here, and then you apply f inverse to that point over there, you go back to this original point. So another way of saying this is that f-- let me do it in another color-- the composition of f inverse with f of some member of the set X is equal to the identity function applied on that item. These two statements are equivalent. So by definition, this thing is going to be your original thing. Or another way of writing this is that f inverse applied to f of a is going to be equal to a. That's what this first statement tells us. If you think of it visually, it's saying you start with an a, you apply f to it, and you get this value right here. That is f of a. I'm saying it equals b, or I said it equalled b earlier on. Then if you apply this f inverse-- and it doesn't always exist-- but if you apply that f inverse to this function, it needs to go back to this. By definition it needs to go back to your original a. It has to be equivalent to just doing this little closed loop right when I introduce you to the identity function. Now that's what this statement is telling us right here. The second statement is saying look, if I apply f to f inverse, I'm getting the identity function on Y. So if I start at some point in Y right there, and I apply f inverse first, maybe I go right here. Let's call that lowercase y So this would be f inverse of lowercase y. Then if I were to apply f to that-- I know this chart is getting very confusing-- if I apply f to this right here, I need to go right back to my original Y. So when I apply f to f inverse of Y this has to be equivalent of just doing the identity function on y. So that's what the second statement is saying. Or another way to write it is that f of f inverse of y, where y is a member of the set capital Y, it has to be equal to Y. You've been exposed to the idea of an inverse before. We're just doing it a little bit more precisely because we're going to start dealing with these notions with transformations and matrices in the very near future. So it's good to be exposed to it in this more precise form. Now the first thing you might ask is let's say that I have a function f, and there does exist a function f inverse that satisfies these two requirements. So f is invertible. The obvious question, or maybe it's not an obvious question is, is f inverse unique? Actually probably the obvious question is how do you know when something's invertible. We're going to talk a lot about that in the very near future. But let's say we know that f is invertible. How do we know, or do we know whether f inverse is unique? To answer that question, let's assume it's not unique. So if it's not unique, let's say that there's two functions that satisfy our two constraints that can act as inverse functions of f. Let's say that g is one of them. So let's say g is a mapping. Remember f is a mapping from X to Y. Let's say that g is a mapping from Y to X such that if I apply f to something and then apply g to it-- so this gets me from X to Y. Then when I do the composition with g, that gets me back into X. This is equivalent to the identity function. This was part of the definition of what it means to be an inverse. I'm assuming that g is an inverse of f. This assumption implies these two things. Now let's say that h is another inverse of f. By definition, by what I just called an inverse, h has to satisfy two requirements. It has to be a mapping from Y to X. Then if I take the composition of h with f, I have to get the identity matrix on the set X. Now that wasn't just part of the definition. It implies even more than that. If something is an inverse, it has to satisfy both of these. The composition of the inverse with the function has to become the identity matrix on x. Then the composition of the function with the inverse has to be the identity function on Y. Let's write that. So g is an inverse of f. It implies this. It also implies-- I'll do it in yellow-- that the composition of f with g is equal to the identity function on y. Then if we do it with h, the fact that h is an inverse of f implies that the composition of f with h is equal to the identity function on y as well. Let me redraw what I drew in the beginning just so we know what we're doing. So if this is a set X right here-- let me do it in a different color-- let's say this right here is the set Y. We know that f is a mapping X to Y. What we're trying to determine is is f's inverse unique. So any inverse, so we're saying that g is a situation that if you take the composition of g with f, you get the identity matrix. So f does that. If you could take g you're going to go back to the same point. So it's equivalent. So taking the composition of g with f-- that means doing f first then g-- this is the equivalent of just taking the identity function in X, so just taking an X and going back to an X. It's equivalent to that. So this is g right here. The same thing is true with h. h should also be. If I start with some element in X and go into Y, and then apply h, it should also be equivalent to the identity transformation. That's what this statement and this statement are saying. Now this statement is saying that if I start with some entry in Y here and I apply g, which is the inverse of f, I'm going to go here. So g will take me there. When I apply f then to that, I'm going to go back to that same element of Y. That's equivalent to just doing the identity function on Y. That's the same thing as the identity function of Y. I could do the same thing here with h. I just take a point here, apply h, then apply f back. I should just go back to that point. That's all of what this is saying. So let's go back to the question of whether g is unique. Can we have two different inverse functions g and h? So let's start with g. Remember g is just a mapping from Y to X. So this is going to be equal to, this is the same thing as the composition of the identity function over x with g. To show you why that's the case, remember g just goes from-- these diagrams get me confused very quickly-- so let's say this is x and this is y. Remember g is a mapping from y to x. So g will take us there. There's a mapping from y to x. I'm saying that this g is equivalent to the identity mapping, or the identity function in composition with this. Because all this is saying is you apply g, and then you apply the identity mapping on x. So obviously you're going to get to the exact same mapping or the exact same point. So these are equivalent. But what is another way of writing the identity mapping on x? What's another way of writing that? Well by definition, if h is another inverse of f, this is true. So I can replace this in this expression with a composition of h with f. So this is going to be equal to the composition of h with f, and the composition of that with g. You might want to put parentheses here. I'll do it very lightly. You might want to put parentheses there. But I showed you a couple of videos ago that the composition of functions, or of transformations, is associative. It doesn't matter if you put the parentheses there or if you put the parentheses there. Actually I'll do that. I'll put the parentheses there at first just so you can understand that this is the same thing as that right there. But we know that composition is associative. So this is equal to the composition of h with the composition of f and g. Now what is this equal to, the composition of f and g? Well it's equal to, by definition, it's equal to the identity transformation over y. So this is equal to h composed with, or the composition of h with, the identity function over y with this right here. Now what is this going to be? Remember h is a mapping from y to x. Let me redraw it. So that's my x and that is my y. h could take some element in y and gives me some element in x. If I take the composition of the identity in y-- so that's essentially I take some element, let me do it this way-- I take some element in y, I apply the identity function, which essentially just gives me that element again, and then I apply h to that. That's the same thing as just applying h to the function to begin with. So just going through this little exercise we've shown, even though we started off saying I have these two different inverses, we've just shown that g must be equal to h. So any function has a unique inverse. You can't set up two different inverses. If you do you'll find that they're always going to be equal to each other. So far we know what an inverse is. We don't know what causes someone to be able to have an inverse or not, but we know if they have an inverse, how to think about it. We also know that that inverse is unique.