Proof: Invertibility implies a unique solution to f(x)=y | Linear Algebra | Khan Academy

I’ve got a function f and it’s
a mapping from the set x to the set y. And let’s just say for the sake
of argument, let’s say that f is invertible. What I want to know is what
does this imply about this equation right here. The equation f of
x is equal to y. I want to know that for every
y that’s a member of our co-domains. So for every y, let me write
this down, for every y that’s a member of my co-domain is
there a unique, in all caps UNIQUE, solution x that’s
a member of our domain. Such that, and I could write
such that well, I’ll just write it out. I was going to write the mathy
way, but I think it’s nicer to write in the actual
words sometimes. Such that f of x
is equal to y. So if we just, let me
just draw everything out a little bit. We have our set x right here. This is x. We have our co-domain y here. We know that f, if you take some
point here, let’s call that a, it’s a member of x and
you apply the function f to it, it will map you to some
element in set y. So that’s f of a right there. This is, so far, what
this tells us. Now I want to look at this
equation here, and I want to know that if I can pick any y
in this set any lower-case y in this set y. So let’s say I pick
something here. Let’s say that’s b. I want to know is there a
unique solution to the equation f of x is equal to b. Is there a unique solution? So one, I guess, you have to
think, is there a solution. So is there a solution to
saying, look is there some x here that if I apply the
transformation f to it that I get there, and I also want
to know is it unique. For example, if this is the only
one that is unique, but it’s not unique if there’s some
other guy, if there is more than one, solution. If there is some other guy
in x that if I apply the transformation I also go to b. This would make it non-unique,
not unique. So what I want to concern
ourselves with in this video is somehow, is invertibility
related to the idea of a unique solution to this for
any y in our co-domain. So let’s just work through our
definitions of invertibility and see if we can get anywhere
constructive. So by definition, f is
invertible implies that there exists this little backward
looking, three-looking thing, this means there exists. I think it’s nice to be exposed
sometimes to the mathy notations Let me just
write that. That means there exists some
function, let’s call it f inverse, that’s a mapping from
y to x, such that, and actually the colons are also the
shorthand for such that, but I’ll write it out. Such that the composition of f
inverse with f is equal to the identity on x. So, essentially, it’s saying
look, if I apply f to something in x then I apply f
inverse to that, I’m going to get back to that point which is,
essentially, equivalent. Or it isn’t just essentially
equivalent, it is equivalent to just applying the identity
function, So that’s i x. So you just get what
you put into it. Such that this inverse function,
the composition of the inverse with the function,
is equal to the identity function. And that the composition of the
function with the inverse function is equal to the
identity function on y. So if you started y and you
apply the inverse, then you apply the function to that,
you’re going to end up back at y at that same point. And that’s equivalent
to just applying the identity function. So this is what invertibility
tells me, this is how I defined invertibility
in the last video. Now we concerned ourselves– so
we are concerned with this equation up here. We’re concerned with the
equation, I will write it in pink, f of x is equal to y. And what we want to know for
any y, or any lower-case cursive y in our big set
y, is there a unique x solution to this. So, what we can do is we know
that f is invertible. I told you that from
the get-go. So given that f is invertible,
we know that there is this f inverse function, and I can
apply that f inverse function. It’s a mapping from y to x. So, I can apply it to
any element in y. So for any y, let’s say that
this is my y right there. So I can apply my f
inverse to that y. And I’m going to go over here
and, of course, y is equal to f of x. These are the exact
same points. So let’s apply our f inverse
function to this. So if I apply the f inverse
function to both sides of the equation, this right here’s an
element in y, and this is the same element in y. Right? They are the same element. Now, if I apply the mapping, the
inverse mapping, to both of that, that’s going to take
me to some element in x. So let’s do that. So, if I take the inverse
function on both sides of this equation, where some element
over here in y, and I’m taking the inverse function to get to
some element in x, what’s this going to be equal to? Well, on the right-hand side,
we could just write the f inverse of y. That’s going to be some
element over here. But what does the left-hand
side of this equation translate to? The definition of this inverse
function is that when you take the composition with f, you’re
going to end up with the identity function. This is going to be equivalent
to– let me write it this way. This is equal to the composition
of f inverse with f of x, which is equivalent to
the identity function being applied to x. And then the identity function
being applied to x is what? That’s just x. This thing right here
just reduces to x. This reduces to x. So, we started with the idea
that f is invertible. We use the definition of
invertibility that there exists this inverse function
right there. And then we essentially apply
the inverse function to both sides of this equation and say,
look you give me any y, any lower-case cursive y in this
set y, and I will find you a unique x. This is the only x that
satisfies this equation. Remember how do I know
it’s the only x? Because this is the only
possible inverse function. Only one inverse function
which is true. I proved that to you
in the last video. That if f is invertible,
it only has one unique inverse function. We tried before to have maybe
two inverse functions, but we saw they have to be
the same thing. So since we only have one
inverse function and it applies to anything in this big
upper-case set y, we know we have a solution. And because it’s only one
inverse function, and functions only map to one value
in this case, then we know this is a unique
solution. So let’s write this down. So we’ve established that if f
is invertible, I’ll do this in orange, then the equation f of
x is equal to y for all. That little v that it looks
like it’s filled up with something, for all y, the member
of our set y, has a unique solution. And that unique solution, if you
really care about it, is going to be the inverse
function applied to y. It might seem like a bit of a
no brainer, but you can see you have to be a little bit
precise about it in order to get to the point you want. Let’s see if the opposite
is true. Let’s see if we assume– let’s
see if we start from the assumption, that for all y that
is a member of our set Y, that the solution, that the
equation f of x, is equal to y has a unique solution. Let’s assume this and see if it
can get us the other way. If given this, we can
prove invertibility. So let’s think about
the first way. So we’re saying that for any y–
let me draw my sets again. So this is my set X and this
is my set Y right there. Now we’re working for the
assumption that you can pick any element in Y right here,
and then the equation right here has a unique solution. Let’s call that unique
solution. Well, we could call
it whatever. But a unique solution x. So you can pick any point here,
and I’ve given you, we’re assuming now, that, look,
you pick a point in Y, I can find you some point
in X such that f of x is equal to y. And not only can I find
that for you, that is a unique solution. So given that, let me define
a new function. Let me define the function s. The function s is a mapping
from y to x. It’s a mapping from y to x and s
of, let’s say s of y, where, of course, y is a member
of our set capital-Y. s of y is equal to the unique
solution in x to f of x is equal to y. Now, you’re saying, hey, Sal, it
looks a little convoluted. But think about it. This is a completely valid
function definition. Right? We’re starting with the idea
that you give me any y here. You give me any member of this
set, and I can always find you a unique solution to
this equation. Well OK, so that means that any
guy here can be associated with a unique solution in the
set X, where the unique solution is the unique
solution to this equation here. So, why don’t I just define a
function that says, look I’m going to associate every member
y with its unique solution to f of x
is equal to y. That’s how I’m defining this
function right here. And, of course, this is
a completely valid mapping from y to x. And we know that this only has
one legitimate value because this, any value y, any
lower-case value y, in this set has a unique solution
to f of x is equal to y. So this can only equal
one value. So it’s well defined. So let’s apply, let’s take some
element here, let me do a good color, let’s
say this is b. And b is a member of y. So let’s find, so let’s just map
it using our new function right here. So let’s take it, and map it,
and this is s of b right here. s of b which is a member of x
Now, we know that s of b is a unique solution by definition. I know it seems a little
circular, but it’s not. We know that s of
b is a solution. So we know that s of b is the
unique solution to f of x is equal to b. Well, if this is the case, if
this is good, we just got this because this is what
this function does. It maps every y to the unique
solution to this equation. Because we said that every
y has a unique solution. So, if this is the case,
then what happens if I take f of s of b. Well, I just said this is the
unique solution to this. So if I put this guy in here,
what am I going to get? I’m going to get b. Or other way of saying this is
that the composition of f with s applied to b is equal to b. Or another way to say it, is
that we take the composition of f with s, this is the same
thing, because if I apply s to b, and then I apply f back to
that, that’s the composition, I just get back to b. That’s what’s happening here. So this is the same thing as
the identity function on y being applied to b. So it’s equal to b. So we can say that the
composition– we can say that there exists, and we know that
this function exists, or that we can always construct this. So we already know
that this exists. This existed by me constructing
it, but I’ve, hopefully, shown you that
this is well defined. That from our assumption that
this always has a unique solution in x for any y here, I
can define this in a fairly reasonable way. So it definitely exists. And not only does it exist,
but we know that the composition of f with this
function that I just constructed here is equal to
the identity function on y. Now let’s do another
little experiment. Let’s take a particular– let
me just draw sets again. This is our set X. Let me take some member
of set X, call it a. Let me take my set
Y right there. And so we can apply the function
to a and we’ll get a member of set Y. Let’s call that right there. Let’s call that f of
a right there. Now, if I apply my magic
function here that always, I can give you any member of set Y
and I’ll give you the unique solution in X to
this equation. So, let me apply that to this. Let me apply s to this. So, if I apply s to this it’ll
give me the unique solution. So, let me write this down. So if I apply s to this, I’m
going to apply s to this– and maybe I shouldn’t point it back
at that, I don’t want to imply that it necessarily
points back at that. So, let me apply s
that, s to this. So what is this going
to point to? What is that point going
to be right there? So that’s going to be s of this
point, which is f of a, which we know is the
unique solution. So, this is equal to the
unique solution to the equation f of x is equal
to this y right here. Or this y right here is
just called f of a. Right? Remember, the mapping s just
mapped you from any member of a to the unique solution
to the equation f of x is equal to that. So this is the mapping from f of
a to the unique, so this s of f of a is going to be a
mapping to the– or this right here, is going to be the
unique solution to the equation f of x is equal
to this member of y. And what’s the remember
y I called? It’s called f of a. Well, you could go say this in
a very convoluted way, but if I were to just, before you
learned any linear algebra, if I said look if I have the
equation f of x is equal to f of a. What is the unique solution
to this equation? What does x equal? x would have to be equal to a. So the unique solution to the
equation f of x is equal to f of a is equal to a. And we know that there’s only
one solution to that because that was one of our starting
assumptions. So this thing is equal to a. Or we could write s of
f of a is equal to a. Or that the composition of s
with f is equal or applied to a is equal to a. Or that the composition of s
with f is just the identity function on the set x. Right? This is a mapping right
here from x to x. So we could write that the
composition of s with f is the identity on x. So, what have we done so far? We started with the idea that
you pick any y in our set capital-Y here and we’re going
to have a unique solution x such that this is true. Such that f of x
is equal to y. That’s what the assumption
we started off with. We constructed this function
s that immediately maps any member here with its unique
solution to this equation. Fair enough. Now from that, we said this
definitely exists. Not only does it exist, but
we figured out that the composition of f with our
constructed function is equal to the identity on the set y. And then we also learned that
s– the composition of s with f is the identity
function on x. Let me write this. So we learned this, and we
also learned that the composition f with s is equal
to the identity on y. And s clearly exists because I
constructed it, and we know it’s well defined because every
y– for every y here there is a solution to this. So given that I was able to find
for my function f, I was able to find a function that
these two things are true. This is by definition. What it means to
be invertible. Remember, so this means
that f is invertible. Remember f being invertible,
in order for f to be invertible, that means there
must exist some function from, so if f is a mapping from to x
to y, invertibility means that there must be some function f
inverse that is a mapping from y to x such that, so I can write
there exists a function, the inverse function composed
with our function should be equal to the identity on x. And the inverse and the function
in the composition of the function, with the inverse
function, should be the identity on y. Well, we just found
a function. It exists, and that
function is s. Where both of these
things are true. We can say that s is
equal to f inverse. So f is definitely invertible. So, hopefully, you found
this satisfying. This proof is very subtle and
very nuanced because we keep bouncing between our
sets X and Y. But what we’ve shown is that if
f, in the beginning part of this video, we show that if f
is invertible then there is for any y a unique solution to
the equation f of x equals y. And the second part of the
video, we showed that the other way is true, That if– let
me put it this way, that if for all Y, a member of
capital Y, there is a unique solution to f of x is equal to
x, then f is invertible. So the fact that both of these
assumptions imply each other, we can write our final
conclusion of the video. That f being invertible, if f,
which is a mapping from x to y is invertible, this is true if
and only if, and we could write that either as
a two-way arrow. Or we could write if for if and
only if so both of these statements imply each other. If and only if for all y, for
every y that is a member of our set y there exists a
unique– I can actually write that like that means there’s
exists a unique x for the– or let me write this way there
exists a unique solution to the equation f of
x is equal to y. So that was our big takeaway
in this video. That invertibility of a function
implies there’s a unique solution to this equation
for any y that’s in the co-domain of our function.

5 thoughts on “Proof: Invertibility implies a unique solution to f(x)=y | Linear Algebra | Khan Academy”

• January 30, 2010 at 3:51 am
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I can spot an analogy, although oversimplistic, of the identity (or 1 in the univariate case) with the wave state of a particle in quantum mechanics.
Both entities have the potential of an infinite variability of instantiations. The identity to an infinite set of f/f-inverse compositions, and the particle to a set of infinite possible trajectories.
Even more, both entities revert to some state under observation. The particle wave collapses to a value; the f-inverse is defined under an assumed f.

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• July 11, 2013 at 8:48 pm
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big help thx ðŸ™‚

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• October 10, 2013 at 3:51 pm
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lost myself at the beginning

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• January 13, 2014 at 12:52 am
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sex? 20:27

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• January 3, 2020 at 11:24 pm
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Absolutely pointless

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