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## Linear algebra

### Course: Linear algebra > Unit 2

Lesson 4: Inverse functions and transformations- 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

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# Showing that inverses are linear

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

## Want to join the conversation?

- 9:45: "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)- 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.(13 votes)

- 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)
- You can multiply by a non-integer of c, in which case, especially if c is irrational, you would be unable to guarantee that you can add vectors to imitate the scalar multiplication(1 vote)

- 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)
- It's so that you can reverse that transformation. So say you had transformed a bunch of lines/vectors/etc and wanted to get back the originals, you would apply the inverse transformation(5 votes)

- At about3:30Sal 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)- No, a function that is invertible does not have to be linear. f(x)=x^3 is invertible and f^-1(x)=x^1/3. This makes sense because x^3 is definitely one-to-one and onto, and (x^3)^1/3=x and (x^1/3)^3=x(4 votes)

- At10:27," 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)
- 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)

- 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)
- I think that's one of the definitions of injective - that "T(a) = T(b) => a = b".(1 vote)

- at10:43does associativity only apply to linear transformations?(2 votes)
- Yes, because only linear transformations can be represented as matrices.(1 vote)

- I'm not sure if this is a valid question,

But at5:56, 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) - 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)
- 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)

- Regarding syntax, at8:33, T^-1(b), b, should have a vector arrow above it, and at18:54, Ax, should also have a vector arrow above the x?(1 vote)

## 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.