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### Course: Linear algebra > Unit 2

Lesson 1: Functions and linear transformations- A more formal understanding of functions
- Vector transformations
- Linear transformations
- Visualizing linear transformations
- Matrix from visual representation of transformation
- Matrix vector products as linear transformations
- Linear transformations as matrix vector products
- Image of a subset under a transformation
- im(T): Image of a transformation
- Preimage of a set
- Preimage and kernel example
- Sums and scalar multiples of linear transformations
- More on matrix addition and scalar multiplication

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# Preimage of a set

Definition of preimage of a set. Created by Sal Khan.

## Want to join the conversation?

- At around 4.45, Sal says that T(T^-1(S)) forms only a subset of S. What confuses me is, if an element in S isn't mapped by the transformation, then that element is not part of S, and is therefore outside S. The preimage of S must, surely, include the preimage of every element in S.(15 votes)
- When Sal defines S he doesn't define it as the result of the transformation, he defines S as just any subset of Y. Therefore this subset S is
*not*limited to those that have a matching element in X.

As a result, there can exist some elements of S that do not have a matching element in the**preimage**of S and these unmatched elements would therefore not appear in the**image**of the**preimage**of S, aka T(T^-1(S)).(31 votes)

- So, just to clarify, the set S is a subset of our CODOMAIN but not necessarily a subset of the IMAGE of our function, right?

And, with regard to his "bonus points" question, the image of our preimage of S under T [ T(T^-1(S)) ] is the intersection of the IMAGE of X under T and S, the subset of our CODOMAIN Y. Correct?(11 votes)- Yes to the first. There are some elements of S which are not necessarily reached, I believe, by the image of the pre image of S under T.

...and yes to the Second!. Both correct as far as I can see!(5 votes)

- Couldn't there be a bunch of preimages? For example, if the domain is R and the codomain is R and the subset S is all R≥0, then the preimage could be x^2, x^4, e^x... because all of these functions map members of R to members of R≥0(3 votes)
- Actually, nevermind. I realize now that the preimage is a set, not a transformation.(3 votes)

- Hi, is there a proof of this T(T^-1(S)) ⊂ S ?

Thanks(2 votes)- Let x be an element of T(T^-1(S)). The image T(V) is defined as the set {k | k=T(v) for some v in V}. So x=T(y) where y is an element of T^-1(S). The preimage of S is the set {m | T(m) is in S}. Thus T(y) is in S, so since x=T(y), we have that x is in S.

Thus we have shown if x is in T(T^-1(S)), then x is in S, so T(T^-1(S)) ⊂ S.(3 votes)

- I think that "T^-1(T(A) = A". Am I right?(2 votes)
- Yep! You can think of inverting a transformation as undoing it.(2 votes)

- Is it safe to say that X is a set of n vectors? Same for Y - a set of m vectors?(2 votes)
- Not really. X can be a set of anything. If it were vectors you would NORMALLY describe it as a set of vectors in Rn, so each vector has n elements. Same with Y, vectors would have m elements.

You could techncally have a set with n vectors mapping onto a set with m vectors, but there wouldn' be a lot that could be done with vector arithmetic, or matrix arithmetic.

Let me know if that did not answer your question fully.(1 vote)

- Ok Sal, fair enough, but what if you find preimages recurseively adinfinitum....

T(T^-1(T(T^-1.........

You are finding infinite subsets that may or may not be smaller than the originals?

I'm trying to think of this functionally where f( f^-1(x) ) would not necessarily be x, but some subset of x?(2 votes) - The problem of what is the co-domain and what is the range of the function comes up again. If T maps to some sub-set of S where S is itself a sub-set of the co-domain, the function defines only this sub-set of S. So how do we define S and the co-domain. They appear to be entirely arbitrary.Any set is a sub-set of the universal set, therefore we can choose anything we like to be the co-domain.(1 vote)
- Yes. However, in general you won't be
**working out**what the co-domain is. You get a function, and the definition tells you what the co-domain is.. Your logic is right though, and if you were creating a function you might have to work out what you wanted the function to have as a co-domain in order for it to make the most sense in context.. maybe choosing something too far removed from the problem will confuse people you try to describe your function to.(2 votes)

- At about4:14Sal says he is talking about the image of our pre-image under S. I thought S was a set. Aren't we taking the image under T, the transformation? I thought when we take a transformation T of S, we call it the transformation of S under T. Am I wrong?(1 vote)
- um, S is a set and T is a transformation. for a transformation T: X -> Y, the preimage of a set S is a set of all x in X such that T(x) = S, where S is a subset of Y, or as written in this case, x = T-1(S)(1 vote)

- It was defined that a transformation's image is the range or all possible outcomes that a set maps onto after being transformed and that , this image might be a subset of the co-domain.

But in this video, he talks about how the pre-image which is just the inverse of the image or range, now maps to not all of the range but a subset of the range .

Am I missing something? that doesn't make any sense(1 vote)

## Video transcript

Let's add some transformation
that maps elements in set X to set Y. We know that we call
X the domain of T. So that's my set X and then my
set that I'm mapping into, set Y, that's the codomain. We know that T is a
transformation that if you take any member of X and you
transform it, you'll associate it with a member of set Y. You'll map it to a
member of set Y. That's what the transformation
or the function does. Now, if we have some subset of
T, let's call A to be some subset of T. So let me draw A like that. This notation right here
just means subset, some subset of T. We've defined the notion of an
image of T of A like that, which is the image of A, of
our subset A, under T. We've defined this to be equal
to the set-- let me write it here-- the set of all-- where if
we take each of the members of our subset, it's the set of
all of their transformations. And, of course, these are going
to be some subset of Y right there. So we essentially take each
of the members of A. This was one of them. You find its transformation. It's that point. You take another member of A. This is all set A right here. Take another member of A. Find its transformation. Maybe it's that point. You keep doing that. Find it's transformation. Maybe it's that point. And then the set of all of those
transformations, maybe it's this blob right here,
we call this the image of A under T. Now, what if we wanted
to think about the opposite problem? What if we were to start with
set Y, which is our codomain, So that's Y, and we were to
have some subset of Y. Let's call our subset of Y S. So S is a subset of
our codomain Y. And I'm curious about what
subset of X maps into S. So I'm curious about this set. I'm curious about the set of all
vectors that are members of my domain such that they're
mapping or the transformation of those vectors ends up
in my subset of S. So what I'm saying is, look,
if I take my domain, there must be some subset of vectors
right here, where if I take any member of this set, it will
map into these guys, and that's what I'm defining
right here. This is equal to that guy. So I'm literally saying what
are all of the members of X where those members of
X all map into S? Now, I want to make a very
subtle nuance here to point out something here. I'm not saying that
every point in S necessarily gets mapped to. For example, maybe there's some
element in S right there that no element in X ever
gets mapped to from our transformation T. That's OK. All I'm saying is that
everything in this set maps to something within S right here. And what we call this set right
here, the notation is the inverse T of S, but
this is equal to the pre-image of S under T. So this is S. This is the pre-image
of S under T. And that makes sense. The image, we go from a subset
of our domain to a subset of our codomain. Preimage, we go from a subset
of our codomain, and we say what subset of our domain
maps into that subset of our codomain? Now let me ask you an
interesting question, and this is kind of for bonus points. What is the image of our
pre-image under S? So if we take this guy, this
is essentially the image of this guy right here, right? This part right here is the
pre-image of S right here. Now, if we take the image of
this, we're saying if we take every member of this, what
vectors do they map into? All of them are going to be
within S, so they're going to map within S, but they don't
necessarily map to everything within S. So this is going to be
some subset of S. So this right here is going
to be some subset of our original S. It's not necessarily equal to
S, but it's a subset of it. And so this is I think the
motivation for where the notation comes from. We can construct a subset of S
by taking the image of the pre-image of S. We can kind of view the image
and the pre-image kind of canceling out, and that's why
the inverse notation was probably introduced. Now this is all very abstract. What I'm going to do in the
next video is actually calculate or determine
the pre-image of some subset of my codomain.