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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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# Linear transformations

Introduction to linear transformations. Created by Sal Khan.

## Want to join the conversation?

- When my teacher says to learn Transformaions: reflections, rotations, translations, dilations.... is this the video I should be watching for that or is Linear transformations something different? if it is could you tell me what that video is called so I can look it up? Thank you so much.. confused a bit here =P(18 votes)
- These linear transformations are probably different from what your teacher is referring to; while the transformations presented in this video are functions that associate vectors with vectors, your teacher's transformations likely refer to actual manipulations of functions.

Unfortunately, Khan doesn't seem to have any videos for transformations, reflections, etc. in his algebra playlist, but the links below might be useful.

WolframMathWorld: http://mathworld.wolfram.com/Transformation.html

Paul's Online Notes: http://tutorial.math.lamar.edu/Classes/Alg/Transformations.aspx

Hope this helps and good luck!

(33 votes)

- Is another name for this 'linear mappings'?(13 votes)
- Linear Transformation = Linear Mapping = Linear Function(16 votes)

- Simple question, (apologies if answered, I'm about 1/2 way through), but, what exactly does "Linear" mean. I understand that it meets those three criterion, but say, in a very abstract sense (and hopefully in laymen's terms), what does it mean? Perhaps it implies continuity? Perhaps it means the transformation won't enter the domain of complex numbers?

Also, can you name a condition or two where 'linearity', that is, the criterion will consistently broken?

I hope I'm clear on the type of answer I'm looking for. Thanks,

TB

EDIT: With a little inductive reasoning, it appears that if a translation is NOT linear, something is being lost or gained either when either the vectors are added together and then transformed, or something is lost or gained when they are transformed then added together.

I guess that something would be lost in transformation, not addition, so if information is lost in transformation then it would still be lost when they are then added together; thus giving a different.

I guess I answered my own question =D

You mentioned squares and exponents. Curious, something inherent in either transforming or adding either squares or exponents is causing a loss of information.

Care to take this logic further?(12 votes)- The textbook definition of linear is: "progressing from one stage to another in a single series of steps; sequential." Which makes sense because if we are transforming these matrices linearly they would follow a sequence based on how they are scaled up or down.(6 votes)

- Why do we need to have two conditions here?

Isn't the vector addition enough? After all, if you can add vector a and a scalar times vector a, then this is the same thing as just multiplying the vector by that scalar + 1, isn't it?(6 votes)- But how would we get a scalar like 1.1 from just adding a vector with itself, or pi for that matter?

This is a great question, and one I used to ask myself. Ultimately, there examples of transformations that satisfy vector addition, but not scalar multiplication, so both conditions for linearity are in fact necessary.(5 votes)

- It would be good if there were more practice problems and quizzes on this unit. It is hard to keep track of all this information without applying it.(8 votes)
- In order for it to be a linear transformation doesn't zero vector have to satisfy the parameters as well? If it is how come it wasn't in the video?(3 votes)
- Let
**v**be an arbitrary vector in the domain. Then T(**0**) = T( 0 ***v**) = 0 * T(**v**) = 0. So you don't need to make that a part of the definition of linear transformations since it is already a condition of the two conditions.(3 votes)

- At4:23, Sal said that component of a vector is scalar..but component of a vector also have their direction (like component along x axis or so)..right? so in that way component of a vector should also be vector, i think...! Well, m confused..plz help..and sorry for the silly out of context question.... :)(2 votes)
- Well, strictly speaking component of a vector, that is just what is written inside a vector cell is a scalar, it has no information in which cell it was written. What you are talking about is vector decomposition, i.e. representing a vector as a sum of axis-aligned vectors, consider example, given vector

v = (3, 4, 5)

it has scalar components - just numbers 3, 4, and 5

it could also be decomposed into a sum, like this

vx = (3, 0, 0)

vy = (0, 4, 0)

vz = (0, 0, 5)

v = vx + vy + vz

look up 'vector basis'(3 votes)

- Is there a third property of a transformation being linear: T(0) = 0? I can't think of when this wouldn't be the case, unless there's a constant in the transformation without a variable.. Wanted to confirm if this is a property or not... Thanks.(3 votes)
- Sal can we find a linear transformation by knowing the basis of its kernal?(2 votes)
- No. Knowing the kernel tells us which basis vectors are sent to 0, but the remaining basis vectors could still be sent anywhere.(3 votes)

- At13:25, Sal mentions that if you're dealing with a linear transformation that involves only a linear combination of different components of inputs, you're "probably" dealing with a linear transformation. But if we're talking about a "linear combination" of components, wouldn't it ALWAYS be a linear transformation?? If not, can someone give an example where a linear combination of components leads to a Non-linear transformation?(1 vote)
- Unfortunately LaTeX does not work in these comment boxes, as otherwise I could have shown you my proof that any transformation consisting of linear combinations is also a linear transformation. Simply put (just to explain the concepts of what would need to be included in the proof), we know that any combination of vectors can be expressed as another vector. Similarly, any combination of constants results in one bigger constant. This means that we can , by proving that T(vector a) = [c1*a1, c2*a2, ..., cn*an] is L.T. for any vector a and for any series of constants c1, c2, ... cn, prove that any transformation including only linear combinations is a transformation that is L.T.

To make the "if T consists of linear combinations of vectors and constants" an "iff ~", all we need to do now is to prove that for any non-linear combination of vectors, but not constants (c1*c2 is still C and thus does not give a different result about whether T is L.T., whatever that result may be).

I made a small mistake by first not seeing that linear combination ONLY involves vectors. Constants are out of the question, therefore I should not have spoken about "linear combinations of vectors >>and constants<<".

I hope that shows that the proof is quite simple, so it is certainly not impossible, even quite easy, to state that all examples will work, as there is a simple proof covering every possible linear combination without loss of generality, by making a few simple but key (and easily forgotten) lemmas and assumptions.(3 votes)

## Video transcript

You now know what a
transformation is, so let's introduce a special kind of
transformation called a linear transformation. It only makes sense that we have
something called a linear transformation because we're
studying linear algebra. We already had linear
combinations so we might as well have a linear
transformation. And a linear transformation,
by definition, is a transformation-- which we
know is just a function. We could say it's from the set
rn to rm -- It might be obvious in the next video why
I'm being a little bit particular about that, although
they are just arbitrary letters -- where
the following two things have to be true. So something is a linear
transformation if and only if the following thing is true. Let's say that we have
two vectors. Say vector a and let's
say vector b, are both members of rn. So they're both in our domain. So then this is a linear
transformation if and only if I take the transformation of
the sum of our two vectors. If I add them up first, that's
equivalent to taking the transformation of each of the
vectors and then summing them. That's my first condition
for this to be a linear transformation. And the second one is, if I take
the transformation of any scaled up version of a vector
-- so let me just multiply vector a times some scalar
or some real number c . If this is a linear
transformation then this should be equal to c times
the transformation of a. That seems pretty
straightforward. Let's see if we can apply these
rules to figure out if some actual transformations
are linear or not. So let me define a
transformation. Let's say that I have the
transformation T. Part of my definition I'm going
to tell you, it maps from r2 to r2. So if you give it a
2-tuple, right? Its domain is 2-tuple. So you give it an x1 and an x2
let's say it maps to, so this will be equal to, or it's
associated with x1 plus x2. And then let's just say it's 3
times x1 is the second tuple. Or we could have written this
more in vector form. This is kind of our
tuple form. We could have written it -- and
it's good to see all the different notations that you
might encounter -- you could write it a transformation of
some vector x, where the vector looks like
this, x1, x2. Let me put a bracket there. It equals some new vector,
x1 plus x2. And then the second component
of the new vector would be 3x1. That's a completely legitimate
way to express our transformation. And a third way, which I never
see, but to me it kind of captures the essence of what
a transformation is. It's just a mapping or
it's just a function. We could say that the
transformation is a mapping from any vector in r2 that looks
like this: x1, x2, to-- and I'll do this notation-- a
vector that looks like this. x1 plus x2 and then 3x1. All of these statements
are equivalent. But our whole point of writing
this is to figure out whether T is linearly independent. Sorry, not linearly
independent. Whether it's a linear
transformation. I was so obsessed with linear
independence for so many videos, it's hard to get it out
of my brain in this one. Whether it's a linear
transformation. So let's test our
two conditions. I have them up here. So let's take T of, let's say
I have two vectors a and b. They're members of r2. So let me write it. A is equal to a1, a2, and
b is equal to b1, b2. Sorry that's not a vector. I have to make sure that
those are scalars. These are the components
of a vector. And b2. So what is a1 plus b? Sorry, what is vector
a plus vector b? Brain's malfunctioning. All right. Well, you just add up
their components. This is the definition
of vector addition. So it's a1 plus b1. Add up the first components. And the second components is
just the sum of each of the vector's second compnents.
a2 plus b2. Nothing new here. But what is the transformation
of this vector? So the transformation of vector
a plus vector b, we could write it like this. That would be the same thing as
the transformation of this vector, which is just a1
plus b1 and a2 plus b2. Which we know it equals
a vector. It equals this vector. Or what we do is for the first
component here, we add up the two components on this side. So the first component here
is going to be these two guys added up. So it's a1 plus a2
plus b1 plus b2. And then the second component
by our transformation or function definition is just 3
times the first component in our domain, I guess
you could say. So it's 3 times the first one. So it's going to be 3 times
this first guy. So it's 3a1 plus 3b1. Fair enough. Now what is the transformation
individually of a and b? So the transformation of a is
equal to the transformation of a -- let me write it this
way -- is equal to the transformation of a1
a2 in brackets. That's another way of
writing vector a. And what is that equal to? That's our definition of our
transformation right up here, so this is going to be equal to
the vector a1 plus a2 and then 3 times a1. It just comes straight out
of the definition. I essentially just replaced
an x with a's. By the same argument, what
is the transformation of our vector b? Well, it's just going to be the
same thing with the a's replaced by the b's. So the transformation of our
vector b is going to be -- b is just b1 b2 -- so it's
going to be b1 plus b2. And then the second component in
the transformation will be 3 times b1. Now, what is the transformation
of vector a plus the transformation
of vector b? Well, it's this vector
plus that vector. And what is that equal to? Well, this is just pure vector
addition so we just add up their components. So it's a1 plus a2
plus b1 plus b2. That's just that component
plus that component. The second component is 3a1 and
we're going to add it to that second component. So it's 3a1 plus 3b1. Now, we just showed you that if
I take the transformations separately of each of the
vectors and then add them up, I get the exact same thing as
if I took the vectors and added them up first and then
took the transformation. So we've met our
first criteria. That the transformation of the
sum of the vectors is the same thing as the sum of the
transformations. Now let's see if this works
with a random scalar. So we know what the transformation of a looks like. What does ca look like,
first of all? I guess that's a good
place to start. c times our vector a is going
to be equal to c times a1. And then c times a2. That's our definition of scalar multiplication time's a vector. So what's our transformation --
let me go to a new color. What is our -- let me do a color
I haven't used in a long time, white. What is our transformation
of ca going to be? Well, that's the same thing as
our transformation of ca1, ca2 which is equal to a new vector,
where the first term -- let's go to our definition
-- is you sum the first and second components. And then the second term is 3
times the first component. So our first term
you sum them. So it's going to be
ca1 plus ca2. And then our second term
is 3 times our first term, so it's 3ca1. Now, what is this equal to? This is the same thing. We can view it as factoring
out the c. This the same thing as c times
the vector a1 plus a2. And then the second
component is 3a1. But this thing right here,
we already saw. This is the same thing as
the transformation of a. So just like that, you see that
the transformation of c times our vector a, for any
vector a in r2 -- anything in r2 can be represented this way
-- is the same thing as c times the transformation of a. So we've met our second
condition, that when you when you -- well I just stated it, so
I don't have to restate it. So we meet both conditions,
which tells us that this is a linear transformation. And you might be thinking,
OK, Sal, fair enough. How do I know that all
transformations aren't linear transformations? Show me something
that won't work. And here I'll do a very
simple example. Let me define my
transformation. Well, I'll do it from r2
to r2 just to kind of compare the two. I could have done it from r to r
if wanted a simpler example. But I'm going to define
my transformation. Let's say, my transformation
of the vector x1, x2. Let's say it is equal to
x1 squared and then 0, just like that. Let me see if this is a
linear transformation. So the first question
is, what's my transformation of a vector a? So my transformation of a vector
a-- where a is just the same a that I did before--
it would look like this. It would look like a1 squared
and then a 0. Now, what would be my
transformation if I took c times a? Well, this is the same thing as
c times a1 and c times a2. And by our transformation
definition -- sorry, the transformation of c times this
thing right here, because I'm taking the transformation
on both sides. And by our transformation
definition this will just be equal to a new vector that
would be in our codomain, where the first term is
just the first term of our input squared. So it's ca1 squared. And the second term is 0. What is this equal to? Let me switch colors. This is equal to c squared
a1 squared and this is equal to 0. Now, if we can assume that c
does not equal 0, this would be equal to what? Actually, it doesn't
even matter. We don't even have to make
that assumption. So this is the same thing. This is equal to c squared times
the vector a1 squared 0. Which is equal to what? This expression right here
is a transformation of a. So this is equal to c squared
times the transformation of a. Let me do it in the
same color. So what I've just showed
you is, if I take the transformation of a vector being
multiplied by a scalar quantity first, that that's
equal to -- for this T, for this transformation that I've
defined right here -- c squared times the transformation
of a. And clearly this statement right
here, or this choice of transformation, conflicts with
this requirement for a linear transformation. If I have a c here I should
see a c here. But in our case, I have
a c here and I have a c squared here. So clearly this negates
that statement. So this is not a linear
transformation. And just to get a gut feel
if you're just looking at something, whether it's going to
be a linear transformation or not, if the transformation
just involves linear combinations of the different
components of the inputs, you're probably dealing with
a linear transformation. If you start seeing things where
the components start getting multiplied by each
other or you start seeing squares or exponents, you're
probably not dealing with a linear transformation. And then there's some functions
that might be in a bit of a grey area, but it
tends to be just linear combinations are going to lead
to a linear transformation. But hopefully that gives you
a good sense of things. And this leads up to what I
think is one of the neatest outcomes, in the next video.