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## Linear algebra

### Course: Linear algebra > Unit 2

Lesson 6: More determinant depth- Determinant when row multiplied by scalar
- (correction) scalar multiplication of row
- Determinant when row is added
- Duplicate row determinant
- Determinant after row operations
- Upper triangular determinant
- Simpler 4x4 determinant
- Determinant and area of a parallelogram
- Determinant as scaling factor

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# Determinant as scaling factor

Viewing the determinant of the transformation matrix as a scaling factor of regions. Created by Sal Khan.

## Want to join the conversation?

- Does this break down once you go beyond 2x2?(12 votes)
- can you guys talk about scales or scal factor or scale factor of a model im having trouble with it(7 votes)
- It reminds me of a derivative in the case of linear functions from the reals to reals or R -> R case. The dy/dx = slope part gives the same feel as Area(T(region))/Area(region) = |det(A)|, where A is T's matrix. Is their a relationship between determinants and derivatives?(5 votes)
- The analogy you're drawing is a little tricky.

Derivatives act not on numbers or vectors but on functions. Thus, the derivative doesn't act on the set of real numbers but instead it maps the set of all possible functions to the set of all possible functions.

The derivative gives the rate of change of a function which in general won't be a linear function, while the determinant gives the change in area/volume/hypervolume of a linear transformation.

The derivative itself is a linear operator (obeying the rules of linearity), while the determinant is not, so this is quite a strong difference between them.

The derivative also generally gives more information than the determinant. From the derivative of a function, you can reproduce the original function up to an added constant. From the determinant, you have almost no capacity to reproduce the original linear operator.

As such, the similarity you've found between the determinant and derivative appears to be quite superficial.

For fun, since the derivative is a linear operator (albeit in the space of functions not numbers), and one where the domain and codomain are equal (meaning the corresponding matrix is square), then it should have a determinant. It's fairly easy to calculate. We can see that dy/dx is not one-to-one, because all constants get mapped to the same value (zero), so dy/dx is not invertible, so the determinant of the derivative is zero.(3 votes)

- is there a way to do it even faster with out all the long work?(3 votes)
- Sure, calculate the original Area and multiply it by the absolute value of the determenant of the image, as specified in the video.(2 votes)

- What is the easiest way to find a scale factor?(3 votes)
- For a transformation T(x) = Ax in R^2, the scaling factor for the area associated with 2 vectors is det(A).(1 vote)

- How do I find scale factor(3 votes)
- For a transformation T(x) = Ax (in R^2 at least) the scaling factor is det(A).(1 vote)

- In this video, Sal plots the transformation using the matrix A on the domain R^2, but I am not sure how the points on the coordinate system have been identified. The column vector elements of the transformation matrix are mere alphabets denoting scalars, so I would presume that the plotting has been an approximation at its best, and arbitrary otherwise, unless someone else can clarify this point. Again, the points such as bK1 and dK2 or aK1 and cK2 in the plot do seem to specify a scaling with respect to the origin, but what is perhaps jarring is the fact that each of these specifications seem to be confined to only a single axis (i.e R^1) with no corresponding coordinate in R^2 (unless they are presumed to be 0), which by itself appears to be a contradiction of the domain space notion to me.(2 votes)
- cant even tell what you're talking about. i'll look at the video.(2 votes)

- If the determinant is scaling factor for area for 2D matrix, does it have a similar meaning for higher dimension matrices ?(3 votes)
- So if the determinant is a scaling factor for the areas of shapes, is the square root of the determinant a scaling factor for the lengths of vectors?(2 votes)
- Check this in R^2 for an arbitrary R^2 matrix A (T(x) = Ax); A = [(a, b) (c, d)] and an arbitrary vector v = (e, f). Does (|Av|)/|v| = ((sqrt(|det(A)|))*|v|)/|v| = sqrt(|det(A)|)? (I doubt it).(2 votes)

- how do you draw a scale factor(2 votes)

## Video transcript

Let me graph our coordinates
in R2. So let's see, let me do
it fairly neatly. I want to do it neater
than that. So that is my vertical
coordinate, this is my horizontal coordinate. Good enough. So let's say I have four points
in R2 specified by four position vectors. So this is R2 right here. So my first position vector
is the zero vector. So it just specifies the point
right there, so it's 0, 0. My second point is specified the
by the vector say k 1, 0, so some constant times 0. Let me write it a little
bit neater than that, a little bit bigger. So it's k 1, 0, so that's my
second position vector, which would specify that point
right there. So let me draw my vector
right there. Let's say my third point
is going to be up here. As you can, see I'm creating
a rectangle. Let's say it's specified
by the position vector. Let me draw the position
vector like that. And that position vector is
going to be the vector k1. That's going to be its
horizontal component and then k2 is its vertical component. If we wanted to graph it on the
coordinates, this would be the k2 coordinate right there. And let's say that the last
point is this point right here, and it is specified by
the position vector 0, k2. And let me define a set. Let me call that set rectangle, say R for rectangle. Let me just call it Rec. Let's say that rectangle is
equal to the set of all of the points when you connect the
points specified by those position vectors. So the rectangle formed by
connecting points specified by these guys here. So if we call this vector a,
this is vector b, this is vector c, and then this is
vector d right here, so rectangle specified
by connecting the points a, b, c and d. Now let's say-- so let me draw
what our set rectangle is going to be. It's going to be all of these
points here, all of these points there, all of those
points there, and then all of those points there, including
our vertices. So that is my rectangle. Now, what is the area within
this rectangle? So this is the area
of rectangle. What's the area within
this rectangle? Well, it's just base
times height. The base has length k1. The distance from here to here
is k1, so it's just k1. And what's the height? It's the distance from here
to here, which is just k2. So it's just k1 times k2. That's fairly straightforward. I haven't shown you anything
that fascinating just yet. But let's say we're going
to transform this set. We're going to transform this
rectangle and we're going to transform it with the
transformation T. Let me pick a nice
color for T. Let's say we have the
transformation T. It is a mapping from R2 to R2,
and it can be represented as-- so when you apply the
transformation T to some vector x, it is equal to the
matrix a, b, c, d times your vector x, just like that. Now, let's see what happens
when we apply the transformation to each
of these points. We saw many videos ago that if
you want essentially the image of our rectangle under this
transformation, you can essentially just get the
transformation of each of the end points and then connect
the dots in your codomain, which is also going to be R2. So we just have to figure out
the transformation of each of these guys. So what is my transformation? I'll do it over here. I'll do it in the colors. What is the transformation
applied to the zero vector, applied to point
a right there? Well, it's just going to be
a times 0 plus b times 0, which is just 0. And then the second term is
going to be c times 0 plus d times 0, which is just 0. That's vector a. Vector b. If I apply the transformation
to vector b, vector b is k1, 0, just like that. If I apply the transformation,
it's going to be a times k-- so the first entry in the
transformed vector or what we're mapping to in our codomain
is going to be a times k1 plus b times
0, so it's just ak1. And then the second entry's
going to be c times k1 plus d times 0. So it's going to
be c times k1. All I'm doing is taking the
matrix vector product of this guy and that guy. And then let's move
on to point c. So the transformation of point
c of k1, k2 is equal to-- or the transformation of
vector c, we could say, is equal to what? The first entry's going to be
a times k1 plus b times k2. a times k1 plus b times k2, and
the second entry's going to be c times k1 plus
d times k2. We have one point left. The transformation, if we take
the transformation of our vector 0, k2, what do we get? We get a times 0 plus b times
k2, which is just b times k2, and then c times 0 plus d times
k2, which is just d times k2, just like that. So let's draw the image of
our rectangle under our transformation. So let me redraw my axes. So this is my vertical
axis right there. This is my horizontal
axis right there. And so a gets mapped to the
zero vector right there. So this is the transformation
of 0, 0 right like that. Now, what does the vector
b get transformed? The vector b gets transformed
to ak1, ck1, so let me draw it like this. The vector b gets transformed
to something that looks like this, where this is the vector--
so this is the transformation of k1 and 0,
which is equal to ak1, ck1. We saw that already. Now, what does this yellow
guy get transformed to? Let me do the blue one first.
So the blue one gets transformed here. Let me finish here. So if we come down here,
this is going to be ak1 right there. If we go to the left like that,
that's going to ck1. That's how we graph it. Now this last guy, this blue
guy, maybe he looks something like this. This is the transformation of
that vector right there by this linear transformation, 0,
k2, which is equal to the vector bk2 and dk2, so this
point right here is bk2 and then this coordinate
right there is dk2. Now this last guy, this yellow
point, when I transformed it, what do I get? I get ak1, plus bk2, so ak1 plus
bk2 will get you right about there, so that's going
to be its x-coordinate. And then its y-coordinate
is ck1 plus dk2. So ck1, and then I'm going to
add that much again to it. So it's going to get me
someplace over here. So I'm going to add this
distance to this whole distance over here, so
I'm going to get up here some point. I'm going to get right
over here. So that vector's going to look
something like that. That is the transformation
of k1, k2. That's the transformation
of that guy. And then if we connect the
dots, remember, the transformation of this set of
this rectangle, or we could say the image of this
rectangle under the transformation, we just take
each of the points that define that rectangle and then
we connect the dots. We saw that awhile ago. So let me draw it this way. This line right here that
connects those two dots will be transformed to this line that
connects these two dots right there. This line-- I was going to do
that in a different color. This line that connects those
two dots is going to be transformed to this line that
connects those two dots. And then we have this line that
connects those two dots is going to be transformed to
this line that connects these two dots right there. And then finally, this line that
connects these two dots is going to be transformed to
this line that connects these two dots right there. Now, just out of curiosity,
what is-- let me just write this. So this rectangle right
here is the image. We could write this as T of our
rectangle, or if we wanted to write it in words, this is
equal to the image of the rectangle under our
transformation T. Now, what is the area of the
image of our rectangle under our transformation? What is the area
of this thing? What is this area right there? Well, we could view this
parallelogram essentially as a parallelogram generated by this
vector and that vector. Or if we wanted to write it
another way, if we had a matrix, if we had some matrix
whose column vectors are this guy and that guy, so let's say,
it's first column vector is this guy, so ak1, ck1, And
this one right here, this was the transformation of k1, 0. And then its second column
vector is this guy right here, which was bk2 and dk2. This parallelogram is a
parallelogram generated by these two column vectors. This was the transformation
right here. This was the transformation
of 0, k2, right? And in the last video, we said
that the area of this parallelogram is equivalent
to-- let me write this right now. So the area of the
parallelogram, which is the same thing as the area of the
image of our original rectangle under our
transformation, the area this parallelogram is equivalent to
the absolute value-- let me call this determinant
right here, let me call this I for image. So it's equal to the absolute
value of the determinant of the matrix whose column
vectors generate that parallelogram. So it should be equal to the
determinant of the absolute value of the determinant of I. We saw that in the last video. That's what the area of this
thing is going to be. Well, what is the determinant
of this? Well, this is going to be equal
to-- we've got to keep our absolute value signs. So the determinant of this is
ak1 times dk2, so we could right this as-- let
me switch colors. We could write it
as k1, k2, ad. All I did is multiply that term
times that term minus this term times that term. So minus k1, k2, bc. That's just the definition
of the determinant of a 2-by-2 matrix. We saw this in the last video. This is nothing new here, or
maybe it might be relatively new if you just watched the last
video, but all I'm saying is, look, if we want to figure
out this parallelogram, this is the parallelogram
generated by that vector and that vector. And if we say that those two
guys are the column vectors of some matrix right there, we saw
on the last video that the area of this parallelogram is
equal to the absolute value of the determinant of
this matrix. Now, what is the equal to? This is equal to-- we can
factor out a k1, k2. So it's the absolute value of
k1, k2 times ad minus bc. Now what is this equal to? Well, ad minus bc, that is
the determinant of our transformation vector. So if we say that T of x-- if we
called that vector-- or our transformation matrix. If we call this matrix A, if we
call it A, this is equal to k1, k2 times the determinant
of A, which is a pretty interesting takeaway. What this is telling us is,
look, if I have some region, or in this case, I just have the
rectangle, but if I have some region in my domain and I
apply a transformation, and let's say my region
has some area. Let's say in this case,
it has some area A. Let me write this
right as area. This was my original area in
my original region, right? And then I apply some
transformation to it where the transformation is equal to some
matrix A times any member of your domain. I apply some transformation
to it. I get some new region. I get the image of this set
under my transformation. Your new area is going to be
equal to the absolute value of the determinant of your
transformation matrix. That's the determinant of the
transformation matrix. That's the transformation
matrix. So it's equal to the
determinant of the transformation matrix times
the area of your original rectangle in this case, right? This was your original
rectangle. You take the absolute value,
just because sometimes you swap these vectors you might
get a negative sign, so you take the absolute value. But that's a really neat idea. The determinant of the
transformation matrix is essentially a scaling
factor on the area of a certain region. Now, I'm not going to prove it
to you here, but you can kind of imagine it. Let's say I have some-- let
me go abstract now. Let's say I have some region in
R2-- let me do a better-- let me do a shape that
you might recognize. Let's say I have something
like that. Let's say it's an
ellipse, right? And this is our domain. It's R2. And let's say that the area
here, the area of this region right here, is equal to A. So area is equal to A. And let's say I have some
transformation T that is a mapping from R2 to R2 And T is
defined, so T applied to some vector x in my domain is equal
to-- well, I already used A, so let's say it's equal to some
matrix B times any vector in my domain. If I take the image of this area
under T, so this is my domain right here. Let me do it in the codomain. Let's say the image of this
under B, let's say it looks something like-- or my image
under T-- let's say it looks something like this. I don't know what it's going to
look like, but let's say it looks something like this. Let's say it looks something
like that right there, so this right here. So if we call this set-- let
me call it E for ellipse. That's this whole thing
right here. This is the ellipse
right there. This is the image of my
ellipse under the transformation T. If I take every point of this
ellipse, I'll construct this right here. Or I guess maybe just to keep
the analogy straight for this example, let's say that the
ellipse is the set of just the boundary, but it also applies
actually if you were to fill up the whole region. So this is boundary gets
mapped to this boundary right here. But it would have worked
just as well if we filled up the region. Since I'm not proving either to
you very rigorously, you'll have to accept it as
a leap of faith. So this is the image of this
boundary under our transformation. And if the area surrounded by
this boundary, or if you could say the area of this region is
A, then the area here, the area of the image of our
ellipse under our transformation is going to be
equal to the absolute value of our area, our original area,
right there times the determinant of our
transformation matrix. If you find it a little bit of
a big leap to go from just an arbitrary rectangle
transformation to maybe these more generalized shapes, you
can imagine-- so T is going from there to there. It's always nice to
draw that arrow. But if you view it as a bit of
a leap to make that jump from rectangles to curves, you can
imagine this shape to be able to be constructed by a bunch
of arbitrary rectangles. So you could essentially fill
this space with a bunch of arbitrary rectangles like
this, and you could keep making them infinitely
small so that you completely fill the space. You completely fill that
space with rectangles. And if you were to map each of
those rectangles using T, each of those rectangles might look
something like this. Maybe they look something
like this. I don't know what they'll
look like. They'll look like a bunch of
parallelograms. Maybe they'll look something like that. I'm not drawing it
that neatly. They'll look like a bunch of
parallelograms, and they will essentially fill this
space right there. So that's a way you can get your
head around the idea of jumping from just the arbitrary
rectangles to these arbitrary curves or
shapes or regions. But this is a pretty neat
outcome, and it's a very interesting way to view
a determinant. A determinant of a
transformation matrix is essentially a scaling factor for
area as you map from one region to another region, or
as we go from one region to the image of that region under
the transformation.