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?

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.

is there a way to do it even faster with out all the long work?

Sure, calculate the original Area and multiply it by the absolute value of the determenant of the image, as specified in the video.

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.

cant even tell what you're talking about. i'll look at the video.

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?

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).

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Linear algebra

Course: Linear algebra > Unit 2

Lesson 6: More determinant depth

Determinant as scaling factor

Name: Determinant as scaling factor
Uploaded: 2011-02-20T16:46:47Z
Description: Viewing the determinant of the transformation matrix as a scaling factor of regions

Google Classroom

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

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FishHead
Posted 10 years ago. Direct link to FishHead's post “Does this break down once...”
Does this break down once you go beyond 2x2?
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Jamesawesome79
Posted 9 years ago. Direct link to Jamesawesome79's post “can you guys talk about s...”
can you guys talk about scales or scal factor or scale factor of a model im having trouble with it
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(7 votes)
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Robert
Posted 7 years ago. Direct link to Robert's post “It reminds me of a deriva...”
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?
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(5 votes)
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- Bernard Field
  Posted 5 years ago. Direct link to Bernard Field's post “The analogy you're drawin...”
  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.
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  (3 votes)
George Horacek
Posted 8 years ago. Direct link to George Horacek's post “is there a way to do it e...”
is there a way to do it even faster with out all the long work?
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(3 votes)
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- michael.relleum
  Posted a year ago. Direct link to michael.relleum's post “Sure, calculate the origi...”
  Sure, calculate the original Area and multiply it by the absolute value of the determenant of the image, as specified in the video.
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  (2 votes)
Urbanqueenkg
Posted 8 years ago. Direct link to Urbanqueenkg's post “What is the easiest way t...”
What is the easiest way to find a scale factor?
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(3 votes)
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- InnocentRealist
  Posted 6 years ago. Direct link to InnocentRealist's post “For a transformation T(x)...”
  For a transformation T(x) = Ax in R^2, the scaling factor for the area associated with 2 vectors is det(A).
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  (1 vote)
Ben Meeks
Posted 8 years ago. Direct link to Ben Meeks's post “How do I find scale facto...”
How do I find scale factor
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(3 votes)
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- InnocentRealist
  Posted 6 years ago. Direct link to InnocentRealist's post “For a transformation T(x)...”
  For a transformation T(x) = Ax (in R^2 at least) the scaling factor is det(A).
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  (1 vote)
Vinod P
Posted 9 years ago. Direct link to Vinod P's post “In this video, Sal plots ...”
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.
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(2 votes)
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- Bob Fred
  Posted 9 years ago. Direct link to Bob Fred's post “cant even tell what you'r...”
  cant even tell what you're talking about. i'll look at the video.
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  (2 votes)
Muhammad Nada
Posted 8 years ago. Direct link to Muhammad Nada's post “If the determinant is sca...”
If the determinant is scaling factor for area for 2D matrix, does it have a similar meaning for higher dimension matrices ?
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(3 votes)
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Kyle Delaney
Posted 8 years ago. Direct link to Kyle Delaney's post “So if the determinant is ...”
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?
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(2 votes)
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- InnocentRealist
  Posted 6 years ago. Direct link to InnocentRealist's post “Check this in R^2 for an ...”
  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).
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  (2 votes)
Breonna Roberts
Posted 6 years ago. Direct link to Breonna Roberts's post “how do you draw a scale f...”
how do you draw a scale factor
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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.