What happens to the area of a square when the perimeter increases by a factor of N?

Video transcript

- [Instructor] We're told that Polygon Q is a scaled copy of Polygon P using a scale factor of one half. Polygon Q's area is what fraction of Polygon P's area? Pause this video and see if you can figure that out. Alright, my brain wants to make this a little bit tangible and once we get some practice, you might be able to do it without drawing pictures but they're saying some arbitrary Polygon Q and P so let's just make a simple one. Since we're talking about area, I like to deal with rectangles since it's easy to think about areas of rectangles. And so, let's see, Polygon Q is a scaled copy of Polygon P. So let's start with Polygon P. And I will do this in red. So Polygon P, let's just say, I'm just gonna create an arbitrary polygon. So let's say that this is, and I'm gonna scale it by one half so I'm gonna make its sides have nice even numbers. So let's say this side right over here is four and this side right over here is equal to eight, this is Polygon P right over here, it's a quadrilateral, it's in fact a rectangle and its area is just going to be four times eight which is 32. Now let's create Polygon Q, and remember, Polygon Q is a scaled copy of P using a scale factor of one half. So we're gonna scale it by one half. So instead of this side being four, it's going to be two and instead of this side over here being eight, the corresponding side in the scaled version is going to be four. So there you go, we've scaled it by one half, and now what is our area going to be? Well our area, and this Polygon Q, and so our area is going to be two times four which is equal to eight. So notice that Polygon Q's area is one fourth of Polygon P's area and that makes sense because when you scale the dimensions of the Polygon by one half, the area is going to change by the square of that. One half squared is one fourth and so the area has been changed by a factor of one fourth or another way to answer this question, Polygon Q's area is what fraction of Polygon P's area? Well it's going to be one fourth of Polygon P's area. And the big takeaway here is if you scale something, if you scale the sides of a figure by one half each, then the area is going to be the square of that and so one half squared is one over four. If it was scaled by one third, then the area would be scaled, or the area would be one ninth. If it was scaled by a factor of two, then our area would have grown by a factor of four. Let's do another example. Here we're told, Rectangle N has an area of five square units. Let me do this in a different color. So Rectangle N has an area of five square units. James drew a scaled version of Rectangle N and labeled it Rectangle P. So they have that right over here. This is a scaled version of Rectangle N. What scale factor did James use to go from Rectangle N to Rectangle P? So let's think about it. They gave us Rectangle P right over here, and let's think about its dimensions. This height is one, two, three, four, five, it's five high, and it is, one, two, three, four, five, six, seven, eight, nine wide, and so its area is equal to 45. Now, Rectangle N had an area of five square units. So our area, let me write this down, so N area to P area, N area to P area, we are multiplying by a factor of nine. If we're going from an area of five square units to 45 square units, you notice N area is five, N's area is five square units, P's area, we just figured out is 45 square units, and so we have it growing by a factor of nine. Now what would be the scale factor if our area grew by a factor of nine? Well we just talked about the idea that area will grow, the factor with which area grows is the square of the scale factor. So one way to think about it is scale factor, scale factor squared is going to be equal to nine, or another way to think about it, our scale factor is going to be equal to three to go from N to P. Now let's verify that we answered their question but I just want us to feel good about it. Let's draw a rectangle that is scaled down from P by a factor of three. So a rectangle if we were to scale it up by a factor of three, we get rectangle P. So its bottom would have length three instead of nine. So it'd be like this. So that would be three, and its height instead of being five, it would be five thirds. Five thirds is one and two thirds, so it'd go about that high, it would look something like that. It would be five thirds. And so our Rectangle N would look like this, and what is its area? Well five thirds times three is indeed five square units. So notice, when we have the area growing by a factor of nine, the scale factor of the size to go from five thirds to five, you multiply by three. To go from three to nine, you multiply by three.

A linear transformation $\vc{T}: \R^n \to \R^m$ [confused?] is a mapping from $n$-dimensional space to $m$-dimensional space. Such a linear transformation can be associated with an $m \times n$ matrix.

If we restrict ourselves to mappings within the same space, such as $\vc{T}: \R^n \to \R^n$, then $\vc{T}$ is associated with a square $n \times n$ matrix. One can calculate the determinant of such a square matrix, and such determinants are related to area or volume. It turns out that the determinant of a matrix tells us important geometrical properties of its associated linear transformation. We'll outline this relationship for one-dimensional, two-dimensional, and three-dimensionional linear transformations.

One-dimensional linear transformations

A one-dimensional linear transformation is a function $T[x] = ax$ for some scalar $a$. To view the one-dimensional case in the same way we view higher dimensional linear transformations, we can view $a$ as a $1 \times 1$ matrix. The determinant of the $1 \times 1$ matrix is just the number $a$ itself. Although this case is very simple, we can gather some intuition about linear maps by first looking at this case.

An example one-dimensional linear transformation is the function $T[x]=3x$. We could visualize this function by its graph, which is a line through the origin with slope 3. However, instead, let's view it as a mapping from the real line $\R$ back onto the real line $\R$. In this case, we think of the function as $x' = T[x]$, which maps the number $x$ on the $x$-axis to the new number $T[x]$ on an $x'$-axis. $T$ takes the number 1 and maps it to 3. $T$ maps 0 to 0 and -1/2 to -3/2. We also use the language that 3 is the image of 1 under the mapping $T$.

We can summarize this mapping by looking how $T$ maps an interval of numbers. For example, $T$ maps the interval $[0,1]$ to the interval $[0,3]$, as illustrated by the following figure. We colored the interval $[0,1]$ and its image $[0,3]$ with a green to red gradient that illustrates how each point in the interval is mapped by $T$. A point of a given color in $[0,1]$ is mapped to a point of the same color in the image $[0,3]$.

The fact that the determinant of the matrix associated with $T$ is 3 means that $T$ stretches objects so that their length is increased by a factor of 3. Since the determinant was positive, $T$ preserves the orientation of objects: in both the interval $[0,1]$ and its image $[0,3]$, red points are to the right of green points.

Another examples is the linear transformation $T[x]=-\frac{1}{2}x$. As shown in the below figure, $T$ maps the interval $[0,1]$ onto the interval $[-\frac{1}{2},0]$. The determinant of the matrix associated with $T$ is $-\frac{1}{2}$. Since the magnitude of this determinant is $\frac{1}{2}$, $T$ shrinks objects to be half their original length. The negative determinant indicates that $T$ reverses the orientation of objects: as indicated by the colors, the right of the interval $[0,1]$ is mapped onto the left of the interval $[-\frac{1}{2},0]$.

In general, the linear transformation $T[x]=ax$ stretches objects to change their length by a factor of $|a|$. If $a$ is positive, $T$ preserves the orientation; if $a$ is negative, $T$ reverses orientation. We will obtain similar conclusions for higher-dimensional linear transformations in terms of the determinant of the associated matrix.

Two-dimensional linear transformations

A two-dimensional linear transformation is a function $\vc{T}: \R^2 \to \R^2$ of the form $$\vc{T}[x,y] = [ax+by,cx+dy] = \left[\begin{array}{cc}a &b\\ c &d\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right],$$ where $a$, $b$, $c$, and $d$ are numbers defining the linear transformation. We can write this more succinctly as $$\vc{T}[\vc{x}] = A\vc{x},$$ where $\vc{x}=[x,y]$ and $A$ is the $2 \times 2$ matrix containing the constants that define the linear transformation, $$A = \left[\begin{array}{cc}a &b\\ c &d\end{array}\right].$$ We will view $\vc{T}$ as mapping objects from the $xy$-plane onto an $x'y'$-plane: $[x',y']=\vc{T}[x,y]$. As in the one-dimensional case, the geometric properties of this mapping will be reflected in the determinant of the matrix $A$ associated with $\vc{T}$.

To begin, we look at the linear transformation $$\vc{T}[x,y] = \left[\begin{array}{cc}-2 &0\\ 0 &-2\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right].$$ As with all linear transformations, it maps the origin $\vc{x}=[0,0]$ back to the origin $[0,0]$. We can get a feel for the behavior of $\vc{T}$ by looking at its action on the standard unit vectors, $\vc{i}=[1,0]$ and $\vc{j}=[0,1]$. $\vc{T}$ maps $[1,0]$ to $[-2,0]$, and it maps $[0,1]$ to $[0,-2]$. It stretched both vectors by a factor of $2$ and rotated them all the way around by $\pi$ radians.

To better visualize the mapping of $\vc{T}$, we can examine how it maps a region in the plane. The below figure shows the mapping of $[x',y']=\vc{T}[x,y]$ on the unit square $[0,1] \times [0,1]$. We colored the quarters of the square in different colors to help visualize how points within the square were mapped. The figure demonstrates that $\vc{T}$ did rotate the square by $\pi$ radians around the origin and stretched each side by a factor of 2.

We claimed this behavior should be reflected in the determinant of the associated matrix $$A = \left[\begin{array}{cc}-2 &0\\ 0 &-2\end{array}\right].$$ In this case, $\det[A] = [-2][-2]-[0][0] = 4$. The determinant is 4 even though it seemed it was streching everything by a factor of 2. And the determinant was positive even though it rotated everything so that points on the right are mapped to points on left and points on the top are mapped to points on the bottom. Shouldn't we have gotten a negative determinant?

The reason for getting a factor of 4 rather than 2 is due to the fact that determinants of $2 \times 2$ matrices reflect area not length. In fact, the absolute value of the determinant of a $2 \times 2$ matrix \begin{align*} \left[ \begin{array}{cc} a_1 & a_2\\ b_1 & b_2 \end{array} \right] \end{align*} gives the area of the parallelogram spanned by the vectors $[a_1,a_2]$ and $[b_1,b_2]$. The mapping $\vc{T}$ stretched a $1 \times 1$ square of area 1 into a $2 \times 2$ square of area 4, quadrupling the area. This quadrupling of the area is reflected by a determinant with magnitude 4.

The reason for a positive determinant is that, in two-dimensions, rotations, even all the way around by $\pi$ radians, is not considered changing the orientation. If we go counterclockwise around the perimeter of the mapped square, we still encounter the colors in the order red, green, yellow, blue. Changing orientation would correspond to taking the square out of the $xy$-plane and flipping it before putting it down in the $x'y'$-plane, as we'll see in the next example.

The linear transformation $$\vc{T}[\vc{x}] = A\vc{x}, \qquad A = \left[\begin{array}{cc}-1 &-1\\ 1 &3\end{array}\right]$$ should change orientation as $\det[A] = [-1][3]-[-1][1] = -2$. It should also increase area by a factor of $|\det[A]| = 2$.

Again, we visualize the transformation $\vc{T}$ by looking at how it maps the unit square $[0,1] \times [0,1]$. As shown below, it maps the square into a parallelogram. Inspection of the paralleogram reveals that it indeed has area 2, so $\vc{T}$ doubles area as claimed. the orientation is also reversed. Moving counterclockwise around the perimeter of the parallelogram leads to the opposite color order red, blue, yelow, green. There is no way to stretch and move the original unit square into the parallelogram without taking it out of the plane and flipping it [or somehow moving the region through itself].

The actual geometric shape and rotation of the square's image is not captured by the determinant. We cannot tell whether or not $\vc{T}$ will map the square into a square, rectangle, rhombus, or other parallelogram from knowledge of just the determinant of its associate matrix. [We know it has to be some type of parallelogram, as all two-dimensional linear transformations do map parallelograms into parallelograms.] The determinant simply tells us how $\vc{T}$ changes area and whether or not it reverses orientation.

You can experiment with these and other linear transformations using the below applet. You can convince yourself that $\vc{T}$ always maps parallelograms onto parallelograms and that the determinant of its associated matrix does capture area stretching and orientation reversing.

Three-dimensional linear transformations

The reflection of geometric properties in the determinant associated with three-dimensional linear transformations is similar. A three-dimensional linear transformation is a function $\vc{T}: \R^3 \to \R^3$ of the form $$\vc{T}[x,y,z] = [a_{11}x+a_{12}y + a_{13}z, a_{21}x+a_{22}y+a_{23}z,a_{31}x+a_{32}y+a_{33}z] = A\vc{x}.$$ where $$A=\left[\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{array}\right]$$ and $\vc{x}=[x,y,z]$. The components $a_{ij}$ of $A$ define the linear transformation.

As above, the determinant $\det[A]$ reflects some of the geometric properties of the mapping $[x',y',z']=\vc{T}[x,y,z]$. Due to the fact that the absolute value of a determinant gives the volume of a parallelepiped, the absolute value $|\det[A]|$ reflects how much $\vc{T}$ expands volume. The sign of $\det[A]$ reflects whether or not $A$ preserves orientation, as above. If the $\det[A]$ is positve, the associated linear transformation preserves orientation in that it only stretches and rotates objects. On the other hand, if $\det[A]$ is negative, the associated linear transformation reverses orientation by also reflecting the object [taking its mirror image].

We illustrate with two examples. The first is the linear transformation associated with the matrix $$A=\left[\begin{array}{rrr}2&1&1\\1&2&-1\\-3&-1&2\end{array}\right].$$ Since $\det[A] = 12$, the linear transformation $\vc{T}[\vc{x}] = A\vc{x}$ expands the volume of objects by a factor of 12. Since the determinant is positive, it preserves the orientation of objects.

The action of $\vc{T}$ on the unit cube $[0,1] \times [0,1] \times [0,1]$ is illustrated in the following applet. $\vc{T}$ rotated the cube and stretched it into a parallelepiped of volume 12. You can confirm that $\vc{T}$ preserved orientation by comparing faces that have the same four colors and checking if the colors have the same order when moving counterclockwise. Since both the cube and the parallelepiped have faces with colors ordered red, yellow, white, magenta as one moves counterclockwise, the linear transformation preserved orientation, as it must given that $\det[A]$ is positive. You can change the cube into other parallelepipeds, and $\vc{T}$ always maps it to another parallelepiped, as it must.

Another example is the linear transformation associated with the matrix $$B=\left[\begin{array}{rrr}3&1&-3\\1&3&-2\\1&1&-3\end{array}\right].$$ Since $\det[B]=-14$, the linear transformation $\vc{T}=B\vc{x}$ stretches volume by the factor $|\det[B]| = 14$. In this case, since $\det[B]$ is negative, the linear transformation reverses orientation. The reversal of orientation can be seen in the below applet illustrating the mapping of the unit cube $[0,1] \times [0,1] \times [0,1]$. $\vc{T}$ maps the cube into a parallelepiped of volume $14$, but also reflects the cube in the process. This reversal of orientation can be observed by noticing that the parallelepiped has a face with the colors in the reverse order red, magenta, white, yellow, when moving counterclockwise.

What happens to the area of a square when its perimeter increases by a factor?

Does area increase if we increase the perimeter of a square?

What is the relationship between area and perimeter of a square?

What happens to the area if the perimeter is doubled?