Equation of unit circle under linear transformation - can't understand role of inverse matrix

Question

In the course I'm following (topic is Geometric Transformations, NOT linear algebra yet) they introduce a way to find the equation of the unit circle under some invertible linear transformation $f(C)$ where $C$ is the unit circle.

So, let $f$ be the linear transformation represented by the following matrix:

$$ A = \begin{pmatrix} 1 & -2\\ 1 & -3\\ \end{pmatrix} $$

And its inverse is

$$ A^{-1} = \begin{pmatrix} 3 & -2\\ 1 & -1\\ \end{pmatrix} $$

Alright, so far so good.

Then the book goes on to say:

Each point $(x, y)$ is the image under $f$ of the point $f^{-1}(x, y)$, which has the position vector $$ A^{-1} \begin{pmatrix} x \\ y \\ \end{pmatrix} = \begin{pmatrix} 3 & -2\\ 1 & -1\\ \end{pmatrix} \begin{pmatrix} x \\ y \\ \end{pmatrix} = \begin{pmatrix} 3x-2y \\ x-y \\ \end{pmatrix} $$ Hence $f$ maps the point $(3x-2y, x-y)$ to the point $(x, y)$. [...]

It follows that: $(3x-2y)^2 + (x-y)^2 = 1$

I don't follow along here. I can't grasp this sentence:

Each point $(x, y)$ is the image under $f$ of the point $f^{-1}(x, y)$ [...]

After finding the inverse matrix, we create a system of equations: $$ \begin{pmatrix} 3x-2y \\ x-y \\ \end{pmatrix} $$

Then we plug that into the equation of the unit circle and then we are done? I understand the mechanical steps, but I can't rationalize for my own understanding why we are doing that.

How is the inverse matrix related to finding the image of the unit circle under a linear transformation?

Why can't I just make use of $f$, the actual transformation and must jump through some hoops to use $f^{-1}$.

Answer 1

Here's one way to look at it. Start with the circle $S^1=\lbrace (x,y):x^2+y^2=1\rbrace$. Apply the function $f$ to $S^1$. The circle is all points $(x,y)$ who satisfy the circle equation, right? So therefore $f(S^1)$ is the set of points $f(x,y)$, where $(x,y)$ satisfies the circle equation. $$\text{Image}=f(S^1)=\lbrace f(x,y): x^2+y^2=1\rbrace=\lbrace (x-2y,x-3y): x^2+y^2=1\rbrace$$ So this says that if you find a pair of real numbers $u$ and $v$ which can be written in this funky way $(u,v)=(x-2y,x-3y)$ AND if $(x,y)$ satisfies the circle equation, then it is in the image. That's kind of bizarre. You'd rather know what the condition on $u$ and $v$ are directly. Well, given $u=x-2y$ and $v=x-3y$ it is not that hard to rewrite the equation $x^2+y^2=1$. In fact, you get $x=3u-2v$ and $y=u-v$, which is what you have.

Let's look at this a different way now, because the above method is kind of the naive thing to do. The one encouraged by your book is not so bad. Lets think of the image as follows: every point in $f(S^1)$ is the image of a point on the circle. Thus, if $(x,y)$ is in the image, then this means that $(u,v)=f^{-1}(x,y)$ is a pair of points satisfying the circle equation. $$u^2+v^2=1$$ But, we know what $f^{-1}$ is. It's just a function like $f$. You compute that $f^{-1}(x,y)=(3x-2y,x-y)$. Thus $(u,v)=(3x-2y,x-y)$ substituting this gives the result.

Answer 2

While $f(A)$ is the collection of the $f$-images of all points of $A$, when it is about transferring the defining property of $A$, we have to use $f^{-1}$, because our starting point is that we are given an arbitrary point $x$ and the question is whether it is in $f(A)$ or not, i.e. is $f^{-1}(x)\in A$ or not, so that we can apply the defining property of $A$ to $f^{-1}(x)$.

To illustrate this general phenomenon with an easier example, let $A:=\{x:x>0\} \subseteq \Bbb R$ and let $f:=x\mapsto x+1$.
Then $f(A)$ is clearly $\{x:x>1\}$, which is the same as $\{x:x-1>0\}=\{x:f^{-1}(x)\in A\}$ but not the same as $\{x:x+1>0 \} =\{x:f(x)\in A\}$.

Answer 3

A linear transform maps the unit circle on an ellipse.

The ellipse corresponding with a matrix $$ A =\ \left[\begin{matrix}a & b\\c &d\\\end{matrix}\right] $$ is defined by:

$$\left(d^2+c^2\right)x^2-2\left(ac+bd\right)xy+\left(a^2+b^2\right)y^2-\left(ad-bc\right)^2=0 $$

The main axes of the ellipse are the singular vectors of the matrix and the length of the axes are the singular values.

The image of the unit-circle by $A$ can be described as:
"the set of all $(u,v)$ that are the image of an $(x,y)$ on the unit circle"
$$\left\{\left[\begin{matrix}u\\v\\\end{matrix}\right]:\left[\begin{matrix}u\\v\\\end{matrix}\right]=A\left[\begin{matrix}x\\y\\\end{matrix}\right]\ and \ x^2+y^2=1\right\}$$

If we apply the inverse transformation on the vectors $(u,v)$ we must end up on the unit circle again:

$$\left\{{\left[\begin{matrix}u\\v\\\end{matrix}\right]:A}^{-1}\left[\begin{matrix}u\\v\\\end{matrix}\right]=\left[\begin{matrix}x\\y\\\end{matrix}\right]\ and \ x^2+y^2=1\right\}$$

All $(u,v)$ on the ellipse are mapped back on the unit-circle by $A^{-1}$

$$\|A^{-1}\left[\begin{matrix}u\\v\\\end{matrix}\right] \| =1$$ We now expand the norm of the vector $A^{-1}\left[\begin{matrix}u\\v\\\end{matrix}\right]$. The norm of a vector $\|w\|=\| w^T w \|$.

$$\left(\|A^{-1}\left[\begin{matrix}u\\v\\\end{matrix}\right]\|\right)^2=\left(A^{-1}\left[\begin{matrix}u\\v\\\end{matrix}\right]\right)^T\left(A^{-1}\left[\begin{matrix}u\\v\\\end{matrix}\right]\right)=1$$

$$\left[\begin{matrix}u\\v\\\end{matrix}\right]^T\left(A^T\right)^{-1}A^{-1}\left[\begin{matrix}u\\v\\\end{matrix}\right]=1$$

$$\left[\begin{matrix}u\\v\\\end{matrix}\right]^T\left({AA}^T\right)^{-1}\left[\begin{matrix}u\\v\\\end{matrix}\right]=1$$

Expanding the expression above leads to the equation of the ellipse corresponding to a matrix $A$.

$$\left(d^2+c^2\right)u^2-2\left(ac+bd\right)uv+\left(a^2+b^2\right)v^2=\left(ad-bc\right)^2$$

Similar reasoning can be followed to calculate the ellipse corresponding to the inverse transform.
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