Linear transformation in the plane

Question

Let $K2$ denote the amount of vectors in the plane originating from Origo in a $(O,i,j)$ - coordinate system. A linear transformation $f: K2 \rightarrow K2$ is illustrated on the figure below: A blue point $P$ with the position vector $\vec{OP}$ is transformed to a red point which is the end point for $f(\vec{OP})$. The red figure is thus the image of $f$ of the blue figure.

Let $e=(i,j)$ denote the standard basis for $K2$ and let $a=(a_1,a_2)$ denote the new basis for $K2$ where $a_1=9i$ and $a_2=2i+2j$

1) Determine the basis transformation matrices $_eM_a$ and $_aM_e$.

2) Analyze the images of $f(a_1)$ and $f(a_2)$ and use them to write the transformation matrix $_eF_a$.

3) Write the transformation matrix $_eF_e$, and show that the vector set $x=(x_1,x_2)$ is a basis for $K2$ consisting of Eigenvectors for $f$. Write the transformation matrix $_xF_x$.

I'm still new to this topic, so I'm kind of struggling to see how I should approach each one.

Answer 1

In general if you have two basis,

$$ \epsilon = \{\epsilon_1,\epsilon_2 \cdots \} ~~~\mbox{and}~~~ e=\{e_1,e_2, \cdots\} $$

And the there exists a transformation between the elements of these two basis

$$ e_i = a_{i1}\epsilon_1 + a_{i2}\epsilon_2 + \cdots $$

Then the coordinates of a vector in the basis $e$ ($x_e$) can be written in terms of its coordinates in the basis $\epsilon$ ($x_\epsilon$) as

$$ x_\epsilon = A^t x_e \tag{*} $$

where $A = (a_{ij})$

1. Now we can solve your problems, all we need to do is to find the matrix $A$, but that one is easy because we know how to write the vectors of $a$ in terms of $e$

   $$ a_1 = 9i, ~~~ a_2 = 2i + 2j ~~~\Rightarrow~~~ A = \left(\begin{array}{cc} 9 & 0 \\ 2 & 2\end{array}\right) $$

   So that using Eq. (*)

   $$ M_{e \to a} = A^t = \left(\begin{array}{cc} 9 & 0 \\ 2 & 2\end{array}\right) $$

   and

   $$ M_{a \to e} = (A^t)^{-1} = \left(\begin{array}{cc} 1/9 & -1/9 \\ 0 & 1/2\end{array}\right) $$

2. I'm going to pick two points in the figure, say the bottom of the front bumper $v_1$ and the bottom of the windshield $v_2$. In the $x$ basis, we now that for the blue figure $v_1^{(\rm blue)} = (0, 2)$ and $v_2^{(\rm blue)} = (1, 3)$.

   Whereas in the red figure $v_1^{(\rm red)} = (0, 6)$ and $v_2^{(\rm blue)} = (-3,9)$, if $v^{\rm(red)} = F v^{\rm(blue)}$ we can see that

   $$ \underbrace{\left(\begin{array}{cc} 0 & -3 \\ 6 & 9\end{array}\right)}_{(v_1^{(\rm red)},~v_2^{(\rm red)})} = F \underbrace{\left(\begin{array}{cc} 0 & 1 \\ 2 & 3\end{array}\right)}_{(v_1^{(\rm blue)},~v_2^{(\rm blue)})} $$

   and therefore

   $$ F = \left(\begin{array}{cc} -3 & 0 \\ 0 & 3\end{array}\right) $$

   you can actually check that this is the correct transformation by transforming the bottom of the rear bumper in the blue image $v = (3,2)$ to the red image, the result is $Fv = (-9, 6)$, which matches perfectly!

   The transformation is then

   $$ F_{\rm blue \to red} = F = \left(\begin{array}{cc} -3 & 0 \\ 0 & 3\end{array}\right) $$

3. Finally, if you notice that $x_1 = 2i - j$ and $x_2 = i + j$, you can apply Eq. (*) again to find the transformation between these two basis.

   For this case

   $$ L = \left(\begin{array}{cc} 2 & -1 \\ 1 & 1\end{array}\right) $$

   therefore

   $$ M_{e \to x} = \left(\begin{array}{cc} 2 & 1 \\ -1 & 1\end{array}\right) $$

   and

   $$ M_{x \to e} = \left(\begin{array}{cc} 1/3 & -1/3 \\ 1/3 & 2/3\end{array}\right) $$