# Thread: linear transformation's effect on unit square

1. ## linear transformation's effect on unit square

The linear transformation is given as $\displaystyle T \begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix}\ x+y \\ x-y\end{bmatrix}$ and I have to show the effect on bothe the unit square as well as a triangle with vertices (-1,0); (0,1); (1,0).

As far as it's effect on the unit circle would it simply a reflection around the x-axis (to account for the -y in x-y) as well as a shift up by y in the x direction? ( I know that doesn't quite make sense..)

For the triangle I'm even more confused, would it still be a refection over the x-axis, but would it also lead to some sort of horizontal shear?

2. ## Re: linear transformation's effect on unit square

Check the vertices. I will help out for the triangle.
$\displaystyle T\begin{bmatrix}-1 \\ 0\end{bmatrix} = \begin{bmatrix}-1+0 \\ -1-0\end{bmatrix} = \begin{bmatrix}-1 \\ -1\end{bmatrix}$ (shifted down one unit).
$\displaystyle T\begin{bmatrix}0 \\ 1\end{bmatrix} = \begin{bmatrix}0+1 \\ 0-1\end{bmatrix} = \begin{bmatrix}1 \\ -1\end{bmatrix}$ (shifted down two units and to the right one unit).
$\displaystyle T\begin{bmatrix}1 \\ 0\end{bmatrix} = \begin{bmatrix}1+0 \\ 1-0\end{bmatrix} = \begin{bmatrix}1 \\ 1\end{bmatrix}$ (shifted up one unit).

The line segment between $\displaystyle (-1,0)$ and $\displaystyle (1,0)$ is given by the line $\displaystyle y=0$. So that edge is going to:
$\displaystyle T\begin{bmatrix}x \\ 0\end{bmatrix} = \begin{bmatrix}x+0 \\ x-0\end{bmatrix} = \begin{bmatrix}x \\ x\end{bmatrix}$ which is the line $\displaystyle y = x$, so a slanted line between (-1,-1) and (1,1).

The line segment between $\displaystyle (-1,0)$ and $\displaystyle (0,1)$ is given by the line $\displaystyle y=x$. So that edge is going to:
$\displaystyle T\begin{bmatrix}x \\ x\end{bmatrix} = \begin{bmatrix}x+x \\ x-x\end{bmatrix} = \begin{bmatrix}2x \\ 0\end{bmatrix}$ So that edge is now horizontal going from (-1,-1) to (1,-1).

The line segment between $\displaystyle (0,1)$ and $\displaystyle (1,0)$ is given by the line $\displaystyle y = -x$. So that edge is going to:
$\displaystyle T\begin{bmatrix}x \\ -x\end{bmatrix} = \begin{bmatrix}x + (-x) \\ x - (-x)\end{bmatrix} = \begin{bmatrix}0 \\ 2x\end{bmatrix}$ so that edge is now a vertical line going from (1,-1) to (1,1). In other words, the triangle is rotated 225 degrees = $\displaystyle \dfrac{5\pi}{4}$ radians and stretched by a factor of $\displaystyle \sqrt{2}$.

Consider the unit circle. Every point of the unit circle is given by $\displaystyle (\cos \theta,\sin \theta)$. So,
$\displaystyle T\begin{bmatrix}\cos \theta \\ \sin \theta\end{bmatrix} = \begin{bmatrix}\cos \theta + \sin \theta \\ \cos \theta - \sin \theta\end{bmatrix}$

Now, points on the unit sphere have the property that $\displaystyle x^2+y^2 = 1$. Now,
\displaystyle \begin{align*}(\cos \theta + \sin \theta)^2 + (\cos \theta - \sin \theta)^2 & = \cos^2\theta + 2\sin \theta \cos \theta + \sin^2\theta + \cos^2\theta - 2\sin\theta \cos\theta + \sin^2\theta \\ & = 2(\cos^2\theta + \sin^2\theta) \\ & = 2\end{align*}

So, you now have a circle of radius $\displaystyle \sqrt{2}$. Seeing a pattern?