Science:Math Exam Resources/Courses/MATH152/April 2011/Question A 14

MATH152 April 2011

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Question A 14
The unit square in $\mathbb {R} ^{2}$ is the set of all points $(x,y)$ with 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. In questions A13 and A14 below, suppose each point in the unit square undergoes a linear transformation that has the given matrix representation $A$. Draw a labelled sketch of the transformed points. $A=\left[{\begin{array}{cc}1&1\\1&1\end{array}}\right]$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Actually, you only need to consider the transformation of 4 special points on the square. Which points are they? The picture of the transformed square follows from linking up these 4 transformed points.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. As per the hint, we only have to consider the transformation of 4 special points. These points are the corners of the square. Linear transformations cannot distort the edges of a figure in such a way that they split an edge into a new corner. Therefore, the transformed figure must have the same amount of corners as the original figure. Because of this, it is easiest to just transform the corners and then connect them to make the new figure. Our transformation matrix is $A={\begin{bmatrix}1&1\\1&1\end{bmatrix}}$ and the four corners of the square are [0,0], [1,0], [0,1], and [1,1]. Firstly notice that the origin will always map to itself because no matrix can multiply [0,0] to produce anything but [0,0]. Therefore, we really only have to transform 3 points. The transformation of [1,0] is, $A{\begin{bmatrix}1\\0\end{bmatrix}}={\begin{bmatrix}1&1\\1&1\end{bmatrix}}{\begin{bmatrix}1\\0\end{bmatrix}}={\begin{bmatrix}1\\1\end{bmatrix}}.$ The transformation of [0,1] is, $A{\begin{bmatrix}0\\1\end{bmatrix}}={\begin{bmatrix}1&1\\1&1\end{bmatrix}}{\begin{bmatrix}0\\1\end{bmatrix}}={\begin{bmatrix}1\\1\end{bmatrix}}$ and the transformation of [1,1] is $A{\begin{bmatrix}1\\1\end{bmatrix}}={\begin{bmatrix}1&1\\1&1\end{bmatrix}}{\begin{bmatrix}1\\1\end{bmatrix}}={\begin{bmatrix}2\\2\end{bmatrix}}.$ Therefore we have ${\color {blue}[0,0]}$ transforms to ${\color {red}[0,0]}$, ${\color {blue}[1,0]}$ transforms to ${\color {red}[1,1]}$, ${\color {blue}[0,1]}$ transforms to ${\color {red}[1,1]}$ and ${\color {blue}[1,1]}$ transforms to ${\color {red}[2,2]}$. The plot to the right shows the original points in ${\color {blue}{\textrm {blue}}}$ while the transformed points are in ${\color {red}{\textrm {red}}}$. We complete the figure by connecting the corners. Notice that in this case it may appear we have lost a vertex but really, as we computed above, two are just overlapping one another. We see that we have turned our square into a line, this is typical of a projection operator (recall that a projection operator maps vectors to a single vector). In this case, if we try to form a projection operator to the vector [1,1] we get (a multiple of) our linear transformation matrix A (try it yourself!)