In 4-1, we worked on the transformation that rotation 90 degree counterclockwise. In this practice, you will write down the transformation matrix for any rotation in 2d and prove that it is a linear transformation.

Consider a transformation matrix $T(\pmb{x}), R^2\rightarrow R^2$:

$$T(\pmb{x}) = \begin{bmatrix} 1 & 0\\ 0 & 2 \end{bmatrix}$$

(1) What is this transformation does to the space, you can try to transform some vector to see.

(2) What kinds of vectors do not change directions after the transformation?

(3) If now the transformation matrix became:

$$T(\pmb{x}) = \begin{bmatrix} 1 & 1\\ 0 & 2 \end{bmatrix}$$

What kinds of vectors do not change directions after the transformation this time? Do we expect the same answer?

Consider following transformation matrix:

$$T_1(\pmb{x}) = \begin{bmatrix} 2 & 0 & 3\\ 0 & 2 & 2 \end{bmatrix}$$

(1) What is the dimension of vectors before and after the transformation?

(2) After such transformation, the space is then transformed by another matrix:

$$T_2(\pmb{x}) = \begin{bmatrix} 2 & 0 \\ 0 & 2 \\ 1 & 2 \end{bmatrix}$$

What is the dimension of vectors after this transformation?

(3)Now combine these two transformation into one transformation $T_3(\pmb{x})$ so that:

$$T_3(\pmb{x}) = T_2(T_1(\pmb{x}))$$

Can you write $T_3(\pmb{x})$ down as a matrix? if so, what is the dimension of this matrix $T_3(\pmb{x})$? (hint: Try to think matrix as columns of vector)

(4)What if you reverse the order of transformation? Say $T_4(\pmb{x}) = T_1(T_2(\pmb{x}))$? Does that gives you the same transformation? What is the result tells you?

Answer sheet

In 4-1, we worked on the transformation that rotation 90 degree counterclockwise. In this practice, you will write down the transformation matrix for any rotation in 2d and prove that it is a linear transformation.

If you want to write down the transformation matrix, you need to find the basis vector of the space and see how they transform. In our case, the basis will be $[1,0]$ and $[0,1]$, if you rotate them counterclockwise by an arbitrary angle $\theta$, you will have:

$$\begin{bmatrix} 1 \\ 0 \end{bmatrix} \rightarrow \begin{bmatrix} \cos\theta\\ \sin\theta \end{bmatrix}$$

$$\begin{bmatrix} 0 \\ 1 \end{bmatrix} \rightarrow \begin{bmatrix} -\sin\theta\\ \cos\theta \end{bmatrix}$$

you can check this by graphing the arms in the unit circle and see the coordinate after the rotation of angle $\theta$

So, our transformation matrix should be:

$$T(\pmb{x}) = \begin{bmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{bmatrix}$$

4-3 Assignment

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