  ## Exercises - Linear Transformations

1. Suppose $M$ is a linear transformation operating on 2-dimensional vectors.

Appealing to the linearity of $M$, we see that

\begin{align} M\begin{pmatrix}-6\\8\end{pmatrix} &= M \left( \begin{pmatrix}-6\\0\end{pmatrix} + \begin{pmatrix}0\\8\end{pmatrix} \right) \\ &= M\begin{pmatrix}-6\\0\end{pmatrix} + M\begin{pmatrix}0\\8\end{pmatrix} \\ &= M\left( -6 \begin{pmatrix}1\\0\end{pmatrix} \right) + M \left( 8 \begin{pmatrix}0\\1\end{pmatrix} \right) \\ &= -6 M\begin{pmatrix}1\\0\end{pmatrix} + 8 M \begin{pmatrix}0\\1\end{pmatrix} \\ &= -6 \begin{pmatrix}4\\5\end{pmatrix} + 8 \begin{pmatrix}-3\\7\end{pmatrix} \\ &= \begin{pmatrix}(-6) \cdot 4\\ (-6) \cdot 5\end{pmatrix} + \begin{pmatrix} 8 \cdot (-3)\\8 \cdot 7\end{pmatrix} \\ &= \begin{pmatrix}(-6) \cdot 4 + 8 \cdot (-3)\\ (-6) \cdot 5 + 8 \cdot 7\end{pmatrix} \\ &= \begin{pmatrix}-48\\26\end{pmatrix} \end{align}

We can apply this same process to find the output of any linear transformation that operates on two-dimensional vectors, provided we know what the linear transformation does to $\begin{pmatrix}1\\0\end{pmatrix}$ and $\begin{pmatrix}0\\1\end{pmatrix}$.

As such, we can abbreviate the process above by writing $M$ as a matrix, and evaluating $M\begin{pmatrix}-6\\8\end{pmatrix}$ in the following way:

$$M\begin{pmatrix}-6\\8\end{pmatrix} = \begin{bmatrix}4 & -3\\5 & 7\end{bmatrix} \begin{pmatrix}-6\\8\end{pmatrix} = \begin{pmatrix}4 \cdot (-6) + (-3) \cdot 8\\5 \cdot (-6) + 7 \cdot 8\end{pmatrix} = \begin{pmatrix}-48\\26\end{pmatrix}$$

2. We know that we can find the matrix form for a linear transformation, $T$, on two-dimensional vectors, if we know the output of $T\begin{pmatrix}1\\0\end{pmatrix}$ and $T\begin{pmatrix}0\\1\end{pmatrix}$. What if we know the output of $T$ when applied to other vectors? Can we still find the matrix form of $T$? Find the matrix form for $T$ if the below facts are known:

$$T \begin{pmatrix}2\\0\end{pmatrix} = \begin{pmatrix}14\\20\end{pmatrix} \quad \quad \textrm{and} \quad \quad T \begin{pmatrix}3\\1\end{pmatrix} = \begin{pmatrix}10\\11\end{pmatrix}$$

3. Show that if $F$ and $G$ are linear transformations that operate on two-dimensional vectors, then the matrix representation of $F-G$ is given by the following:

$$\begin{bmatrix}a & b\\c & d\end{bmatrix}- \begin{bmatrix}e & f\\g & h\end{bmatrix} = \begin{bmatrix}a-e & b-f\\c-g & d-h\end{bmatrix}$$

4. Suppose $F$ is a linear transformation that operates on three-dimensional vectors. Let us adopt the convention of writing such transformations $F$ as matrices in the following form

$$F = \left[ \begin{array}{ccc} a & d & g\\ b & e & h\\ c & f & i \end{array} \right]$$

to imply that

$$F \begin{pmatrix}1\\0\\0\end{pmatrix} = \begin{pmatrix}a\\b\\c\end{pmatrix}, \quad F \begin{pmatrix}0\\1\\0\end{pmatrix} = \begin{pmatrix}d\\e\\f\end{pmatrix}, \quad \textrm{ and } \quad F \begin{pmatrix}0\\0\\1\end{pmatrix} = \begin{pmatrix}g\\h\\i\end{pmatrix}$$

1. Use the linearity of $F$ to determine $F\begin{pmatrix}x\\y\\z\end{pmatrix}$.

2. If $F$ and $G$ are the following linear transformations that operate on three-dimensional vectors, find the matrix representation of their sum and composition.

$$F = \left[ \begin{array}{ccc} a & d & g\\ b & e & h\\ c & f & i \end{array} \right] \quad, \quad G = \left[ \begin{array}{ccc} j & m & p\\ k & n & q\\ l & o & r \end{array} \right]$$

3. Use the matrix representations found above to find the matrix representation of the following two linear transformations.

$$\left[ \begin{array}{ccc} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{array} \right] + \left[ \begin{array}{ccc} 1 & -1 & 2\\ 2 & -3 & 5\\ -1 & 2 & 7 \end{array} \right]$$
$$\left[ \begin{array}{ccc} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{array} \right] \circ \left[ \begin{array}{ccc} 1 & -1 & 2\\ 2 & -3 & 5\\ -1 & 2 & 7 \end{array} \right]$$

5. We should convince ourselves that rotation by $\theta$ degrees is indeed a linear transformation before proceeding. Do both properties of a linear transformation hold? Recall that $F$ is a linear transformation (operating on vectors) if and only if, for all scalars $c$ and vectors $\bar{x}$ and $\bar{y}$ we have:

1. $F(c\bar{x}) = cF(\bar{x})$
2. $F(\bar{x}+\bar{y}) = F(\bar{x}) + F(\bar{y})$

1. If we stretch a vector $\bar{x}$ by a factor of $c$ first, and then rotate it clockwise by $\theta^{\circ}$, is the result the same as rotating the vector $\bar{x}$ first and then stretching it by a factor of $c$?

2. If we add two vectors $\bar{x}$ and $\bar{y}$ (in the normal way, "head-to-tail") and then rotate the vector representing their sum, is the result the same as first rotating the individual vectors $\bar{x}$ and $\bar{y}$ and then adding them together?

Geometrically, we should be able to quickly say "yes" to both questions. As such, we can now focus our attention on finding a matrix form for such a linear transformation. Suppose we wish our matrix to rotate vectors by $25^{\circ}$ counter-clockwise. Recall, that the first and second columns of the matrix form for a linear transformation (on 2-dimensional vectors) indicate what that transformation does to the vectors $\begin{pmatrix}1\\0\end{pmatrix}$ and $\begin{pmatrix}0\\1\end{pmatrix}$, respectively

If our transformation is a rotation counter-clockwise of 25 degrees, notice that

$$\begin{pmatrix}1\\0\end{pmatrix} \rightarrow \begin{pmatrix} \cos 25^\circ \\ \sin 25^\circ\end{pmatrix} \quad \textrm{and} \quad \begin{pmatrix}0\\1\end{pmatrix} \rightarrow \begin{pmatrix}-\sin 25^\circ\\ \cos 25^\circ \end{pmatrix}$$

As such, the matrix form we seek is:

$$\begin{bmatrix} \cos 25^\circ & -\sin 25^\circ\\ \sin 25^\circ& \cos 25^\circ\end{bmatrix}$$

Following similar logic, the matrix form for a counter-clockwise rotation by any angle $\theta$ is given by

$$\begin{bmatrix} \cos \theta & -\sin \theta\\ \sin \theta & \cos \theta \end{bmatrix}$$

With the rotation matrix found, rotating a particular vector is easy...

Suppose we wish to rotate the vector $\begin{pmatrix}3\\4\end{pmatrix}$ by $25^\circ$. We just find

$$\begin{bmatrix} \cos 25^\circ & -\sin 25^\circ\\ \sin 25^\circ& \cos 25^\circ\end{bmatrix} \begin{pmatrix}3\\4\end{pmatrix} = \begin{pmatrix}3\cos 25^\circ - 4 \sin 25^\circ \\ 3 \sin 25^\circ + 4 \cos 25^\circ \end{pmatrix} \approx \begin{pmatrix}1.028\\4.893\end{pmatrix}$$

Other rotations are found in a similar manner...