The real numbers (any numbers on the number line) are an example of a vector space.

The real numbers are a one-dimensional vector space.

The set of all vectors that look like \(\begin{bmatrix} x\\y \end{bmatrix}\) form a two-dimensional space.

The key is to think of the function as transforming the entire space at once, instead of one each number at a time.

Example:

\(f(x) = 3x\) multiplies every number we plug into it by 3. It*stretches* the whole space by a factor of 3.

\(f(x) = 3x\) multiplies every number we plug into it by 3. It

Example:

\(f\left( \begin{bmatrix}x\\y\end{bmatrix} \right) = \begin{bmatrix}2x\\y\end{bmatrix}\) stretches the entire plane horizontally by a factor of 2. Any images in the plane are stretched too!

\(f\left( \begin{bmatrix}x\\y\end{bmatrix} \right) = \begin{bmatrix}2x\\y\end{bmatrix}\) stretches the entire plane horizontally by a factor of 2. Any images in the plane are stretched too!

Any linear transformation can be represented by a matrix.

That is, we need $$ f\left( \begin{bmatrix}0\\0\end{bmatrix} \right) = \begin{bmatrix}0\\0\end{bmatrix} $$

Any linear transformation can be represented by a matrix.

Example:

\(f\left( \begin{bmatrix}x\\y\end{bmatrix} \right) = \begin{bmatrix}2x\\y\end{bmatrix}\) is a linear transformation, because $$ f\left( \begin{bmatrix}0\\0\end{bmatrix} \right) = \begin{bmatrix}2(0)\\0\end{bmatrix}= \begin{bmatrix}0\\0\end{bmatrix} $$

\(f\left( \begin{bmatrix}x\\y\end{bmatrix} \right) = \begin{bmatrix}2x\\y\end{bmatrix}\) is a linear transformation, because $$ f\left( \begin{bmatrix}0\\0\end{bmatrix} \right) = \begin{bmatrix}2(0)\\0\end{bmatrix}= \begin{bmatrix}0\\0\end{bmatrix} $$

Any linear transformation can be represented by a matrix.

Example:

\(f\left( \begin{bmatrix}x\\y\end{bmatrix} \right) = \begin{bmatrix}x+2\\y\end{bmatrix}\) is NOT a linear transformation, because $$ f\left( \begin{bmatrix}0\\0\end{bmatrix} \right) = \begin{bmatrix}0+2\\0\end{bmatrix}= \begin{bmatrix}2\\0\end{bmatrix} $$

\(f\left( \begin{bmatrix}x\\y\end{bmatrix} \right) = \begin{bmatrix}x+2\\y\end{bmatrix}\) is NOT a linear transformation, because $$ f\left( \begin{bmatrix}0\\0\end{bmatrix} \right) = \begin{bmatrix}0+2\\0\end{bmatrix}= \begin{bmatrix}2\\0\end{bmatrix} $$

The results give us the first and second column of our matrix, respectively.

Example: We'll find the matrix of \(f\left( \begin{bmatrix}x\\y\end{bmatrix} \right) = \begin{bmatrix}2x\\y\end{bmatrix}\)

First find where the vectors \([1,0]\) and \([0,1]\) are sent to: $$ {\scriptsize f\left( \begin{bmatrix}1\\0\end{bmatrix} \right) = \begin{bmatrix}2(1)\\0\end{bmatrix}= \begin{bmatrix}2\\0\end{bmatrix} \ \mbox{ and } \ \ f\left( \begin{bmatrix}0\\1\end{bmatrix} \right) = \begin{bmatrix}2(0)\\1\end{bmatrix}= \begin{bmatrix}0\\1\end{bmatrix} }$$ then build the matrix: $$ \begin{bmatrix}2&0\\0&1\end{bmatrix}$$

First find where the vectors \([1,0]\) and \([0,1]\) are sent to: $$ {\scriptsize f\left( \begin{bmatrix}1\\0\end{bmatrix} \right) = \begin{bmatrix}2(1)\\0\end{bmatrix}= \begin{bmatrix}2\\0\end{bmatrix} \ \mbox{ and } \ \ f\left( \begin{bmatrix}0\\1\end{bmatrix} \right) = \begin{bmatrix}2(0)\\1\end{bmatrix}= \begin{bmatrix}0\\1\end{bmatrix} }$$ then build the matrix: $$ \begin{bmatrix}2&0\\0&1\end{bmatrix}$$

Example: What does this transformation do?
$$ \begin{bmatrix}2&0\\0&1\end{bmatrix}$$

The x-coordinate of every point is moved twice as far away from the origin as it was. The y-coordinate doesn't change.

Example: The whole plane is stretched out, horizontally, by a factor of 2:
$$ \begin{bmatrix}2&0\\0&1\end{bmatrix}$$

Example: The stretching affects everything in the plane, too:
$$ \begin{bmatrix}2&0\\0&1\end{bmatrix}$$

Example: We'll find the matrix of \(f\left( \begin{bmatrix}x\\y\end{bmatrix} \right) = \begin{bmatrix}2x+y\\2y\end{bmatrix}\)

First find where the vectors \([1,0]\) and \([0,1]\) are sent to: $$ {\tiny f\left( \begin{bmatrix}1\\0\end{bmatrix} \right) = \begin{bmatrix}2(1) + 0 \\2(0) \end{bmatrix}= \begin{bmatrix}2\\0\end{bmatrix} \ \mbox{ and } \ \ f\left( \begin{bmatrix}0\\1\end{bmatrix} \right) = \begin{bmatrix}2(0) + 1\\2(1)\end{bmatrix}= \begin{bmatrix}1\\2\end{bmatrix} }$$ then build the matrix: $$ \begin{bmatrix}2&1\\0&2\end{bmatrix}$$

First find where the vectors \([1,0]\) and \([0,1]\) are sent to: $$ {\tiny f\left( \begin{bmatrix}1\\0\end{bmatrix} \right) = \begin{bmatrix}2(1) + 0 \\2(0) \end{bmatrix}= \begin{bmatrix}2\\0\end{bmatrix} \ \mbox{ and } \ \ f\left( \begin{bmatrix}0\\1\end{bmatrix} \right) = \begin{bmatrix}2(0) + 1\\2(1)\end{bmatrix}= \begin{bmatrix}1\\2\end{bmatrix} }$$ then build the matrix: $$ \begin{bmatrix}2&1\\0&2\end{bmatrix}$$

Example: This matrix does something a little more complicated. It stretches and 'shears'.
$$ \begin{bmatrix}2&1\\0&2\end{bmatrix}$$

Example: We have an image, and want to reflect it (across a horizontal line) so it is upside down.

Example: Picture the image in the plane, along the horizontal axis. We want to flip the image over this axis.

Example: What happens to the points/vectors \(\color{blue}{(1,0)}\) and \(\color{red}{(0,1)}\) when we flip over the horizontal axis?

\(\color{blue}{(1,0)}\) doesn't move, but \(\color{red}{(0,1)}\) ends up at \((0,-1)\)

Example: What happens to the points/vectors \(\color{blue}{(1,0)}\) and \(\color{red}{(0,1)}\) when we flip over the horizontal axis?

\(\color{blue}{(1,0)}\) doesn't move, but \(\color{red}{(0,1)}\) ends up at \((0,-1)\)

So the function of our reflection does this:
$$ {\scriptsize f\left( \begin{bmatrix}1\\0\end{bmatrix} \right) = \begin{bmatrix}1\\0\end{bmatrix} \ \mbox{ and } \ \ f\left( \begin{bmatrix}0\\1\end{bmatrix} \right) = \begin{bmatrix}0\\-1\end{bmatrix} }$$
and our matrix is
$$ \begin{bmatrix}1&0 \\ 0&-1\end{bmatrix}$$
Example: Let's rotate the triangle by 90 degrees (quarter turn) counterclockwise:

Example: Again, see where the vectors \(\color{blue}{(1,0)}\) and \(\color{red}{(0,1)}\) go:

Example: Again, see where the vectors \(\color{blue}{(1,0)}\) and \(\color{red}{(0,1)}\) go.

This time, they both changed: $$ {\scriptsize f\left( \begin{bmatrix}1\\0\end{bmatrix} \right) = \begin{bmatrix}0\\1\end{bmatrix} \ \mbox{ and } \ \ f\left( \begin{bmatrix}0\\1\end{bmatrix} \right) = \begin{bmatrix}-1\\0\end{bmatrix} }$$ so our matrix is $$ \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}$$

This time, they both changed: $$ {\scriptsize f\left( \begin{bmatrix}1\\0\end{bmatrix} \right) = \begin{bmatrix}0\\1\end{bmatrix} \ \mbox{ and } \ \ f\left( \begin{bmatrix}0\\1\end{bmatrix} \right) = \begin{bmatrix}-1\\0\end{bmatrix} }$$ so our matrix is $$ \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}$$

$$ \begin{bmatrix}a&0\\0&1\end{bmatrix}$$

will stretch everything horizontally by a factor of \(a\).

$$ \begin{bmatrix}1&0\\0&b\end{bmatrix}$$

will stretch everything vertically by a factor of \(b\).

$$ \begin{bmatrix}a&0\\0&b\end{bmatrix}$$

will stretch everything horizontally by a factor of \(a\) and vertically by a factor of \(b\).

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

will rotate the plane (around the origin) by an angle of \(\theta\), counterclockwise.

$$ \begin{bmatrix}-1&0\\0&1\end{bmatrix}$$

will reflect the plane over the y-axis.

$$ \begin{bmatrix}1&0\\0&-1\end{bmatrix}$$

will reflect the plane over the x-axis.

$$ \begin{bmatrix}-1&0\\0&-1\end{bmatrix}$$

will reflect the plane over the x-axis and y-axis... hey, that's a rotation!

(In fact, any two reflections, over any lines, will be some rotation).

$$ \begin{bmatrix}-1&0\\0&-1\end{bmatrix}$$

will reflect the plane over the x-axis and y-axis... hey, that's a rotation!

(In fact, any two reflections, over any lines, will be some rotation).

$$ \begin{bmatrix}0&0\\0&1\end{bmatrix}$$

will project the plane onto the y-axis. It sets all x-coordinates to 0, and keeps y-coordinates where they were.

$$ \begin{bmatrix}1&0\\0&0\end{bmatrix}$$

will project the plane onto the x-axis. It sets all y-coordinates to 0, and keeps x-coordinates where they were.

$$ \begin{bmatrix}0.5&0.5\\0.5&0.5\end{bmatrix}$$

will project the plane onto the line $y=x$, the diagonal line through the origin at an angle of \(45^\circ\) to the x-axis.

$$ \begin{bmatrix}1&a\\0&1\end{bmatrix}$$

will shear horizontally by a factor of \(a\).

$$ \begin{bmatrix}1&0\\a&1\end{bmatrix}$$

will shear vertically by a factor of \(a\).

Example:
$$ \begin{bmatrix}1 & 2\\3 & 6\end{bmatrix} $$
is a projection, since
$$ (1)(6) - (2)(3) = 0$$

Example:
$$ \begin{bmatrix}1 & 3\\3 & 6\end{bmatrix} $$
is NOT a projection, since
$$ (1)(6) - (3)(3) = -3 \neq 0$$

The Linear Transformation Applet