Rotating straight line through origin

Question

Q: Find a linear function $f: \mathbb{R}^2\to\mathbb{R}^2$ that rotates every straight line through the origin through a 45 degree angle. What does $f$ do to the line at $x+y=1$?

Ok so I'm struggling to pick up the concepts here and then I tend to overthink them. Would I be using the $v=tw+z$ line formula here where w would have something to do with the slope being $1$? I feel like I'm very wrong. Any help is appreciated.

Also, my professor didn't pick a book for this course so I have no real reference.

Answer 1

I would say "A linear map that rotates the plane $45^\circ$ around the origin" is a fine description on its own, as long as you also specify which direction the plane is rotated. If your teacher insists on a formula description, you could work your way from $f(1, 0) = \left(\frac{\sqrt2}2, \frac{\sqrt2}{2}\right)$ and $f(0,1) = \left(-\frac{\sqrt2}2, \frac{\sqrt2}2\right)$ (assuming positive, i.e. counterclockwise rotation), and by linearity conclude that $$ f(x, y) = \left((x-y)\frac{\sqrt2}2, (x+y)\frac{\sqrt2}2\right) $$ If your teacher goes so far as to require a matrix representation, well, just put $f(1, 0)$ and $f(0,1)$ (in that order) as the columns of a $2\times 2$ matrix. (For when you're starting on rotation matrices, you will recognize $\frac{\sqrt2}2$ as $\cos 45^\circ$ and $\pm \frac{\sqrt2}2$ as $\pm\sin 45^\circ$.) (In hindsight, that sentence would've made more sense for any angle other than $45^\circ$.)

As for what happens to the line, here are a few suggestions:

Solution 1: pick two points on the line (for instance $(0,1)$ and $(1, 0)$, or whichever other pair is your favourite), transform them according to $f$, and find the equation for the line going through the two new points.

Solution 2: your line is parametrised as $(0,1) + t(1, -1)$ for $t\in \Bbb R$. By linearity, we have $$ f((0,1) + t(1, -1)) = f(0,1) + t\cdot f(1, -1). $$

Solution 3: your line is at a $45^\circ$ angle, and its distance from the origin is $\frac{\sqrt2}{2}$. Rotating it $45^\circ$ around the origin would let it keep its distance from the origin, but now it's vertical / horizontal (depending on which direction $f$ rotates the plane).

Answer 2

There are different ways to model a straight line $L$ through the origin.

One way is using a direction vector $d=(d_1, d_2)$ (which can be any vector different from the null vector) and have the points of the line created by all endpoints of the scalar multiples of $d$: $$ L = \{ t d \mid t \in \mathbb{R} \} $$

In two dimensions we can describe the lines by a normal vector $n$ of the line, the endpoints of all vectors perpendicular to that normal vector form the line: \begin{align} L &= \{ u \mid u \perp n \} \\ &= \{ u \mid n \cdot u = 0 \} \\ &= \{ u = (x, y) \mid n_x x + n_y y = 0 \} \end{align}

The description as functions $$ y(x) = m x $$ is slightly less general, as a vertical line would need infinite slope.

A rotation of a vector in the plane by an angle $\alpha$ around the origin can be described by a rotation matrix $$ R_\alpha= \begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix} $$

Its first column is the rotated vector $(1,0)^T$, its second column the rotated vector $(0,1)^T$.

For the above descriptions it could be applied to a direction vector and its scalar multiples or to a normal vector of a line.

For $\alpha=45^\circ$ it would be $$ R = \begin{pmatrix} 1/\sqrt{2} & -1/\sqrt{2} \\ 1/\sqrt{2} & 1/\sqrt{2} \end{pmatrix} $$ and applied as $$ f(u) = R u \iff \\ f(x,y) = R \, (x,y)^T = ((x-y)/\sqrt{2}, (x+y)/\sqrt{2})^T $$ for any vector $u$.

The line $x + y = 1$ can be written as $$ (x, y)^T = (x, 1-x)^T = x(1,-1)^T + (0,1)^T \quad (x \in \mathbb{R}) $$ so it does not intersect the origin.

We can apply the rotation to those vectors: $$ f(x, 1-x) = ((x-(1-x))/\sqrt{2}, (x+(1-x))/\sqrt{2})^T \\ = ((2x-1)/\sqrt{2}, 1/\sqrt{2})^T \quad (x \in \mathbb{R}) \\ = (t, 1/\sqrt{2})^T \quad (t \in \mathbb{R}) $$ which is the horizontal line $y = 1/\sqrt{2}$.

The image above shows the line $x+y=1$ in red colour, and the rotated line in green colour.

Answer 3

Work in the complex plane.

By multiplying by $\sqrt{i}=\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}$ eight times we are equivalently multiplying by $(\sqrt{i})^8=1$ with no excess in rotation. Because we get back to where we started (as $1$ is the multiplicative identity) this is equivalent to a $360$ degree rotation about the origin. Hence multiplying by $\sqrt{i}$ is equivalent to rotating $\frac{360}{8}=45$ degrees. So to rotate the "point" $x+yi$ simply multiply by $(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})$.

$$(x+yi)(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})$$

$$=(x-y)\frac{\sqrt2}2+(x+y)\frac{\sqrt2}2 i$$

This tells us to replace $x$ by $(x-y)\frac{\sqrt2}2$ the real part of our number. And replace $y$ with $(x+y)\frac{\sqrt2}2$, the imaginary part of our number. This is going to rotate the point $45$ degrees so,

$$f(x,y)=\left((x-y)\frac{\sqrt2}2, (x+y)\frac{\sqrt2}2\right)$$

Note $(\frac{\sqrt2}2+i \frac{\sqrt2}2)(1+0i)=(\frac{\sqrt2}2+i \frac{\sqrt2}2)$ so this is counterclockwise. In particular the value we used for $\sqrt{i}=e^{i \frac{\pi}{4}}$. To rotate clockwise first rotate clockwise $90$ degrees to get $(x,y) \to (y,-x)$ then rotate $45$ counterclockwise to get $f(y,-x)$. Or instead multiply $x+yi$ by $e^{-\frac{\pi}{4}i}$.