2
$\begingroup$

Suppose that the transformation T from $\mathbb R^{2}$ to $\mathbb R^{2}$ rotates 30 degrees counterclockwise about (5,5). For which $(c,d)$ is the transformation $T(x-c,y-d)$ linear?

Solution (My attempt or Ideas) I know linear transformations in $\mathbb R^{2}$ are represented as matrices. So first, I attempt to find the rotation matrix that rotates space counterclockwise by 30 degrees about the point $(5,5)$.

First, I translate the origin to the point and get:= \begin{bmatrix}1&0&5\\ 0& 1&5\\0&0&1\end{bmatrix} Then, now I rotate space counterclockwise by 30 degrees. and get: \begin{bmatrix}\sqrt(3)/2&-1/2&0\\ 1/2& \sqrt(3)/2&0\\0&0&1\end{bmatrix} Now, I translate back and get \begin{bmatrix}1&0&-5\\ 0& 1&-5\\0&0&1\end{bmatrix}

Multiplying all three matrices together, I get: (multiplying the first matrix I wrote* second matrix * third :\begin{bmatrix}\sqrt(3)/2&-1/2&1/2(15-5\sqrt(3))\\ 1/2& \sqrt(3)/2&1/2(5-5\sqrt(3))\\0&0&1\end{bmatrix}.

So now that I have this (hopefully this is correct), I'm not sure how exactly to proceed to find the (c,d) such that the transformation $T(x-c,y-d)$ is linear. Should I use (does it have something to do with the properties) the properties of linear transformation, i.e.: $T(v_1+v_2)$=$T(v_1)$+$T(v_2)$ and $T(av)=aT(v)$? Also, I'm slightly confused (my linear algebra is rusty), but I thought linear transformations from $\mathbb R^{2}$ to $\mathbb R^{2}$ should be represented as 2 x 2 matrices, but what I have above is a 3 x 3 matrix. This makes me worried that I am thinking of the question incorrectly.

Any help would be much appreciated. Thank you very much.

$\endgroup$
2
$\begingroup$

Ashwin's method is great for working out all the details, but I thought I'd mention a low-down sneaky short cut to the solution.

One thing we know about linear transformations is that they must map the origin to the origin, i.e. $$ (0,0) = T(0-c, 0-d) = T(-c,-d). $$ So you could just find the point that $T$ maps to $(0.0)$ -- say by geometric methods and a bit of trigonometry -- and then take its negative to find $c$ and $d$.

The big problem with this approach is that, while it tells you what the values of $c$ and $d$ must be for the map to be linear, it doesn't establish that $(x,y) \mapsto T(x-c, y-d)$ actually is linear. It looks like your question includes showing that the resulting "shifted $T$" map is linear so this method isn't enough, but you might want to see what result it gives just to check your longer computations against this answer.

$\endgroup$
  • $\begingroup$ Great, thanks. So If I find T(x,y) to be( ((sqrt(3)/2)*(x-5)-1/2*(y-5)+5, 1/2(x-5)+(sqrt(3/2))*(y-5)+5). Do I just set the first and second components equal to 0 and solve the system of equations for x and y? $\endgroup$ – kemb Mar 19 '18 at 6:15
  • $\begingroup$ Yes, that will work. I had suggested using geometry to solve this, but since you've got an algebraic expression for $T$ you can do as you described. Just remember that the $x$ and $y$ that you find that map to $(0.0)$ are the negatives of the values of $c$ and $d$ that you want. $\endgroup$ – JonathanZ Mar 19 '18 at 6:27
  • $\begingroup$ Got it, thank you so much. $\endgroup$ – kemb Mar 19 '18 at 6:28
  • $\begingroup$ And once you've found $c$ and $d$ you need to substitute $x-c$ for $x$ and $y-d$ for $y$ in your expression for $T$ and verify (after you've simplified it) that the resulting transformation is linear. Do you know how to recognize a linear map? (You said your linear algebra is rusty.) $\endgroup$ – JonathanZ Mar 19 '18 at 6:33
  • $\begingroup$ Do I show that it is linear using the two properties that I highlighted T(v1+v2)=T(v1)+T(v2), T(kv1)=kT(v1) (i.e. check that these two properties are satisfied)? $\endgroup$ – kemb Mar 19 '18 at 6:35
2
$\begingroup$

What you have so far indeed does not work -- you have the immediate issue that transformations of $\mathbb{R}^2$ are represented as $2\times 2$ matrices, not $3\times 3$. Moreover, you can't represent the "shift" from $(0,0)$ to $(5,5)$ as a $2\times 2$ matrix -- any linear transformation will send $(0,0)$ to $(0,0)$.

Let's instead think of a function of two variables. We have the first translation function, which is $f(x,y)=(x-5,y-5)$. This takes $(5,5)$ to $(0,0)$, where we know how to rotate it. The second is the rotation matrix. Normally we write it like this: $\frac{1}{2}\begin{pmatrix} \sqrt 3 & -1 \\ 1 & \sqrt 3\end{pmatrix}$ but it's easier to think of it as a function $A$: $A(x,y)=(\frac{\sqrt 3}{2} x-\frac{1}{2}y, \frac{1}{2}x+\frac{\sqrt 3}{2}y)$. Now we have the inverse function $f^{-1}(x,y)=(x+5,y+5)$, which moves everything back to $(5,5)$.

So we can compose all of these and understand $T=f^{-1}\circ A\circ f$. You can check if a function $\phi$ is a linear function by seeing if $\phi(u_1+v_1,u_2+v_2)=\phi(u_1,u_2)+\phi(v_1,v_2)$, and $\phi(ku_1,ku_2)=k\phi(u_1,u_2)$. One of the natural consequences of this is that $\phi(0,0)=0$. From here, you should be able to set up a system of equations for $c$ and $d$ to see which $c$ and $d$ satisfy these properties.

$\endgroup$
  • $\begingroup$ Thanks, this really helps. I'm however, not sure of a couple of things. Does our system of equations for c and d explicitly come from the properties highlighted, i.e.:$T(u_1+v_1,u_2+v_2)=T(u_1,u_2)+T(v_1,v_2)$, and $T(ku_1,ku_2)=kT(u_1,u_2)$? I'm not exactly sure how to set up a system. Thanks. $\endgroup$ – kemb Mar 19 '18 at 5:28
  • $\begingroup$ Yes. My notation is unclear here -- I'll edit it. I think that a good way to think about this is that $T$, the rotation, is $f^{-1}\circ A\circ f$, and then we have a function $\phi_{c,d}(x,y)=T(x-c,y-d)$. So you want to find the values $c,d$ such that $\phi_{c,d}$ has the linearity properties. $\endgroup$ – Ashwin Trisal Mar 19 '18 at 5:38
  • $\begingroup$ I sort of get it. So by composing, is our T, which is $f^-1 o A o f$ equal to $T(x,y)= (sqrt(3)/2) *(x-5)-1/2*(y-5) +5, 1/2(x-5)+ (sqrt(3)/2)*(y-5) +5)$? $\endgroup$ – kemb Mar 19 '18 at 5:53
  • $\begingroup$ Yes, that looks right to me. $\endgroup$ – Ashwin Trisal Mar 19 '18 at 5:57
  • $\begingroup$ So now to set up the equations for linearity do I do: T(x-c,y-d)=T(x,y)+T(-c,-d) and T(k(x-c),k(y-d))=kT(x-c,y-d)? I'm not too sure, thank you. $\endgroup$ – kemb Mar 19 '18 at 6:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.