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.


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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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