Find the standard matrix for the composition of the following two linear operators on $\Bbb R^2$: A reflection about the line $y = x$, followed by a rotation counterclockwise of $60^\circ$.

I'm not sure if it's the standard matrix of reflection about y=x multiply by the standard matrix of rotation,and then plug in θ=60°.

This is my work so far:

$$ T(x, y) = (y, x)\\ T(\vec e_1) = T(1, 0) = (0, 1)\\ T(\vec e_2) = T(0, 1) = (1, 0) $$so the standard matrix for the reflection transformation is: $\left[\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right]$

The standard matrix for rotation transformation: $\left[\begin{smallmatrix}\cos \theta & \,-\sin\theta\\\sin\theta & \cos\theta\end{smallmatrix}\right]$ $$ \left[\begin{matrix}0&1\\1&0\end{matrix}\right]\left[\begin{matrix}\cos \theta & \,-\sin\theta\\\sin\theta & \cos\theta\end{matrix}\right] = \left[\begin{matrix}\sin \theta & \cos\theta\\\cos\theta & \,-\sin\theta\end{matrix}\right] $$ Let the matrix $A$ be the standard matrix after two transformations $$ A = \left[\begin{matrix}\cos 60^\circ & \,-\sin60^\circ\\\sin60^\circ & \cos60^\circ\end{matrix}\right] = \begin{bmatrix}\frac{1}{2} & -\frac{\sqrt3}{2}\\\frac{\sqrt{3}}{2} & \frac12\end{bmatrix} $$

  • 1
    $\begingroup$ Seems to be OK. $\endgroup$
    – zoli
    Jun 22, 2015 at 6:31

1 Answer 1


There are two mistakes that I can see. The most important one is this: If you first want to apply trasformation $T$ and then transformation $R$, the matrix for the total transformation isn't $TR$, it's $RT$. The reason is that given a vector $\vec v$, you first apply apply $T$ to get the vector $T\vec v$. Then you apply $R$ to this vector, and get $R(T\vec v) = (RT)\vec v$. So you need to do the matrix multiplication the other way (first reflection, then rotation means rotation times reflection).

The other, and more minor, mistake is that when calculating $A$, you've inserted the rotation matrix, not the resulting product matrix. So the $A$ you have is the matrix for rotating $60^\circ$, and nothing more.

  • $\begingroup$ Oh,I see! And is the answer :[cos60° -sin60° sin60° cos60°]×[0 1 1 0]=[-3^(1/2)/2 1/2 1/2 3^(1/2)/2] (enter from left to right,from up to down)? $\endgroup$
    – JimfyWinsy
    Jun 22, 2015 at 7:43
  • $\begingroup$ @JimfyWinsy $\left[ \begin{smallmatrix}-\frac{\sqrt3}{2} & \frac{1}{2}\\\frac{1}{2} & \frac{\sqrt3}{2}\end{smallmatrix} \right]$ is what I got too. $\endgroup$
    – Arthur
    Jun 22, 2015 at 7:53

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