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T is a reflection on the $x$ axis and after a reflection on the line $y = x$
Show that T is a rotation and give the angle.

So my matrix of transformation would be $$\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$$

and I know that a matrix of rotation is $$\begin{bmatrix} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x) \end{bmatrix}$$

By logic I know that the angle would be $90^\circ$, but there is an approach of how can I demonstrate it? And how to show that T is a rotation?

Thank you

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When linear transformations are expressed through the same matrix, they are the same. In your case, you showed that the matrix $$ A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} $$ corresponds to

  1. reflection around the $x$-axis followed by reflection around $y=x$;
  2. rotation of $90^\circ$.

Therefore these transformations are identical.

If you are asking about why these come out the same, think about what happens to the point $(x,y)$ under each one geometrically. The first way flips it around $x$ axis to map $(x,y) \mapsto (x,-y)$. Then you flip around $x=y$ which basically exchanges the coordinates, you your final map is $$ (x,y) \mapsto (x, -y) \mapsto (-y, x). $$ Meanwhile, directly rotating $(x,y)$ by $90^\circ$ clockwise does exactly the same thing. To convince yourself of that, note that you map $$ (x,0) \mapsto (0,x) \text{ and } (0,y) \mapsto (-y, 0) $$ and since the transformation is linear, $$ T(x,y) = T(x,0) + T(0,y) = (0,x) + (-y,0) = (-y,x) $$ as desired.

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