0
$\begingroup$

Let T : R 3 → R 3 be the linear transformation corresponding to rotating by π/4 clockwise around the z-axis, and then reflecting over the x-axis. Since T is a linear transformation, it corresponds to left multiplication by some matrix A.

I am familiar with linear transformation problems that give you specific vectors, but this one doesn't. I know I have to use the standard basis vectors to determine the transformed vectors which make up the final matrix. I just don't know how to go about doing that though. Thanks

$\endgroup$
  • 2
    $\begingroup$ Find out how $T$ affects the standard basis. $\endgroup$ – copper.hat Oct 12 '17 at 0:00
  • $\begingroup$ ...and put the result as columns into the matrix $A$. Done. $\endgroup$ – amsmath Oct 12 '17 at 0:01
  • $\begingroup$ So since it is being rotated around the z axis, would the third standard basis vector be transformed? Like: [0, 0, 1] ---> [0, 0, pi/4] $\endgroup$ – Jennifer Hall Oct 12 '17 at 0:01
  • $\begingroup$ No. Rotation leaves it invariant. The reflection then sends it to [0,0,-1]. $\endgroup$ – amsmath Oct 12 '17 at 0:03
  • $\begingroup$ So would the matrix A just be the identity matrix with -1 in the bottom right? That seems too simple to me. @amsmath $\endgroup$ – Jennifer Hall Oct 12 '17 at 0:13
0
$\begingroup$

A linear transformation "corresponds to left multiplication by a matrix" in a given basis. That is why copper hat refers to the "standard basis". <1, 0, 0> rotated $\pi/4$ radians around the z axis becomes $<\sqrt{2}/2, \sqrt{2}/2, 0>$. Then reflecting over the x-axis it becomes $<\sqrt{2}/2, -\sqrt{2}/2, 0>$. <0, 1, 0> rotates to $<-\sqrt{2}/2, \sqrt{2}/2, 0>$. Then reflecting over the x axis, it becomes $<-\sqrt{2}/2, -\sqrt{2}/2, 0>$. Finally, <0, 0, 1> which is on the z-axis "rotates" to itself, <0, 0, 1>, but then reflecting in the x-axis it becomes <0, 0, -1>.

As amsmath suggests, use those vectors as columns in the matrix.$

$\endgroup$
  • $\begingroup$ So the reason why you took the solution of cos(pi/4) instead of just pi/4 is because we are looking for the distance of the vector? $\endgroup$ – Jennifer Hall Oct 12 '17 at 0:23

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.