# Calculating a Matrix Given Linear Transformation

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

• Find out how $T$ affects the standard basis. – copper.hat Oct 12 '17 at 0:00
• ...and put the result as columns into the matrix $A$. Done. – amsmath Oct 12 '17 at 0:01
• 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] – Jennifer Hall Oct 12 '17 at 0:01
• No. Rotation leaves it invariant. The reflection then sends it to [0,0,-1]. – amsmath Oct 12 '17 at 0:03
• So would the matrix A just be the identity matrix with -1 in the bottom right? That seems too simple to me. @amsmath – Jennifer Hall Oct 12 '17 at 0:13

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>.