Science talk:Math Exam Resources/Courses/MATH152/April 2016/Question A 21

According to the lecture note,

http://www.math.ubc.ca/~elyse/152/2018/7TransformationsRandomWalks.pdf

I think students also understand geometric picture of the projection operator. Actually, if you think this problems geometrically, it is easier to get the solution because you don't need to compute the projection matrix. So it would be great if anyone add a new solution based on this.

Hello,

The matrix in Hint 2 is correct (my bad :-( ). However, it appears that it would be hard for a student to figure out how to derive the matrix in Hint 2. Perhaps an easier way (assuming they have covered change of basis matrices) is to simply state that under the change of bases (conjugation by an appropriate rotation), the two matrices become

${\begin{pmatrix}1&0\\0&0\end{pmatrix}}$

and

${\begin{pmatrix}0&0\\0&1\end{pmatrix}}$

Best, Brian

Hello,

The formula given in hint 2 is incorrect. The first column of the matrix is correct, but the second column is not (permute the coordinates of the first column, and after that, take the negative of the first coordinate).

Best,

Brian

This is for MATH152/April 2016/Question A 21