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

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## Contents

Thread title | Replies | Last modified |
---|---|---|

New solution based on geometric interpretation of projection | 0 | 07:09, 24 March 2018 |

Hint and solution correct - but possibly not suitable for first years | 0 | 02:47, 13 March 2018 |

Incorrect Hint and Solution | 1 | 00:01, 13 March 2018 |

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

and

Best, Brian