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Find the standard matrix of the given linear transformation from ${\bf R}^2$ to ${\bf R}^2$

Projection onto the line $y=2x$

So basically, I got the standard matrix to be $$ \left( \begin{array}{cc} 1 & 0 \\ 2 & 0 \\ \end{array} \right), $$ my question is: is this actually right? I have been looking online and it appears that maybe I am wrong?

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  • $\begingroup$ You can always try a simple test case and see if you get the right answer. For instance, if you were to project $(0,1)$ onto the this line, would the correct answer be $(0,0)$ as suggested by your solution? $\endgroup$ – AnonSubmitter85 Apr 6 '14 at 0:37
  • $\begingroup$ Yeah it should be, since 2(0)=0, therfore it would be (0,0) $\endgroup$ – A A Apr 6 '14 at 0:41
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    $\begingroup$ I think you are mistaken. The projection of $(0,1)$ onto this line is definitely not $(0,0)$. $\endgroup$ – AnonSubmitter85 Apr 6 '14 at 1:54
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The line has direction vector $v = (1,2)$. The projection of $(x,y) \in {\bf R}$ onto the line is given by $$ proj_v(x,y) = \left(\frac{(x,y)\cdot v}{v\cdot v}\right) v = \frac{x + 2y}{5}v. $$ The standard matrix for this linear map is thus $$ [proj_v(1,0)' \ \ proj_v(0,1)'] = \left[ \begin{array}{cc} 1/5 & 2/5 \\ 2/5 & 4/5 \\ \end{array} \right] = \frac{1}{5}\left[ \begin{array}{cc} 1 & 2 \\ 2 & 4 \\ \end{array} \right]. $$

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  • $\begingroup$ How did the last step work? Why did you do proju(1,0) proju(0,1), is there a reason you used (1,0) and (0,1)? $\endgroup$ – A A Apr 6 '14 at 0:50
  • $\begingroup$ The standard matrix of any linear transformation $T$ (w.r.t. to the standard basis) is given by taking $T(1,0)$ as the first column and $T(0,1)$ as the second column. $\endgroup$ – user139388 Apr 6 '14 at 0:52
  • $\begingroup$ How did you find the direction vector of the line? $\endgroup$ – A A Apr 6 '14 at 0:59
  • $\begingroup$ Since we are talking about the line $y=2x$, for a "run" of $1$ you need a "rise" of $2$. The vector $(1,2)$ achieves this (notice we could have used $(2,4)$ also, for instance). $\endgroup$ – user139388 Apr 6 '14 at 1:05

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