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Give a $3 \times 3$ matrix that maps all points in $\mathbb{R}^3$ onto the line $[x,y,z] = t[a,b,c]$ and does not move the points that are on that line. Prove your matrix has these properties.

Can someone verify if I am doing this correctly?

I first find a matrix that takes the standard basis to a basis that has $\begin{bmatrix}a \\ b \\ c \end{bmatrix}$ in it:

$\begin{bmatrix} a & 0 & 0 \\ b & 1 & 0 \\ c & 0 & 1 \end{bmatrix} = A$

now I choose a matrix that projects $\mathbb{R^3}$ onto the given line:

$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = B$

Now I need to invert $A$ to go back to the standard basis and so $A^{-1} B A$ will project all of $\mathbb{R}^3$ onto the given line.

Multiplying that out:

$\begin{bmatrix}1/a & 0 & 0\\-b/a & 1 & 0 \\-c/a & 0 & 1\end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} a & 0 & 0 \\ b & 1 & 0 \\ c & 0 & 1 \end{bmatrix}$

$\begin{bmatrix} 1 & 0 & 0\\-b & 0 & 0 \\-c & 0 & 0 \end{bmatrix}$

To show that this maps points in $\mathbb{R}^3$ to $[x, y, z] = t[a, b, c]$

$\begin{bmatrix} 1 & 0 & 0\\-b & 0 & 0 \\-c & 0 & 0 \end{bmatrix}\begin{bmatrix} x \\ y\\z\end{bmatrix} = \begin{bmatrix}x \\ -bx \\-cx \end{bmatrix} = x\begin{bmatrix}1\\-b\\-c\end{bmatrix}$

I apologize if some of my explanations don't make sense. I am trying to solve this the way my tutor showed me but I may have misunderstood some of his explanations.

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  • $\begingroup$ The matrix above does not project onto the line $\operatorname{sp} \{ (a,b,c)^T \}$ unless $a=-1$. If you apply the matrix to the point $(a,b,c)^T$ you get $(a,-ba,-bc)$. $\endgroup$ – copper.hat Nov 21 '13 at 16:14
  • $\begingroup$ Also, you need $a \neq 0$ for the $A$ matrix to be invertible. $\endgroup$ – copper.hat Nov 21 '13 at 16:17
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You didn't invert correctly. You need a $1$ in the $(2,2)$ and $(3,3)$ positions. Also, if you multiply reverse the order (multiply by $A$ on left and $A^{-1}$ on right) you get $(x,bx/a,cx/a)$. This defines the same line $(ax,bx,cx)$.

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  • $\begingroup$ Oops. I don't think it affects my answer though does it? $\endgroup$ – user109609 Nov 21 '13 at 6:48
  • $\begingroup$ It doesn't. There is something awry with your method. I think I found your error. I'll edit it above. $\endgroup$ – Doc Nov 21 '13 at 6:51
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Since it maps all the vectors into directions of single vector, hence it must be rank 1; in particular following solution will work

$\left[\begin{array}{ccc} \gamma a & \alpha a & \beta a\\ \gamma b & \alpha b & \beta b\\ \gamma c & \alpha c & \beta c \end{array}\right]\left[\begin{array}{c} x\\ y\\ z \end{array}\right]=(\gamma x+\alpha y+\beta z)\left[\begin{array}{c} a\\ b\\ c \end{array}\right] $

Now you need to choose $\gamma\ ,\alpha \ \& \beta $ such that

$\left[\begin{array}{ccc} \gamma a & \alpha a & \beta a\\ \gamma b & \alpha b & \beta b\\ \gamma c & \alpha c & \beta c \end{array}\right]\left[\begin{array}{c} a\\ b\\ c \end{array}\right]=\left[\begin{array}{c} a\\ b\\ c \end{array}\right] $

Which should be straight forward to find.

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  • $\begingroup$ Following your method, I found $\alpha = 1/a$, $\gamma = 0$, $\beta = 0$. Is this correct? Also I'm not sure what to do with this result. $\endgroup$ – user109609 Nov 21 '13 at 6:51
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Here is a way, just project the point $x$ directly onto the line spanned by $p=(a,b,c)^T$.

Given $x$, solve $\min_\lambda \|x-\lambda p\|^2$ for $\lambda$. This gives $\|x-\lambda p\|^2 = \|x\|^2 + \lambda^2 \|p\|^2- 2 p^T x $. Differentiating and setting the derivative to zero gives $\lambda = \frac{1}{\|p\|^2} p^T x $. Hence the projection of $x$ onto the line is $\frac{1}{\|p\|^2} (p^T x) p$. Hence we have the projection $\Pi x = \frac{1}{\|p\|^2} (p^T x) p = \frac{1}{\|p\|^2} p(p^T x) = \frac{1}{\|p\|^2} pp^T x$, from which we see that $\Pi = \frac{1}{\|p\|^2} pp^T$.

Grinding through the details gives $\Pi = \frac{1}{a^2+b^2+c^2}\begin{bmatrix} a^2 & ab & ac \\ ab & b^2 & bc \\ ac & bc & c^2\end{bmatrix} $

This is the unique orthogonal projection onto the line.

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  • $\begingroup$ Don't we all know that the projection of $x$ on $\boldsymbol Rp$ is $\frac{\langle x,p\rightangle}{\|p\|^2}p$? $\endgroup$ – Michael Hoppe Nov 21 '13 at 12:13
  • $\begingroup$ @MichaelHoppe: I'm not sure I follow your comment - that is exactly what is in the answer above? $\endgroup$ – copper.hat Nov 21 '13 at 15:53
  • $\begingroup$ Of course, but your derivation of the projection is too complicated in my opinion. $\endgroup$ – Michael Hoppe Nov 21 '13 at 15:55

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