Finding a $3 \times 3$ Matrix that maps points in $\mathbb{R}^3$ onto the a given line 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.
 A: 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.
A: 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.
A: 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 $(ax,bx,cx)$.
