Finding matrix linear transformation Question:
Find the $3 \times 3$ matrix $ A$, associated with the linear transformation that projects vectors in $\mathbb{R^3}$ (orthogonally) onto the plane $x+y+z=0$.
I was given this question just as a review question for my class.  I took linear algebra over 2 years ago, so my memory is really fuzzy.
I was given the hints to find the matrix $A$ by thinking of it as a composition of a rotation, simpler projection, then another rotation.  
Thoughts:
I know the normal vector of the plane is $<1,1,1>$. I think this may be a useful fact.  Also after googling a bit I found that the matrix .  Perhaps, I could work with this, although I honestly don't really know exactly how to proceed. I know there are many projections onto this plane, but I don't know exactly how to find an orthogonal one. Any help would be much appreciated.
 A: All you have to do is to find where the transformation sends the
basic unit vectors
$e_1=\pmatrix{1\\0\\0}$, $e_2$ and $e_3$ to, since these will be
the columns of your transformation matrix.
A normal vector to the plane $x+y+z=0$ is $n=\pmatrix{1\\1\\1}$
so $e_1$ will be sent to $e_1+\alpha n=\pmatrix{1+\alpha\\\alpha\\\alpha}$ for some $\alpha$. What then should $\alpha$ be?
A: A projection matrix onto a plane in 3-D Euclidean space always has the form $$\mathbf P = \mathbf I - \frac{\mathbf n \mathbf n^T}{\mathbf n^T \mathbf n},$$
where $\mathbf n$ is the normal vector on the plane. Let's see why this is true. What do we expect if we apply $\mathbf P$ to $\mathbf n$? We should get zero right? Well, as you can verify we do, since $$\mathbf P \mathbf n = \mathbf n - \frac{\mathbf n \mathbf n^T\mathbf n}{\mathbf n^T \mathbf n} = \mathbf n - \mathbf n  = 0.$$ What happens to any other vector $\mathbf a$? We get $$\mathbf P \mathbf a = \mathbf a - \mathbf n\frac{ \mathbf n^T\mathbf a}{\mathbf n^T \mathbf n},$$ which is exactly the projection formula: subtract from $\mathbf a$ the piece of $\mathbf n$ that is as long as the projection of $\mathbf a$ onto $\mathbf n$. If you draw this in 2-D, it should become clear quite quickly. 
Here is an illustration:

To project $\mathbf a$ on a "plane" (in 2-D, a line $\mathbf p$), subtract from $\mathbf a$ the blue vector, which is just like the normal vector $\mathbf n$ but scaled to the length of $\mathbf a$ projected onto $\mathbf n$.
Now, with this formula, since you already have your $\mathbf n$, the rest should be easy...
