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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 enter image description here. 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.

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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: Projection in 2-D 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...

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  • $\begingroup$ Thanks for the answer. In the formula is n a column vector, and I the identity matrix? I'm not exactly sure how to calculate nnT/ nTn. I don't really know how to divide matrices. Thanks. $\endgroup$ – kemb Jan 19 '18 at 0:31
  • $\begingroup$ @kemb: Yes, $\mathbf n$ is a column vector and $\mathbf I$ is an identity matrix. Since $\mathbf n$ is a column vector, $\mathbf n^T \mathbf n$ is a scalar, in fact, equal to the squared Euclidean norm. So the division is, of course, just a scalar division. $\endgroup$ – Florian Jan 19 '18 at 15:23
  • $\begingroup$ @kemb: did you suceed with answering the question? Is anything still unclear? If the answer was helpful to you, consider accepting it as correct. $\endgroup$ – Florian Jan 22 '18 at 13:49
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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?

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  • $\begingroup$ Hello, Thanks. I'm not entirely sure what you are getting at. What does a represent and how do we find a? $\endgroup$ – kemb Jan 18 '18 at 5:23
  • $\begingroup$ If I understand you correctly, is a=0? $\endgroup$ – kemb Jan 18 '18 at 5:29

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