Show that $f$ is an orthogonal projection

Question

$A=M(f,B)=\frac{1}{3}\begin{pmatrix} 1 & -1 & 1\\ -1& 1 & -1\\ 1& -1 & 1 \end{pmatrix}$

Show that $f$ is an orthogonal projection on a line to be determined. How to solve ?

Answer 1

Hint:

It is geometrically obvious that the direction of the line must be an eigenvector of this transformation because any vector on this line must be projected to itself. Find this eigenvector first (there should be only one non-zero eigenvalue based on our geometric intuition). So, find the eigenvector associated with the non-zero eigenvalue of $A$ first. You can normalize the eigenvector if you want. In this particular case, the non-zero eigenvalue is equal to $1$. (I just checked it).

Once you've found it, let's call it $v$, we can decompose our space as follows $$\mathbb{R}^3 = \langle v \rangle \oplus\langle v \rangle ^ {\perp}$$ To show that it's an orthogonal transformation, we should show that it sends $\langle v \rangle ^ {\perp}$ to $0$. Find an orthogonal basis for $\langle v \rangle ^ {\perp}$ and show that $$\ker{A}=\langle v \rangle ^ {\perp}$$

Note that since our matrix $A$ is symmetric, its eigenvectors are orthogonal. So, to find a basis for $\langle v \rangle ^ {\perp}$, you can just find the eigenvectors associated to the eigenvalue $\lambda=0$.

Edit: After some calculations as I explained, you can see that $v=\frac{1}{\sqrt{3}}(1,-1,1)$ and $\langle v \rangle ^ {\perp}=\langle n,m\rangle$ where $$n=\frac{1}{\sqrt{6}}(1,2,1), m=\frac{1}{\sqrt{2}}(1,0,-1)$$

It is easily seen that $Av=v$ and $An=Am=0$.

Answer 2

Let $\textbf{a} = (1,-1,1)$. Then for any $\textbf{v} \in \mathbb{R^3}$, we have $$ A \mathbf{v} = \frac{\textbf{a} \cdot \textbf{v}}{\| \textbf{a}\|^2} \textbf{a}, $$ which one can recognize as the formula for orthogonal projection onto the line spanned by $\textbf{a}$. (To see that the projection is orthogonal, one can compute the dot product of $\mathbf{v} - A \mathbf{v}$ with $\mathbf{a}$.)