Matrix $A$ projects vectors orthogonally to the plane $y=z$. Find $A$.

Question

A $3 \times 3$ matrix $A$ is associated with the transformation that projects vectors in $\mathbb{R}^3$ (orthogonally) onto the above plane $y = z$. Find the matrix A by determining where the vectors $(1,0,0) , (0,1,0)$ and $(0, 0, 1)$ go.

So when projecting the vector $(1,0,0)$ onto the plane, we should probably get $(1,0,0)$ as it is already on the plane. When projecting $(0,1,0)$ and $(0,0,1)$, I am getting $(0,0.5,0.5)$ and $(0,0.5,0.5)$.

Is the matrix $$A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0.5 & 0.5 \\ 0 & 0.5 & 0.5 \\ \end{pmatrix} ?$$

Answer 1

Yes, the answer is correct.

Projection of $(1,0,0)$ is clearly itself.

To find projection of $(0,1,0)$ onto the plane $y=z$, we minimize $x^2+(y-1)^2+y^2$, and it is $(0,0.5,0.5)$.

Similarly for the other unit basis.

Answer 2

Vector $\mathrm v := \begin{bmatrix} 0 & 1 & -1\end{bmatrix}^\top$ is orthogonal to the plane defined by $y - z = 0$. Thus, the projection matrix that projects onto the line spanned by $\rm v$ is

$$\rm P := \dfrac{\,\,v v^\top}{\| v \|_2^2} = \begin{bmatrix} 0 & 0 & 0\\ 0 & \frac{1}{2} & - \frac{1}{2}\\ 0 & - \frac{1}{2} & \frac{1}{2}\end{bmatrix}$$

and the projection matrix that projects onto the orthogonal complement of the aforementioned line, i.e., the given plane itself, is the following

$$\rm I_3 - P = \begin{bmatrix} 1 & 0 & 0\\ 0 & \frac{1}{2} & \frac{1}{2}\\ 0 & \frac{1}{2} & \frac{1}{2}\end{bmatrix}$$