### Instruction: Find the matrix $\mathbf{P}$ such that for any vector $\mathbf{v},$ $\mathbf{P} \mathbf{v}$ is the projection of $\mathbf{v}$ onto the $yz$-plane.### Response: To project a vector onto the $yz$-plane, I need to eliminate the $x$-component of the vector and keep the $y$ and $z$ components unchanged.
That means that the matrix $\mathbf{P}$ should have the following effect: $\mathbf{P} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ y \\ z \end{bmatrix}$.
One way to achieve this is to have the matrix $\mathbf{P}$ be a diagonal matrix with zeros on the diagonal except for the $y$ and $z$ entries, which should be one.
That is, $\mathbf{P} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
To check if this works, I can multiply $\mathbf{P}$ with a generic vector $\mathbf{v} = \begin{bmatrix} a \\ b \\ c \end{bmatrix}$ and see if the result is the projection onto the $yz$-plane.
$\mathbf{P} \mathbf{v} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 0 \\ b \\ c \end{bmatrix}$.
Yes, this is what I wanted.
The $x$-component is zero and the $y$ and $z$ components are unchanged.
Therefore, the matrix $\mathbf{P}$ that projects any vector onto the $yz$-plane is $\mathbf{P} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.