### 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 vector $\begin{pmatrix} 2 \\ -2 \\ -1 \end{pmatrix}.$.### Response: Let's call the vector $\begin{pmatrix} 2 \\ -2 \\ -1 \end{pmatrix}$ vector $\mathbf{w}$.
The projection of $\mathbf{v}$ onto $\mathbf{w}$ is $\frac{\mathbf{v} \cdot \mathbf{w}}{\|\mathbf{w}\|^2} \mathbf{w}$.
So the matrix $\mathbf{P}$ should take in a vector $\mathbf{v}$ and return $\frac{\mathbf{v} \cdot \mathbf{w}}{\|\mathbf{w}\|^2} \mathbf{w}$.
Let's compute $\|\mathbf{w}\|^2$.
It's $2^2 + (-2)^2 + (-1)^2 = 9$.
So $\mathbf{P}$ should take in a vector $\mathbf{v}$ and return $\frac{1}{9} \mathbf{v} \cdot \mathbf{w} \mathbf{w}$.
The matrix $\mathbf{P}$ should therefore be $\frac{1}{9} \mathbf{w} \mathbf{w}^T$.
Let's compute that. $\mathbf{w} \mathbf{w}^T = \begin{pmatrix} 2 \\ -2 \\ -1 \end{pmatrix} \begin{pmatrix} 2 & -2 & -1 \end{pmatrix} = \begin{pmatrix} 4 & -4 & -2 \\ -4 & 4 & 2 \\ -2 & 2 & 1 \end{pmatrix}$.