### Instruction: Let $P$ be the plane passing through the origin with normal vector $\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}.$  Find the matrix $\mathbf{R}$ such that for any vector $\mathbf{v},$ $\mathbf{R} \mathbf{v}$ is the reflection of $\mathbf{v}$ through plane $P.$.### Response: To reflect a vector through a plane, I need to subtract its projection onto the normal vector twice.
That means, if $\mathbf{n}$ is the normal vector, $\mathbf{R} \mathbf{v} = \mathbf{v} - 2 \frac{\mathbf{v} \cdot \mathbf{n}}{\mathbf{n} \cdot \mathbf{n}} \mathbf{n}.$
I can write this as a matrix multiplication by using the identity matrix $\mathbf{I}$ and the outer product $\mathbf{n} \mathbf{n}^T.$
So, $\mathbf{R} \mathbf{v} = (\mathbf{I} - 2 \frac{\mathbf{n} \mathbf{n}^T}{\mathbf{n} \cdot \mathbf{n}}) \mathbf{v}.$
Now, I just need to plug in the given normal vector and simplify.
The normal vector is $\mathbf{n} = \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$, so $\mathbf{n} \mathbf{n}^T = \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} \begin{pmatrix} 1 & 1 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 1 & -1 \\ 1 & 1 & -1 \\ -1 & -1 & 1 \end{pmatrix}.$
The dot product of the normal vector with itself is $\mathbf{n} \cdot \mathbf{n} = 1^2 + 1^2 + (-1)^2 = 3.$
So, the matrix $\mathbf{R}$ is $\mathbf{R} = \mathbf{I} - 2 \frac{\mathbf{n} \mathbf{n}^T}{\mathbf{n} \cdot \mathbf{n}} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} - \frac{2}{3} \begin{pmatrix} 1 & 1 & -1 \\ 1 & 1 & -1 \\ -1 & -1 & 1 \end{pmatrix} = \frac{1}{3} \begin{pmatrix} 1 & -2 & 2 \\ -2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix}.$