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How do you construct a matrix in $\mathbb{R}^3$ that reflects about the plane $y=z$? And is there a way to construct a reflection matrix about any plane in general?

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Any plane can be specified by a unit vector perpendicular to the plane. If the vector is $v \in\mathbb{R}^3$, then the matrix that reflects about the plane is

$$ R_v = I - 2 vv^T. $$

It is easy to check that $R_v$ flips the sign of any vector which is a multiple of $v$ and acts as identity on any vector perpendicular to $v$. See Householder transformation for more details.

In particular, the plane $y=z$ is perpendicular to $(0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})^T \in \mathbb{R}^3$, so the matrix is

$$ R_{y=z}=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} - 2\cdot\begin{pmatrix} 0 & 0 & 0 \\ 0 & \frac{1}{2} & -\frac{1}{2} \\ 0 & -\frac{1}{2} & \frac{1}{2} \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} $$

which leaves the $x$ coordinate unchanged and swaps the $y$ and $z$ coordinates as expected.

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To answer the first question. Look what the reflection does to the unit vectors in $\mathbb{R}^{3}$. $(1, 0, 0)$ will be mapped to $(1, 0, 0)$, $(0, 1, 0)$ will be mapped to $(0, 0, 1)$ and $(0, 0, 1)$ will be mapped to $(0, 1, 0)$. This defines the linear mapping and therefore will tell you how the matrix looks like.

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To answer the second part of your question, I offer a visual way to show that such a reflection matrix exists for any plane.

Suppose there exists a linear transformation $L$ that reflects all (column, let's use column for consistency) vectors about a certain plane.

Pick a vector $\vec{v}_1$ that lies on the plane. Next, pick another vector $\vec{v}_2$ that lies on the plane, such that there does not exist $k$ is real, $k\vec{v}_1 = \vec{v}_2$. (If you're a bit more hardworking, you can find a $\vec{v}_2$ that's perpendicular to $\vec{v}_1$.) It should be pretty easy for you to find $\vec{v}_1$ and $\vec{v}_2$ because the plane is usually defined as $\vec{r} = a + b\vec{v}_1 + c\vec{v}_2$, where $a$, $b$, and $c$ are real. We know that $L\vec{v}_1 = \vec{v}_1$ and $L\vec{v}_2 = \vec{v}_2$ because they lie on the plane, and vectors that lie on the plane remain the same after being transformed by $L$.

Next, we find $\vec{v}_3$ that's perpendicular to the plane. Since we already have $\vec{v}_1$ and $\vec{v}_2$ on the plane, we can easily define $\vec{v}_3$ as the cross product of $\vec{v}_1$ and $\vec{v}_2$, as the cross product must be perpendicular to both $\vec{v}_1$ and $\vec{v}_2$. It will be tough* to find $L\vec{v}_3$, but once you get it, you're almost done!

As $\vec{v}_1$, $\vec{v}_2$ and $\vec{v}_3$ are linearly independent, $\begin{pmatrix}\vec{v}_1 & \vec{v}_2 & \vec{v}_3\end{pmatrix}$ is invertible.

$$L\begin{pmatrix}\vec{v}_1 & \vec{v}_2 & \vec{v}_3\end{pmatrix} = \begin{pmatrix}\vec{v}_1 & \vec{v}_2 & L\vec{v}_3\end{pmatrix}$$

$$L = \begin{pmatrix}\vec{v}_1 & \vec{v}_2 & L\vec{v}_3\end{pmatrix} \begin{pmatrix}\vec{v}_1 & \vec{v}_2 & \vec{v}_3\end{pmatrix}^{-1}$$

*Simply find the distance from $\vec{v}_3$ to the plane, and then subtract $2$ times of that vector.

I'm not very good in linear algebra, but I picture that it could be solved in this manner. Do correct me if I'm wrong.

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