# Finding the standard matrix of a reflection operator

I was working on an exercise in which I am given a vector $(2,-1, 2)$ and I am supposed to find the standard matrix $A$ of the reflection operator $T$ on $\mathbb{R}^3$ such that $T(v)=-v$.

Here's my attempt at the problem:

$$2x_1-x_2+2x_3=0$$

I use it to get $(1,2,0)$ and $(0,-2,1)$ (the two columns of the new matrix $C$) as part of the basis. After that however; I'm confused what I should do. Does anyone have any ideas? I'm not clear so any tips directing me in the right direction would be much appreciated. Thanks.

• What text are you using, and how did they define "standard matrix"? – J. M. is not a mathematician Apr 27 '11 at 6:32
• For instance for a linear transformation T:Rn to Rm, there exists a unique matrix, A called the standard matrix of T. For example Ax=b for certain number of solutions for every b. It's the definition found on most textbooks and around the internet. – EuropaDust Apr 27 '11 at 6:39
• apologies for not figuring out the formatting yet – EuropaDust Apr 27 '11 at 6:40
• Then why wouldn't the negative of the identity work? – J. M. is not a mathematician Apr 27 '11 at 6:41
• as the standard matrix? I think the standard matrix must be unique to the vector. But I'm not sure. – EuropaDust Apr 27 '11 at 6:46

The vector you give defines a reflection about a plane whose normal vector is $v$. For a unit vector $u$, you can find the reflection about the plane to which $u$ is normal through the following formula:
$$x = 2 (u \cdot x) u - x$$
Try to express this in matrix notation, using the fact that $u \cdot x = u^Tx$ and $x = Ix$.
• I think that you meant to write $x\mapsto x-2(u\cdot x)u$. If $u$ is not a unit vector use $x\mapsto x-2((u\cdot x)/(u\cdot u))u$ instead :-) – Jyrki Lahtonen Jun 7 '11 at 17:50
Surely the matrix $A = \begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix}$ (or in index notation, $A_{ij} = -\delta_{ij}$) will transform any vector $\mathbf{v}$ to $-\mathbf{v}$, since it will reverse all its components.