# Finding a basis such that the $\mathcal{B}$-matrix is diagonal for orthogonal projection and reflection

Find a basis of the transformations such that the $\mathcal{B}$-matrix is diagonal.

1. Orthogonal projection $T$ onto the plane $3x_1+x_2+2x_3=0$ in $\mathbb{R^3}$.

2. Reflection $T$ about the plane $x_1+2x_2+x_3=0$ in $\mathbb{R^3}$.

In both cases, the $\mathcal{B}$-matrix will be diagonal iff $T(v_i)=kv_i$ for all vectors $v_i$ in the basis.

For number 1, we know the plane is defined by $\left( \begin{array}{ccc}3 \\ 1\\2 \end{array} \right)$, and we choose our basis to be three vectors $v1 = \left( \begin{array}{ccc}3\\1\\2 \end{array} \right), v_2 = \left( \begin{array}{ccc}2\\0\\-3 \end{array} \right), v_3 = \left( \begin{array}{ccc}1\\-3\\0 \end{array} \right)$ so that $v_2$ and $v_3$ are perpendicular to $v_1$. Then we have $T(v_1) = v_1$, but I'm not sure how to calculate $T(v_2)$ and $T(v_3)$.

For number 2, we know the plane is defined by $\left( \begin{array}{ccc}1\\2\\1 \end{array} \right)$ and we again consider three vectors $v1 = \left( \begin{array}{ccc}1\\2\\1 \end{array} \right), v_2 = \left( \begin{array}{ccc}1\\0\\-1 \end{array} \right), v_3 = \left( \begin{array}{ccc}2\\-1\\0 \end{array} \right)$ so that $v_1$ is parallel to the plane and $v_2,v_3$ are perpendicular to the plane. In this case, I know $T(v_1)=v_1$, but again I'm not sure how to calculate $T(v_2)$ and $T(v_3)$.

In either case, is my initial work a correct procedure? Also, how do we calculate $T(v_2)$ and $T(v_3)$ in either case for $\mathbb{R^3}$. It doesn't seem so clear to see this geometrically as it is in $\mathbb{R^2}$.

For 1: $v_2$ and $v_3$ are constructed to be in the plane under question. Thus, the projection leaves these vectors as they are, $T(v_2)=v_2$ and $T(v_3)=v_3$. (Expanded: These vectors are already in the plane on which $T$ projects. The projection onto a plane takes vectors outside the plane and throws them onto the plane. If such a vector is already on a plane, then the projection does nothing.)

Moreover, $v_1$ is the direction normal to the plane, hence $T(v_1)=0$.

For 2: With similar considerations, one finds $T(v_1)=-v_1$, $T(v_2)=v_2$ and $T(v_3)=v_3$.

(Expanded: Reflection changes the direction of incoming rays. If the ray is perpendicular to the plane (or the mirror) it gets thrown back and its direction changes: $T(v_1)=-v_1$.)

• Can you please elaborate? I'm trying to imagine these geometrically and I'm failing to see what you describe. For the first one, I understand why $T(v_1)=0$, but I don't understand why $T(v_2)=v_2$ and $T(v_3)=v_3$. Also for the second one, why is $T(v_1)=-v_1$? Why is it not $T(v_1)=v_1$ if we're just reflecting the plane on itself (thus keeping it the same)? And I honestly don't understand the $T(v_2)$ nor $T(v_3)$ here either. Oct 14 '14 at 10:14
• Actually, I'm not sure I understand why $T(v_1)=0$. Isn't the orthogonal of something onto itself just that something? Oct 14 '14 at 10:29
• en.wikipedia.org/wiki/Projection_%28linear_algebra%29
– daw
Oct 14 '14 at 11:10
• en.wikipedia.org/wiki/Reflection_%28mathematics%29
– daw
Oct 14 '14 at 11:11
• Can you explain why $v_1$ is perpendicular to the plane in both cases? I thought $v_1$ was parallel to the planes. Similarly, why are $v_2,v_3$ parallel to the plane? Oct 15 '14 at 8:15