Find vector such that rotation matrix has required form

Question

Let $v_1 = \frac{1}{3}(1,2,-2)^T, v_2 = \frac{1}{3}(2,1,2)^T$. Let $\Phi \colon \mathbb R^3 \to \mathbb R^3$ be the rotation by $\pi/2$ within the plane $E: v_1 \cdot x = 0$. Determine $v_3 \in \mathbb R^3$ such that the transformation matrix $A'$ of $\Phi$ with respect to the basis $\{v_1, v_2, v_3\}$ has the form $$A' = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}.$$ I have made the ansatz that $A'v_1 = v_1, A' v_2 = v_3$ and $A'v_3 = -v_2$ but this did not lead my anywhere. I feel like there is a simple solution besides calculating the transformation matrix with repspect to the standard basis. Any hints?

Answer 1

\begin{eqnarray} A'v_1 &=& \frac{1}{3}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix}1\\2\\-2\end{pmatrix}\\ &=& \frac{1}{3}\begin{pmatrix}1\\2\\2\end{pmatrix}\neq v_1\ , \end{eqnarray} so your statement that $\ A'v_1= v_1\ $ isn't true. You're confounding operations on the coordinates of the vectors $\ v_1, v_2, v_3\ $ with respect to the basis, which is what $\ A'\ $ is required to do, with operations on the vectors themselves, which is what $\ \Phi\ $ does. The coordinate vectors of $\ v_1, v_2, v_3\ $ with respect to the basis $\ v_1, v_2, v_3\ $ are just the standard basis vectors $\ e_1, e_2, e_3\ $, and what you actually have is $\ A'e_1=e_1\ $, $\ A'e_2=e_3\ $ and $\ A'e_3=-e_2\ $. The second of these equations tells you that $\ v_3=\Phi v_2\ $.

A hint for determining $\ v_3\ $ is that $\ v_2\ $ is perpendicular to $\ v_1\ $. Thus, since $\ \Phi\ $ represents a rotation through an angle $\ \frac{\pi}{2}\ $ in the plane $\ E\ $ of vectors perpendicular to $\ v_1\ $, it follows that $\ v_3\ $ must be perpendicular to both $\ v_1\ $ and $\ v_2\ $.