I am stuck on the following linear algebra problem :
Consider $\mathbb{R}^3$ and its canonical basis $\{ i,j,k \}$ and let $r$ denote the 90° angle rotation around the axis given by the vector $u := i+j$. I am asked to find an orthonormal basis (ONB) $\{ a,b,c \}$ of $\mathbb{R}^3$ such that the matrix of $r$ with respect to the basis $\{ a,b,c \}$ has the following form :
\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}
Isn't this the matrix of the 90° angle rotation around the x-axis ? Anyway I thought that I should tried to put the vector $u$ in the place of the x-axis so I tried the following :
\begin{align} a &=\frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \end{align}
a unit vector colinear with the axis of rotation. From $a$ I figured out that the tangent plan to $a$ through 0 is given by : $$ P: x+y=0. $$
I noticed that
\begin{align} b &=\frac{1}{\sqrt{2}} \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} \end{align}
lies in $P$ and is unitary. Now I find a third unit vector by computing the vector product of $a$ and $$ and diving the result by its norm :
\begin{align} a \times b &= \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \end{align}
Is it correct to assume that the matrix of $r$ with respect to the basis $\{ a,b,c \}$ is
\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}
?
Thanks for your help.