= Solution
{c}
The two nonzero proper <invariant subspaces> are
$$
\boxed{U=\operatorname{span}\{e_3\},\qquad W=\operatorname{span}\{e_1,e_2\}.}
$$
Both are preserved by the <rotation>. To show there are no others, note that
$$
P=\frac{I+T^2}{2},\qquad P(x,y,z)=(0,0,z)
$$
is the <linear projection> onto $U$, and $I-P$ is the <linear projection> onto $W$. Every <invariant subspace> $L$ is preserved by $I,T^2$, and hence by both of these <linear projections>. Therefore
$$
L=(L\cap U)\oplus(L\cap W).
$$
Because $U$ has <dimension> one, $L\cap U$ is either $0$ or $U$. If $L\cap W$ contains a nonzero $v=(a,b,0)$, it also contains $Tv=(-b,a,0)$. These two <vectors> are <linearly independent>, because their two-dimensional <determinant> is $a^2+b^2>0$. They <span> $W$, so $L\cap W$ is either $0$ or $W$. The only possibilities for $L$ are therefore $0,U,W,V$.
Finally every $(x,y,z)$ has the unique decomposition $(0,0,z)+(x,y,0)$, with one summand in each of $U,W$. Hence
$$
\boxed{V=U\oplus W.}
$$
This also illustrates how <invariant subspaces of a quarter-turn in three dimensions> can be classified by <linear projections>.
Back to article page
