Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics I / 3

Choose the rotation taking $e_1$ to $e_2$, $e_2$ to $-e_1$ and $e_3$ to itself. The columns of the matrix of a linear map are these images of the standard basis, soThis fixes the orientation used throughout this question.

Using the convention ${\chi }_T(x)=\det (xI-A)$, the characteristic polynomial isSince $x^2+1>0$ for every $x\in \mathbb{R}$, the only eigenvalue over the real numbers is $1$. Solving $(A-I)v=0$ givesThus the only eigenspace over the real numbers is the axis of the rotation.

The two nonzero proper invariant subspaces areBoth are preserved by the rotation. To show there are no others, note thatis 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. ThereforeBecause $U$ 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 $U,W$. HenceThis also illustrates how invariant subspaces of a quarter-turn in three dimensions can be classified by linear projections.

Over the complex numbers, the characteristic polynomial splits into three distinct roots of a polynomial, $1,i,-i$. Direct multiplication gives the eigenvectorsThey are linearly independent: the last two have two-dimensional determinant $2i\ne 0$ in the first two coordinates, and the first has nonzero third coordinate. ThusThis is a basis of eigenvectors of ${\mathbb{C}}^3$, proving that $T^{\prime }$ has a diagonalizable matrix.