Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics I / 4 / b

If $T^2=I$, every $v\in V$ has the decompositionwhose first term belongs to $E_1$ and second to $E_{-1}$. Their intersection is zero by ii, so $V=E_1\oplus E_{-1}$ is a direct sum. Choosing bases of the two eigenspaces gives a diagonal matrix with entries $1$ and $-1$.

Since $T^2\ne T$ while $T^2=I$, we have $T\ne I$. Thus $\dim E_{-1}$ is $1$, $2$, or $3$. Reordering the basis gives exactlyFor $-I_3$, $T$ is point inversion through the origin. For $\mathrm{diag}(1,1,-1)$, it fixes the plane $E_1$ pointwise and reverses the complementary line $E_{-1}$: if $v=p+\ell$ in this direct sum, then $Tv=p-\ell$. For $\mathrm{diag}(1,-1,-1)$, it fixes the line $E_1$ pointwise and reverses the complementary plane $E_{-1}$.

These are instances of oblique symmetry of complementary subspaces, allowing oblique eigenspaces. They are a Euclidean reflection in a hyperplane and a rotation in three dimensions through $\pi$ about a line, respectively, when the two eigenspaces are perpendicular. The hypotheses do not require an orthogonal transformation or an orthonormal basis, so that additional metric property cannot be inferred just from the displayed diagonal forms.

For this tripotent linear operator, the relation $T^3=T$ gives $T(T^2-I)=0$. Since $T^2\ne I$, there is $v$ with $(T^2-I)v\ne 0$, and that nonzero vector belongs to $\ker T$. Hence $n_T\ge 1$. Also $T\ne 0$, because $T=0$ would imply $T^2=T$. The rank-nullity theorem in dimension three therefore givesNow suppose $n_T=2$. Then $\mathrm{im}T$ is a line. On this line, $T^2$ is the identity operator: if $y=Tx$, then $T^{2}y=T^{3}x=Tx=y$. Consequently $T$ acts on this line by a scalar $c$ satisfying $c^2=1$. If $c=1$, then $T^{2}x=Tx$ for every $x$, contrary to $T^2\ne T$. Thus $c=-1$.

Choose a nonzero $w\in \mathrm{im}T$, so $Tw=-w$. It does not belong to $\ker T$. A basis $k_1,k_2$ of $\ker T$, followed by $w$, is therefore a basis of $V$, in which