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

Choose an ordered basis $\mathcal{B}$ of $V$ and let $A=[T{]}_{\mathcal{B}}$. Define the characteristic polynomial byIf a different basis gives $B=P^{-1}AP$ for an invertible matrix $P$, multiplicativity of the determinant givesThus the definition is independent of the chosen basis. The alternate convention $\det (A-xI)$ differs by the constant factor $(-1{)}^{\dim V}$ and gives the same roots.

A real number $\lambda$ is an eigenvalue of $T$ if $Tv=\lambda v$ for some nonzero vector $v\in V$. Its eigenspace isIt is a vector subspace because it contains $0$ and $T(av+bw)=aTv+bTw=\lambda (av+bw)$ whenever $v,w\in E_{\lambda }$ and $a,b\in \mathbb{R}$.

If $v\in E_{\lambda }\cap E_{\mu }$, then $\lambda v=Tv=\mu v$, so $(\lambda -\mu )v=0$. Since $\lambda \ne \mu$, this forces $v=0$. Therefore

The definition of an eigenvalue and the kernel criterion for a singular square matrix giveFor the second equivalence, a linear map from a finite-dimensional vector space to itself has zero kernel exactly when its rank of a linear map is the full dimension, by the rank-nullity theorem, and therefore exactly when it is invertible. This proves both directions.

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