= Solution
{c}
If $T^2=I$, every $v\in V$ has the decomposition
$$
v=\frac{v+Tv}{2}+\frac{v-Tv}{2},
$$
whose first term belongs to $E_1$ and second to $E_{-1}$. Their intersection is zero by <4/a/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 exactly
$$
\boxed{-I_3,\qquad\operatorname{diag}(1,1,-1),\qquad
\operatorname{diag}(1,-1,-1).}
$$
For $-I_3$, $T$ is \b[point inversion through the origin]. For $\operatorname{diag}(1,1,-1)$, 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 $\operatorname{diag}(1,-1,-1)$, 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.
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