Eigenspace decomposition of a linear involution ($P_{\pm }=(I\pm F)/2$)

Let $F$ be a linear map on a vector space over a field of characteristic different from two, with $F^2=I$. ThenThe projections satisfy $P_{+}{P}_{-}=0$ and $P_{+}+P_{-}=I$, giving existence and uniqueness of the decomposition. In characteristic two, the two signs coincide and the conclusion need not hold.

For complementary vector subspaces $U,W$ of a real vector space, the rule $T(u+w)=u-w$ is a well-defined linear map satisfying $T^2=I$. It fixes $U$ pointwise and reverses $W$. In a basis formed by adjoining bases of $U$ and $W$, its matrix is $\mathrm{diag}(I,-I)$. Conversely every real linear map with $T^2=I$ has this form, with $U=\ker (T-I)$ and $W=\ker (T+I)$, as follows from $v=(v+Tv)/2+(v-Tv)/2$.

In a Euclidean space, such a linear map is an orthogonal transformation if and only if $U$ and $W$ are perpendicular. Necessity follows from $⟨u,w⟩=⟨Tu,Tw⟩=-⟨u,w⟩$; sufficiency follows by expanding the squared Euclidean norm of $u+w$. In three dimensions, a fixed plane with reversed complementary line is a possibly oblique reflection, and a fixed line with reversed complementary plane is a possibly oblique axial symmetry. Their diagonal forms alone do not imply a Euclidean reflection in a hyperplane or a Euclidean half-turn.

For a linear operator with $A^2=I$, every even power is $I$ and every odd power is $A$. Separating the absolutely convergent matrix exponential series into those powers proves the displayed identity without a Hermiticity assumption. If $A$ is also a Hermitian operator, the exponential is a unitary operator.