Minimal polynomial of projection on a plane?

Question

If $g: \mathbb{R}^3 \to \mathbb{R}^3$ is the projection on a plane, what is the minimum polynomial of $g$?

Related to this, what is the minimum polynomial of a reflection with respect to a plane?

Answer 1

For the question of minimal polynomial, all that matters about a projection is that it is diagonalisable with eigenspace for $\lambda=1$ (the set of fixed vectors, the image of the projection) and for $\lambda=0$ (the kernel, the subspace parallel to which the projection is done) only. The whole space being the direct sum of these eigenspace means that the polynomial $(X-1)X$ is an annihilating polynomial (the factor $X-1$ vanishes on the fixed vectors, the factor $X$ on the kernel, so all vectors are annihilated by $(X-1)X$), and the fact that these eigenspaces each have nonzero dimension means that neither $X-1$ not $X$ is in itself an annihilating polynomial (it would kill only one of the eigenspaces). Therefore $(X-1)X=X^2-X$ is the minimal polynomial of any projection for which both image and kernel have nonzero dimension. Technically any operator $P$ satisfying $P^2=P$ is a projection, but this includes the special cases $P=I$ and $P=0$ that do not have $X^2-X$ as minimal polynomial (but rather $X-1$ respectively $X$).

Note that the minimal polynomial does not "see" any properties of the image and kernel subspaces, apart from each having nonzero dimension and them being complementary subspaces. Notably it is unimportant whether this is an orthogonal projection or not, and whether it is with respect to a plane or to a line in $\Bbb R^3$; it could also be a projection to some nonzero proper subspace, and parallel to some complementary subspace, in some other vector space than $\Bbb R^3$.

For a reflection the situation is similar, by instead of then kernel one has an eigenspace for $\lambda=-1$: then vectors in the subspace parallel to which the reflection is taking place are sent to their opposites. The minimal polynomial is then $(X-1)(X+1)=X^2-1$.

Answer 2

Hints:

• Any projection $P$ satisfies $P^2=P$, so it's minimal polynomial divides $x^2-x$.
• Any reflection $R$ satisfies $R^2=1$, so it's minimal polynomial divides $x^2-1$.