Can you diagonalize the matrix $\mathrm P$?

Question

Let be $P\in\mathbb{C}^{n\times n}$ so that $P^2=P$. The question is, when is $P$ diagonalizable? I got that the eigenvalues of $P$ can either be 0 or 1 and if there exists a diagonal matrix $D$ so that $D=S^{-1}PS$ then you get $D^2=D$. But I really don't know what to do from there on.

Answer 1

Every matrix satisfying $P^2 = P$ is diagonalizable.

Note that a matrix $P$ is diagonalizable if and only if every vector can be written as a linear combination of $P$'s eigenvectors. Our matrix $P$ is diagaonlizable because every $x$ can be written in the form $$ x = (x - Px) + Px $$ and both $(x - Px)$ and $Px$ are necessarily either the zero vector or eigenvectors of $P$.

Answer 2

Hint: A linear operator $ A $ over a finite dimensional vector space $ V $ is diagonalizable if and only if its minimal polynomial splits as the product of distinct linear factors.

Here is a proof sketch for the sake of completion: Let the minimal polynomial $ f(T) $ be a product of distinct linear factors, as $ f(T) = \prod_i (x - \lambda_i) $. Define $ h_i(T) = f(T)/(x - \lambda_i) $, then the ideal generated by the $ h_i(T) $ in the polynomial ring above the ground field is the entire ring, hence we may pick polynomials $ k_i(T) $ such that

$$ \sum_i k_i(T) h_i(T) = 1 $$

Note that substituting $ A $ for $ T $ gives

$$ \sum_i k_i(A) h_i(A) = I $$

Let $ v \in V $ be any vector, and right multiply both sides by $ v $ to obtain

$$ \sum_i k_i(A) h_i(A) v = v $$

Now, note that $ (A - \lambda_i I) k_i(A) h_i(A) v = f(A) k_i(A) v = 0 $, so $ k_i(A) h_i(A) v \in V_{\lambda_i} $. Since $ v $ was arbitrary, we conclude that any vector can be written as the linear combination of the eigenvectors of $ A $, and the result follows. The converse direction is obvious.

Answer 3

Since the minimal polynomial $x^2-x$ has all distinct roots, there are no eigenvalues with algebraic and geometric multiplicities different. The matrix is diagonalizable. A basis in which it is diagonal will consist of a basis of the null space of $P$ and a basis of the null space of $I-P$.