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There is a problem that says for a matrix P such that $P^2 = P$, then P is diagonalizable.

However, I am kind of lost at how can we know that this matrix is diagonalizable, when we can't even show that it is invertible?

A related problem is "for a matrix A, if $(I - A)^k = 0$ for a positive integer k, then A is invertible". I have problems with this statement either. From the condition $(I - A)^k = 0$ I agree that we can conclude that one of the eigenvalues of A is 1, but we cannot rule out the possibility that A has 0 as eigenvalue. Why is this statement true? (By the way, A is a 3 by 3 matrix, though I don't think it matters here).

Thanks

Caprikuarius
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    If $P^2-P=0$, then the matrix satisfies the polynomial $t^2-t$; so the minimal polynomial divides this polynomials. If $(I-A)^k = 0$, then $0$ cannot be an eigenvalue: if $v$ were an eigenvector of $0$, then $(I-A)(v) = v$, and then what is $(I-A)^k(v)$? – Arturo Magidin Sep 03 '19 at 03:14
  • You don't need a matrix to be invertible for it to be diagonalizable! And a matrix that satisfies $P^2=P$ may indeed not be invertible, but that is irrelevant. The zero matrix is not invertible, but is certainly diagonalizable. – Arturo Magidin Sep 03 '19 at 03:15

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Notice that $P(P-I)=0$, so that the minimal polynomial of $P$ divides the polynomial $t(t-1)$. A matrix $A$ is diagonalisable exactly when its minimal polynomial splits into linear factors (see this question for a proof), which is certainly true for factors of $t(t-1)$. Thus $P$ must be diagonalisable.

YiFan Tey
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