A linear transformation satisfying $P^2 = P$ Prove that a linear transformation $P \colon V \to V$ of a finite dimensional vector space satisfies $P^2 = P$ if and only if there exists a basis with respect to which $P$ can be written as a block matrix 
$$P = \begin{bmatrix} I & 0 \\ 0 & 0 \end{bmatrix}.$$
Hence determine the minimal and characteristic polynomials of $P$.
I don't know what to do with this problem at all, so some help would be great, Thanks.
 A: Well, if $P$ satisfies $P^2=P$, then it is the root of the polynomial $x^2-x=x(x-1)$.
Show that $P$ restricted to ${\rm im\,}P$ is the identity, and that
$$V=\ker P\oplus{\rm im\,}P\,.$$
Then choose arbitrarily bases both in $\ker P$ and in ${\rm im\,}P$.
Here $\ker P$ is the $P$-invariant subspace corresponding to the factor $x$ of the minimal polynomial, and ${\rm im\,} P=\{x:Px=x\}$ is the $P$-invariant subspace corresponding to the factor $x-1$.
A: One direction of the above is easy.
We have $x=Px+(I-P)x$, so we see that $V = {\cal R}P + {\cal R}(I-P)$. If we set $Px+(I-P)y = 0$, then multiplying across by $P$ shows $Px=0$, and multiplying by $I-P$ shows that $(I-P)y = 0$, hence the sum is a direct sum (that is, any $x\in V$ uniquely decomposes into the sum of an element of ${\cal R}P$ and an element of $ {\cal R}(I-P)$). We write this as $V = {\cal R}P 
\oplus {\cal R}(I-P)$.
Now note that ${\cal R}(I-P) = \ker P$, and so $V = \ker P \oplus {\cal R}P $.
Note that both $\ker P $ and $ {\cal R}P $ are $P$ invariant, in fact, $P$ is the zero operator on $\ker P$ and the identity operator on ${\cal R}P $. This gives the block form above.
The characteristic polynomial can be found immediately from the above block form. Since $P = \begin{bmatrix} I & 0 \\ 0 & 0 \end{bmatrix}$ in the appropriate basis, we have $\lambda I-P = \begin{bmatrix} (\lambda-1)I_r & 0 \\ 0 & \lambda I_k \end{bmatrix}$, and so $\det(\lambda I-P) = (\lambda-1)^r\lambda^k$.
The minimal polynomial can be determined by computing $P(I-P)$.
