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A little time ago I proved this exercise:

If $B$ is a matrix $n\times n$ and $B^2=B$ ($B$ is idempotent) then the matrix $(I_n-B)$ is idempotent. Also show that the eigenvalues of $B$ are $\{0,1\}$ if $B\neq 0$ and $B\neq I_n$.

Now, what can I say about the Jordan form of $B$?

Thanks for your help.

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For linear algebra, a linear transformation (matrix) $B$ such that $B^2=B$ is always a projection. The only eigenvalues are $0$ and $1$, and the space decomposes as $\mathrm{null}(B)\oplus\mathrm{Im}(B)$. In particular, $B$ is always diagonalizable, so the Jordan canonical form of $B$ is diagonal, with $\mathrm{nullity}(B)$ zeros and $\mathrm{rank}(B)$ ones.

To verify this: note that $\mathrm{null}(B)$ is the eigenspace of $0$, and $\mathrm{Im}(B)$ is the eigenspace of $1$ (since $\mathbf{z}\in\mathrm{Im}(B)$ implies $B(\mathbf{z})=\mathbf{z}$). Since the dimension $\mathbf{F}^n$ equals the sum of the dimensions of the eigenspaces, $B$ is diagonalizable.

(Note: The fact that if $B$ is an idempotent then so is $1-B$ holds in far more generality: it's true in any ring (since $(1-B)(1-B) = 1-B-B+B^2 = 1-2B+B = 1-B$; in fact, this is one of the key ingredients in showing that a direct decomposition of a ring corresponds to the existence of a nontrivial central idempotent).

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Thank you very much! –  Hiperion Mar 23 '12 at 15:24
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