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Which Jordan cells $J(\lambda,k)$ are idempotent? And how can I use that to determine the Jordan canonical form of any square idempotent matrix?

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Hint. It seems the following.

You can easily calculate the square of a Jordan cell. And when you will do it you will easily found which Jordan cells are idempotents and which Jordan canonical forms of a square idempotent matrix can be.

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Hints:

Idempotent matrices over a field can only have eigenvalues lying in $\{0,1\}$, and every idempotent matrix is diagonalizable.

To confirm what this suggests, take the block and decompose it into $\lambda I+I'$, where $I'$ denotes the matrix with $1$ on the superdiagonal.

Computing $(\lambda I+I')^2=\lambda I+I'$, you will discover that $\lambda^2=\lambda$, so that $\lambda\in \{0,1\}$. You will also need to notice that $(I')^2$ is zero on the places where $I'$ is nonzero.

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  • $\begingroup$ So the Jordan form has eigenvalues along the main diagonal and 1 along the superdiagonal, right? Since the matrix only has 0 and 1 eigenvalues, doesn't that mean that the only Jordan cells which are idempotent are either the 1x1 cells 0 or 1, or a block such as $\begin{bmatrix}0&1\\0&1\end{bmatrix}$. $\endgroup$ – user116084 Dec 16 '13 at 20:09
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    $\begingroup$ @user116084 This $2 \times 2$ block is not a Jordan block, because such blocks have constant diagonals. $\endgroup$ – Vedran Šego Dec 16 '13 at 20:10
  • $\begingroup$ @Vedran Ah, so in this case, the only idempotent Jordan cells would be the 1x1 0 or 1 matrices? $\endgroup$ – user116084 Dec 16 '13 at 20:17
  • $\begingroup$ @user116084 Yes: Jordan blocks bigger than $1\times 1$ do not play well with the $(\lambda I+I')^2=\lambda I+I'$ equation. $\endgroup$ – rschwieb Dec 16 '13 at 20:25
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    $\begingroup$ @user116084 A matrix is diagonalizable if and only if its Jordan blocks are of order $1$. $\endgroup$ – Vedran Šego Dec 16 '13 at 20:26

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