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prove that for an idempotent matrix, trace=rank

samuel11
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  • How do you know the trace is an integer? I only know the $\chi^2$-distribution with an integer parameter. – Patrick Da Silva Oct 26 '15 at 23:38
  • @PatrickDaSilva: I doubt that an integer argument for $tr(A)$ is implied. The chi-squared distribution is for a sum of $k$ squares of normally distributed independent standard normal random variables, but here $tr(A)$ must be playing the role of the sum of squares. The degrees of freedom $k$ must depend on the dimension of $A$. – hardmath Oct 26 '15 at 23:51
  • @samuel11: You completely changed the question from one about probability to one about linear algebra. Please refrain from doing this in the future. If you wish, you may self-delete a question if it has no Accepted or up-voted answer. A new question deserves a separate post. – hardmath Oct 27 '15 at 22:15

1 Answers1

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An idempotent matrix is trivially diagonalizable. If two matrices are similar they have the same rank and trace. So it suffices to assume your matrix is diagonal.

$$D^2=D \iff d_{ii}^2=d_{ii} \iff d_{ii}\in \{0,1\}$$

The number of nonzero diagonal entries in $D$ (the rank) is therefore equal to the trace.

David P
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