If $A \in M(n,\mathbb{C})$ with $A^2 =A$ , then $rk(A)= tr(A)$ .
Is it true if yes then how can anyone explain?
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Alberto Andrenucci
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Ni TiSh
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Rearrange to find $A^2-A=0$ i.e. $A(A-I)=0$. Now... what do you know about minimal polynomials and eigenvalues and how they relate to the rank and to the trace? – JMoravitz Jun 20 '17 at 13:39
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Here's a MathJax tutorial :) – Shaun Jun 20 '17 at 13:40
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How to delete this question .I was not aware that someone already asked it. It down voted me with 4 points – Ni TiSh Jun 20 '17 at 16:21
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Notice that $A^2=A$ means that $A(A-1)=0$ so you have that the polynomial $x(x-1) \in I(A)$. The only eigenvalues you have are $0$ and $1$ and surely the matrix is diagonalizable. So you can say that:
$$A \sim \begin{pmatrix} 1 \\ & \ddots \\ &&1 \\ &&&0 \\ &&&&\ddots \\ &&&&&0\end{pmatrix}$$ So clearly you see that the rank is equal to the trace. (In fact, they are invariant under conjugation).
Alberto Andrenucci
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Any idempotent matrix is diagonalizable, with eigenvalues $0$ and $1$. Since trace and rank are invariant under conjugation, the claim follows.
Francesco Polizzi
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