Is trace equal to rank for idempotent matrices?

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For an idempotent matrix A, it is true that trace(A) equals rank(A), as the eigenvalues of such matrices are either 0 or 1, making them diagonalizable. This relationship stems from the fact that the trace counts the number of 1s (which correspond to the rank) in the diagonalized form. The discussion also highlights a need for resources that delve into the connections between trace and rank, particularly in the context of regression analysis. Recommendations for books on matrix analysis or inference that cover these topics in detail are sought. Understanding these concepts is crucial for applying linear algebra in practical scenarios.
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Hi

Is it true that for an idempotent matrix A (satisfying A^2 = A), we have

trace(A) = rank(A)

Where can I find more general identities or rather, relationships between trace and rank? I did not encounter such things in my linear algebra course. I'm taking a course on regression analysis this semester and that's where I ran into it.

I'd appreciate if someone could point me to a book on matrix analysis or inference where these things would be mentioned in some detail. For some reason, the more "practically relevant" results were not covered in my freshman math courses.

Thanks in advance.
Cheers
Vivek.
 
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Hint 1: an idempotent matrix is diagonalizable.
Hint 2: the eigenvalues of an idempotent matrix are either 0 or 1.
 
radou said:
Hint 1: an idempotent matrix is diagonalizable.
Hint 2: the eigenvalues of an idempotent matrix are either 0 or 1.

Thanks..yes, I thought of the identity matrix and it all made sense.
 

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