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I'm trying to prove the above statement, and I had a look at this site.

This ends the proof with the following statement

the rank is the number of non-zero eigenvalues

But, this is what makes me confusing, because it's not all the case that rank of a matrix equals the number of non zero eigenvalues as far as I know. This shows a counter example of this.

Any comment about the rank of idempotent matrix would be grateful.

Feng
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jason 1
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    If $A^2=A$, then $A$ is diagonalizable. Hence in this case, the rank is the number of non-zero eigenvalues? – Feng Oct 01 '23 at 02:25
  • Similar matrices have same rank so when the matrix is diagonalizable its rank equals the number of non zero eigenvalues and idempotents are diagonalizable. – Tony Pizza Oct 01 '23 at 06:39

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From $A^2=A$ you get that the only possible eigenvalues of $A$ are $0$ and $1$.

From the fact that $A^2=A$ you get that any vector $x$ can be written as $x=Ax + (I-A)x$ and that this representation is unique. In other words, the range of $A$ and the range of $I-A$ are complementary subspaces.

By the Rank-Nullity Theorem, $\operatorname{rank}A=\dim\ker (I-A)$. And $\ker (I-A)$ is precisely the eigenspace corresponding to the eigenvalue $1$. Similarly, $\operatorname{rank}(I-A)=\dim\ker A$.

If we let $e_1,\ldots,e_m$ be an ortonormal basis of $\ker A$ and $e_{m+1},\ldots,e_n$ an orthonormal basis of $\ker (I-A)$, then $$ \operatorname{Tr}(A)=\sum_k\langle Ae_k,e_k\rangle=\sum_{k=m+1}^n \langle Ae_k,e_k\rangle=\sum_{k=m+1}^n \langle e_k,e_k\rangle=\sum_{k=m+1}^n1=\operatorname{rank}(A). $$

Martin Argerami
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