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$M_n(F)$ is the set of all $n\times n$ matrices over field $F$. Suppose $A\in M_n(F)$. Show that: $$A^2=A \iff \text{rank}(A)+\text{rank}(A-I)=n$$ Actually I'm somehow new to linear algebra and I'm not familiar with its techniques. I tried to find row/column spaces of $A$ and $A^2$ or $A-I$ and find their basis. But I wasn't that much successful. Also I tried to connect this problem to linear transformations, but I counldn't find out how.

Any help ot hints is so much appreciated.

Mason Rashford
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  • We can use eigenvalues of $B=A^2$ and $A$. Because both matrices are equal to each other and eigenvalues of $A^n$ are $\lambda_i^n$ (when $\lambda_i$ are eigenvalues of $A$) then the only possible eigenvalues of $A$ are $0$ and $1$. For $A-I$ they would be correspondingly $-1$ and $0$. – Widawensen Nov 17 '23 at 13:16
  • @Widawensen Thanks for your comment. But as it is one of my homeworks, I'm not allowed to use eigenvalues. Can you come up with other argument though, please? – Mason Rashford Nov 17 '23 at 13:17
  • On one hand, since $(I-A)A=0$, we have $\operatorname{range}(A)\subseteq\ker(I-A)$ and hence $\operatorname{rank}(A)\le n-\operatorname{rank}(I-A)$. On the other hand, $n=\operatorname{rank}(I)=\operatorname{rank}\big(A+(I-A)\big)\le\operatorname{rank}(A)+\operatorname{rank}(I-A)$. Combine the two inequalities to finish. – user1551 Nov 17 '23 at 16:59

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$A^2=A\Leftrightarrow A(A-I)=0\Leftrightarrow \text{Ker}(A)\oplus\text{Ker}(A-I)=F^n$ see Name for "the kernel lemma"? hence using $\text{rank}(M)+\text{dim(Ker(}M))=n$ for any matrix $M$ in $M_n(F)$ provides the equivalence you want.

Fred Bernard
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Assuming you know how to diagonalize a matrix:
Notice that $A^2=A$ if and only if $A(A-I)=0$, which is equivalent to the minimal polynomial of $A$ dividing $x(x-1)$. So $A$ has to be diagonalizable and its eigenvalues are all included in the set $\{0,1\}$. Assume that the multiplicity of $0$ as an eigenvalue is $k$ and therefore of $1$, as an eigenvalue, is $n-k$. Since $k=\dim(\ker(A))=n-\operatorname{rank}(A)$ and $n-k=\dim(\ker(A-I))=n-\operatorname{rank}(A-I)$ we have that $$\operatorname{rank}(A)+\operatorname{rank}(A-I)=(n-k)+k=n$$

Dennis Gulko
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  • I highly appreciate your answer. But as this question is one of my howeworks, I should use concepts we're taught in the class. We're studying Hoffman and Kunze book and we've studied until chapter 3.5. So we are not allowed to use minimal polynomials and eigenvalues. But we know about similar matrices and nullity. Do you have any other arguments that doesn't use these concepts in mind? I would be really thankful if you add it. – Mason Rashford Nov 17 '23 at 13:21
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First, eigenvalues of such matrix $A$ can only $\in \{0,1\}$, and $A$ is diagonalizable. For details please refer to this link, or this wiki page. The proofs there are not very involved.

Note that $\text{rank}(A)=$ how many nonzero eigenvalues (counting multiplicities) of $A$ are $1$. $\text{rank}(A-I)=$ how many eigenvalues (counting multiplicities) of $A$ are $0$.

Given $A$ has only $0,1$ as eigenvalues and $A$ has $n$ eigenvalues, the desired equation is deduced.

SouthChinaSeaPupil
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