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suppose $n \in \mathbb{N}$ and $A$ is a $n \times n$ idempotent matrix, meaning $A^2 = A$.

show that:

$$ \textrm{rank}(A) + \textrm{rank}(A - I) = n $$


I think we should first use rank-nullity theorem: $$ \textrm{rank}(A) + \textrm{dim}(\textrm{null}(A)) = n $$

then show that: $$ \textrm{dim}(\textrm{null}(A)) = \textrm{rank}(A - I) $$

but I don't know how to show the second phrase.

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3 Answers 3

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Since $A(A-I) = 0$, the eigenvalues of $A$ are $0,1$, and the matrix is diagonalizable. The rank of $A$ is the number of non-zero eigenvalues, which is the multiplicity of the eigenvalue $1$. Similarly, the rank of $A-I$ is the number of eigenvalues different from $1$, which is the multiplicity of the eigenvalue $0$. The sum of these multiplicities must be $n$.

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  • $\begingroup$ If $M=\left[\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right]$, then $\operatorname{rank}M=1$. However, $M$ has no non-zero eigenvalues. $\endgroup$ Commented Oct 31, 2021 at 22:55
  • $\begingroup$ Fortunately, in our case the matrix is diagonalizable, so this problem is avoided. $\endgroup$ Commented Oct 31, 2021 at 22:57
  • $\begingroup$ Yes, I know that. But you did not mention it in your answer. $\endgroup$ Commented Oct 31, 2021 at 22:59
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    $\begingroup$ No. I am calling your attention to the fact that your answer is incomplete. $\endgroup$ Commented Oct 31, 2021 at 23:01
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    $\begingroup$ The converse is also true. For any matrix A if rank A + rank (A-I)=n, then A is idempotent, where A is n by n. $\endgroup$ Commented Nov 1, 2021 at 8:08
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It can be shown that $F^n\cong \text{im}(A)\oplus\text{im}(I-A)$ (where $F$ is the base field, and $\text{im}$ stands for image.) Then the rank property folows directly.

To show this for any $v\in F^n$, $v = A(v) + (I-A)(v)$, hence $F^n\cong \text{im}(A)+\text{im}(I-A)$. Now it's enough to show the two images have trivial interection.

If $A(x) = (I-A)y$, then apply $A$ to both sides, $A^2(x) = A(I-A)y$, $A(x) = (A-A^2)y = (A-A)y = \vec 0$.

We can also show the second phrase directly. It's enough to show that $\text{im }(I-A)=\ker A$. Indeed, if $x = (I-A)y\in\text{im }(I-A)$, then $A(x) = A(I-A)(y) = 0$, so $x\in\ker A$. If $x\in \ker A$, then $(I-A)(x) = Ix-Ax = x$, hence $x\in\text{im }(I-A)$.

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If $v\in Range(I-A)$, $v=(I-A)w$, then $Av=A(I-A)w=0$ so $Range(I-A)\subseteq Null(A)$. Conversely if $v\in Null(A), Av=0$ then $(I-A)v=v$ and $v\in Range(I-A)$. Thus $Range(I-A)=Null(A)$. Hence $$rank(I-A)=dim(Range(I-A))=dim(Null(A))=nul(A).$$

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