Find Rank of given matrix Let $$A= 
\left[\begin{matrix}
        a & b & c \\
        p & q & r \\
        x & y & z \\
        \end{matrix}
\right]$$ be an idempotent matrix with rank 2. Then the rank of
$$B= \left[
        \begin{matrix}
        a-1 & b & c \\
        p & q-1 & r \\
        x & y & z-1 \\
        \end{matrix}
\right]$$ is?
Since the matrix $B$ can be written as $B=A-I$. And one of property of idempotent matrix says that the trace of an idempotent matrix = rank of matrix.
this implies that $a+q+z=2$?
and therefore, rank of $(a-1)+(q-1)+(z-1)=2-3=-1$?
But how rank cannot be negative. Where am I going wrong?
 A: Notice that $B$ is not idempotent. Since $B=A-I$ and $A^2=A$ we have:
$$BB=(A-I)(A-I)=A^2-2AI+I^2=A-2A+I=I-A=-B$$
$B$ is idempotent if $B=I-A$ but not if $B=A-I$. So the assumption that the rank of $B$ equals its trace is incorrect with the given definition of $B$.
A: BB = (A-I)(A-I) = AA - 2AI + II = A - 2A + I = I - A = -B, so B is not idempotent. In fact its trace is -1, and its rank is 1 (invoking, e.g., the result given in Seber, A Matrix Handbook for Statisticians, 8.79 [p. 171]).
A: Since $A$ is idempotent, then $A^2  = A$. This means that:
$$A^2 - A = 0 \Rightarrow A(A-I) = 0 \Rightarrow AB = 0.$$
For the Sylvester’s rank inequality, we have that:
$$\text{rank}(A) + \text{rank}(B) - n \leq \text{rank}(AB) \Rightarrow \\
2 + \text{rank}(B) - 3 \leq 0 \Rightarrow\\
\text{rank}(B) \leq 1,$$
where $n=3$ is the dimension of the matrices involved.
This means that $\text{rank}(B) = 0$ or $\text{rank}(B) = 1$.
It's clear that $\text{rank}(B) = 0$ if and only if $B = 0$. But in this case, we get:
$$0 = B = A-I \Rightarrow A = I,$$
which is a contradiction because $\text{rank}(A) = 2$. Then, we conclude that $$\text{rank}(B) = 1.$$ 
A: We have $A^2-A=O$, which mean the eigenvalues of $A$ must be $0$ or $1$.
Using Sylvester inequality gives:
$$rank(A)+rank(A-I)\leqslant rank(A(A-I))+3$$
Thus $rank(A-I)\leqslant 1$
If $rank(A-I)=0$ then $A=I$, which is against $rank(A)=2$
So $rank(A-I)=1$
