Find Rank of given matrix

Question

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?

Answer 1

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 $$\text{rank}(B) = 1.$$

Answer 2

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 $B$.

Answer 3

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]).

Answer 4

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 $rank(A-I)=1$