Prove: $\det(I+A) = 2^{\text{rank}(A)}$ if $A$ is a square idempotent matrix. Find $(I+A)^{-1}$ such that the expression doesn't have inverses.

Question

To prove:

$\det(I+A)$ = $2^{\operatorname{rank}(A)}$ if $A \in$ $\mathbb{R}^{n\times n}$ and $A^{2}=A$.

Find an expression for $(I+A)^{-1}$ such that it does not involve inverses.

Is there any way I could use Jordan Block? I am not sure where I need to start in order to solve it.

Answer 1

Idempotent matrices are diagonalizable to a matrix of the form $$B=\begin{pmatrix}I_m \\ & 0_{n-m}\end{pmatrix}$$ where $m=\text{rank }A$, $I_m$ is an identity matrix, and $0_m$ is a matrix of all zeroes. (This can be proved using the Jordan normal form and inspection of each block individually.) That is, $A=SBS^{-1}$, so $I+A=SS^{-1}+SBS^{-1}=S(I+B)S^{-1}$ where $$I+B=\begin{pmatrix}2I_m \\ & I_{n-m}\end{pmatrix}$$ and so $\det(I+A)=\det(S(I+B)S^{-1})=2^m=2^{\text{rank }A}$, as desired.

Hint for the second part: if $A$ was a formal variable you could write down a power series (think of a series for $\frac{1}{1+x}$). But because of $A$'s idempotency, any power series $\sum a_nA^n$ simplifies to one of the form $aI+bA$ (for $a_n,a,b\in\mathbb{C}$). Try using this form to find an inverse to $I+A$.

Answer 2

If $p$ is a polynomial with non-zero constant term and $p(B)=0$ you can use $p$ to find $B^{-1}$. So it seems reasonable to try setting $B=I+A$ and seeing what $p$ we can find. $$0=A^2-A=(B-I)^2-(B-I)=B^2-3B+2I,$$or $$2I=B(3I-B),$$so $$B^{-1}=\frac12(3I-B)=I-\frac12A.$$