Assume $\displaystyle A$ is idempotent matrix of order n,in other words:$\displaystyle A^2 = E_{n}$,and $\displaystyle det(A) > 0$.show that:$\displaystyle det(A + E_{n}) \not=0$
I am not sure. I am reading a book of Linear Algebra. this problem is in that book. if n is odd ,and $\displaystyle A^2 = E_{n}$,I have proved $\displaystyle det(A + E_{n}) \not=0$. when n is even ,i have no idea.
If$\displaystyle A^2 = A$,I am not sure whether the conclusion was right?
thanks very much.
assuming that $\displaystyle E_n$ is the identity matrix (this is an unusual notation for the identity matrix and you should've mentioned what it means!) this still can't be the question because it's false!
a trivial counter-example: $\displaystyle n$ any even number and $\displaystyle A = -E_n.$
for odd values of $\displaystyle n$ the problem is still false. a counter example is $\displaystyle A=\begin{bmatrix}-1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}.$ we have $\displaystyle A^2=I, \ \det A= 1 > 0$ but $\displaystyle \det(A + I)=0.$
yes, and we only need $\displaystyle \det A \neq 0.$ the proof is very simple: $\displaystyle A(A+I)=A^2 + A=2A.$ thus: $\displaystyle (\det A) \det (A+I)=2^n \det A$ and hence $\displaystyle \det(A+I)=2^n,$ because $\displaystyle \det A \neq 0.$
If $\displaystyle A^2 = A$,and $\displaystyle det(A) > 0$. can we get $\displaystyle det(A + E_{n})\not=0$?