1. Problem of Idempotent matrix

Assume $\displaystyle A$ is idempotent matrix of order n,in other words:$\displaystyle A^2 = E_{n}$,and $\displaystyle det(A) > 0$.show that:$\displaystyle det(A + E_{n}) \not=0$

2. Originally Posted by Xingyuan
Assume $\displaystyle A$ is idempotent matrix of order n,in other words:$\displaystyle A^2 = E_{n}$,and $\displaystyle det(A) > 0$.show that:$\displaystyle det(A + E_{n}) \not=0$
wait a second! idempotent means $\displaystyle A^2=A$ not $\displaystyle A^2=I.$ are you sure you wrote the question correctly?

3. 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.

4. Originally Posted by Xingyuan
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.
well, this is not very helpful! post the question exactly as you've seen it. do not add anything to it! then i'll help you to solve it.

5. OK, Assume $\displaystyle A^2 = E_{n}$,and $\displaystyle det(A) > 0$,show that :$\displaystyle det(A + E_{n}) \not= 0$

thanks very much !

6. Originally Posted by Xingyuan
OK, Assume $\displaystyle A^2 = E_{n}$,and $\displaystyle det(A) > 0$,show that :$\displaystyle det(A + E_{n}) \not= 0$

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 $\displaystyle A = -E_n.$

7. Oh,yes .If n is odd .the conclusion is correct.thanks very much.

If $\displaystyle A^2 = A$,and $\displaystyle det(A) > 0$. can we get $\displaystyle det(A + E_{n})\not=0$?

8. Originally Posted by Xingyuan

Oh,yes .If n is odd .the conclusion is correct.thanks very much.
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 $\displaystyle \det(A + I)=0.$

If $\displaystyle A^2 = A$,and $\displaystyle det(A) > 0$. can we get $\displaystyle det(A + E_{n})\not=0$?
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 $\displaystyle \det A \neq 0.$

9. Thanks Very Much!!