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

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- May 22nd 2009, 10:12 PMXingyuanProblem 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:$\displaystyle det(A + E_{n}) \not=0$(Headbang)

- May 22nd 2009, 10:16 PMNonCommAlg
- May 22nd 2009, 10:51 PMXingyuan
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. - May 22nd 2009, 10:55 PMNonCommAlg
- May 22nd 2009, 11:08 PMXingyuan
OK, Assume $\displaystyle A^2 = E_{n}$,and $\displaystyle det(A) > 0$,show that :$\displaystyle det(A + E_{n}) \not= 0$

thanks very much ! - May 23rd 2009, 12:30 AMNonCommAlg
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 $\displaystyle A = -E_n.$ - May 23rd 2009, 01:45 AMXingyuan
Oh,yes .(Itwasntme)If n is odd .the conclusion is correct.thanks very much.

If $\displaystyle A^2 = A$,and $\displaystyle det(A) > 0$. can we get $\displaystyle det(A + E_{n})\not=0$? - May 23rd 2009, 02:24 AMNonCommAlg
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 $\displaystyle \det(A + I)=0.$

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

- May 24th 2009, 05:32 AMXingyuan
Thanks Very Much!!