# Problem of Idempotent matrix

• May 22nd 2009, 10:12 PM
Xingyuan
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:$\displaystyle det(A + E_{n}) \not=0$(Headbang)
• May 22nd 2009, 10:16 PM
NonCommAlg
Quote:

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:$\displaystyle det(A + E_{n}) \not=0$(Headbang)

wait a second! idempotent means $\displaystyle A^2=A$ not $\displaystyle A^2=I.$ are you sure you wrote the question correctly?
• May 22nd 2009, 10:51 PM
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.
• May 22nd 2009, 10:55 PM
NonCommAlg
Quote:

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.
• May 22nd 2009, 11:08 PM
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 !
• May 23rd 2009, 12:30 AM
NonCommAlg
Quote:

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 $\displaystyle A = -E_n.$
• May 23rd 2009, 01:45 AM
Xingyuan
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 AM
NonCommAlg
Quote:

Originally Posted by Xingyuan

Oh,yes .(Itwasntme)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 $\displaystyle \det(A + I)=0.$

Quote:

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 $\displaystyle \det A \neq 0.$
• May 24th 2009, 05:32 AM
Xingyuan
Thanks Very Much!!