Problem of Idempotent matrix

**Xingyuan** · May 22nd 2009, 10:12 PM

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

**NonCommAlg** · May 22nd 2009, 10:16 PM

Originally Posted by Xingyuan

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

wait a second! idempotent means $A^2=A$ not $A^2=I.$ are you sure you wrote the question correctly?

**Xingyuan** · May 22nd 2009, 10:51 PM

I am not sure. I am reading a book of Linear Algebra. this problem is in that book. if n is odd ,and $A^2 = E_{n}$ ,I have proved $det(A + E_{n}) \not=0$ . when n is even ,i have no idea.
If $A^2 = A$ ,I am not sure whether the conclusion was right?

thanks very much.

**NonCommAlg** · May 22nd 2009, 10:55 PM

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 $A^2 = E_{n}$ ,I have proved $det(A + E_{n}) \not=0$ . when n is even ,i have no idea.
If $A^2 = A$ ,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.

**Xingyuan** · May 22nd 2009, 11:08 PM

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

thanks very much !

**NonCommAlg** · May 23rd 2009, 12:30 AM

Originally Posted by Xingyuan

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

thanks very much !

assuming that $E_n$ 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: $n$ any even number and $A = -E_n.$

**Xingyuan** · May 23rd 2009, 01:45 AM

Oh,yes .

If n is odd .the conclusion is correct.thanks very much.

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

**NonCommAlg** · May 23rd 2009, 02:24 AM

Originally Posted by Xingyuan

Oh,yes .

If n is odd .the conclusion is correct.thanks very much.

for odd values of $n$ the problem is still false. a counter example is $A=\begin{bmatrix}-1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}.$ we have $A^2=I, \ \det A= 1 > 0$ but $\det(A + I)=0.$

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

yes, and we only need $\det A \neq 0.$ the proof is very simple: $A(A+I)=A^2 + A=2A.$ thus: $(\det A) \det (A+I)=2^n \det A$ and hence $\det(A+I)=2^n,$ because $\det A \neq 0.$

**Xingyuan** · May 24th 2009, 05:32 AM

Thanks Very Much!!

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