acidhoony Posted March 16, 2011 Share Posted March 16, 2011 now i'm using apostol calculus book user, the problem says that given n*n matrix A with real entries such that A^2=-I I=UNIT MATRIX THEN PROVE THAT det A = 1 i know that (det A)^2=1 so, det A = +1 or -1 but i cannot prove that why -1 is not how can we prove this thing without using cayley-hamiltion Th. ;because it's proof is either complicated i think this problem may not using cayley hamiltion Th. please solve this problem~ Link to comment Share on other sites More sharing options...

the tree Posted March 17, 2011 Share Posted March 17, 2011 If there is a proof not reliant on Cayley Hamilton, then I can't see it. I imagine it would be acceptable to take that theorem as a given for the sake of homework. Link to comment Share on other sites More sharing options...

DrRocket Posted March 17, 2011 Share Posted March 17, 2011 now i'm using apostol calculus book user, the problem says that given n*n matrix A with real entries such that A^2=-I I=UNIT MATRIX THEN PROVE THAT det A = 1 i know that (det A)^2=1 so, det A = +1 or -1 but i cannot prove that why -1 is not how can we prove this thing without using cayley-hamiltion Th. ;because it's proof is either complicated i think this problem may not using cayley hamiltion Th. please solve this problem~ Let [math] A= \left ( \begin{array} {cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right )[/math] Then [math]A^2 = I[/math] and det[math]A=-1[/math] Link to comment Share on other sites More sharing options...

acidhoony Posted March 17, 2011 Author Share Posted March 17, 2011 hello thank you for reply. DrRocket.. but.. i said that A^2 is not I BUT -I PLEASEEEGIVE ME ANOTHER PROOF.. Link to comment Share on other sites More sharing options...

DrRocket Posted March 17, 2011 Share Posted March 17, 2011 hello thank you for reply. DrRocket.. but.. i said that A^2 is not I BUT -I PLEASEEEGIVE ME ANOTHER PROOF.. [math]A^2= \-I[/math] Note: [math](det \ A)^2=det(-I)=(-1)^n \ \Rightarrow \ n[/math] is even. Note also that the conclusuin that n is even and [math] det \ A \ = \ \pm 1 [/math] would not hold over the complex field, so the proof must take advantage of the fact that we are working over the reals. Sketch of proof: 1. Using Jordan canonical form A can be assumed to be block diagonal, with 2x2 blocks. This reduces the problem to the case n=2. 2. So now [math] A \in (O2)[/math]. [math]A[/math] is thus a composition of a rotation and, if [math]det \ A \ = \ -1[/math], an orientation changing transformation, which can be taken to be . [math] C= \left ( \begin{array} {cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right )[/math] Note that [math]C[/math] does not commute with rotations, but yet that any element in [math]O2[/math] can be written as [math]B[/math] or [math]BC[/math] where [math]B[/math] is some rotation. 3. Now examine the possibilities [math]A= BC[/math] where [math]B[/math] is a rotation and show that [math](BC)^2 \ne \ -I[/math] . 4. That the shows that [math] A [/math] is a rotation through [math]\frac {\pi}{2}[/math] or [math]\frac {3 \pi}{2}[/math] and hence has determinant 1. Link to comment Share on other sites More sharing options...

acidhoony Posted March 18, 2011 Author Share Posted March 18, 2011 [math]A^2= \-I[/math] Note: [math](det \ A)^2=det(-I)=(-1)^n \ \Rightarrow \ n[/math] is even. Note also that the conclusuin that n is even and [math] det \ A \ = \ \pm 1 [/math] would not hold over the complex field, so the proof must take advantage of the fact that we are working over the reals. Sketch of proof: 1. Using Jordan canonical form A can be assumed to be block diagonal, with 2x2 blocks. This reduces the problem to the case n=2. 2. So now [math] A \in (O2)[/math]. [math]A[/math] is thus a composition of a rotation and, if [math]det \ A \ = \ -1[/math], an orientation changing transformation, which can be taken to be . [math] C= \left ( \begin{array} {cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right )[/math] Note that [math]C[/math] does not commute with rotations, but yet that any element in [math]O2[/math] can be written as [math]B[/math] or [math]BC[/math] where [math]B[/math] is some rotation. 3. Now examine the possibilities [math]A= BC[/math] where [math]B[/math] is a rotation and show that [math](BC)^2 \ne \ -I[/math] . 4. That the shows that [math] A [/math] is a rotation through [math]\frac {\pi}{2}[/math] or [math]\frac {3 \pi}{2}[/math] and hence has determinant 1. i have another solution please read me dr rocket we haver real matrix A so we have eigen vector which is unreal. and this eigen vector is conjugated because the polynomier of matrix A is "poly"nomier it self and every coefficienet is "real" every eigen value is conjugated!! and A^2 = - I => A^2X=-X =AAX=AaX=a^2x (here a is eigenvalue of matrix A) so we know that (eivenvalue of A )^2 = -1 matrix A is 2k*2k matrix either so we have a1 ................ak, a(k+1),a(k+2),,,,,a2k and they are a k pair of conjugated eigenvalue eigenvalue's product is 1 i didn't prove it well but i think you will be understood it. Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now