linear algebra help

[acidhoony]

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~

 

 

[the tree]

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.

[DrRocket]

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]

[acidhoony]

hello

 

thank you for reply. DrRocket..

 

but..

 

i said that A^2 is not I BUT -I

 

PLEASEEEGIVE ME ANOTHER PROOF..

[DrRocket]

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.

[acidhoony]

[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.