# linear algebra help

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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.

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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.

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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.

Let $A= \left ( \begin{array} {cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right )$

Then $A^2 = I$ and det$A=-1$

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hello

but..

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

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hello

but..

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

$A^2= \-I$

Note: $(det \ A)^2=det(-I)=(-1)^n \ \Rightarrow \ n$ is even.

Note also that the conclusuin that n is even and $det \ A \ = \ \pm 1$ 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 $A \in (O2)$. $A$ is thus a composition of a rotation and, if $det \ A \ = \ -1$, an orientation changing transformation, which can be taken to be

.

$C= \left ( \begin{array} {cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right )$

Note that $C$ does not commute with rotations, but yet that any element in $O2$ can be written as $B$ or $BC$ where $B$ is some rotation.

3. Now examine the possibilities $A= BC$ where $B$ is a rotation and show that $(BC)^2 \ne \ -I$ .

4. That the shows that $A$ is a rotation through $\frac {\pi}{2}$ or $\frac {3 \pi}{2}$ and hence has determinant 1.

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$A^2= \-I$

Note: $(det \ A)^2=det(-I)=(-1)^n \ \Rightarrow \ n$ is even.

Note also that the conclusuin that n is even and $det \ A \ = \ \pm 1$ 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 $A \in (O2)$. $A$ is thus a composition of a rotation and, if $det \ A \ = \ -1$, an orientation changing transformation, which can be taken to be

.

$C= \left ( \begin{array} {cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right )$

Note that $C$ does not commute with rotations, but yet that any element in $O2$ can be written as $B$ or $BC$ where $B$ is some rotation.

3. Now examine the possibilities $A= BC$ where $B$ is a rotation and show that $(BC)^2 \ne \ -I$ .

4. That the shows that $A$ is a rotation through $\frac {\pi}{2}$ or $\frac {3 \pi}{2}$ and hence has determinant 1.

i have another solution

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.

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