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#1
I've just started it, find it ....not so easy. Plz someone can help. The question is find matrix C so that C^4 = I.
0
14 years ago
#2
(Original post by BCHL85)
I've just started it, find it ....not so easy. Plz someone can help. The question is find matrix C so that C^4 = I.
What wrong with C=I? Or for variety C=-I?
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#3
(Original post by RichE)
What wrong with C=I? Or for variety C=-I?
Sorry, I forgot that C is 2x2 matrix.
I just done C^2 = I. which gives the matrix
a b
c -a
where |C| = -1 (a^2 + bc = 1).
But then question asks C^4 = I, and I don't know how to do. It's not just I and -I, i think
0
14 years ago
#4
(Original post by BCHL85)
Sorry, I forgot that C is 2x2 matrix.
I just done C^2 = I. which gives the matrix
a b
c -a
where |C| = -1 (a^2 + bc = 1).
But then question asks C^4 = I, and I don't know how to do. It's not just I and -I, i think
I'm sorry I thought you wanted a C not all of them.

But you've missed out some solutions to C^2 = I.

Clearly C = I or C = -I satisfies this equation but aren't of the form you've given. But these are the only ones you've missed.
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#5
(Original post by RichE)
I'm sorry I thought you wanted a C not all of them.

But you've missed out some solutions to C^2 = I.

Clearly C = I or C = -I satisfies this equation but aren't of the form you've given. But these are the only ones you've missed.
Yeah, right, cuz I supposed b and c were not 0
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14 years ago
#6
its a rotation..... 90 degrees....
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#7
(Original post by anchemis)
its a rotation..... 90 degrees....
Hmm, can't get you.
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14 years ago
#8
M^n = I is always a rotation where n is any integer... the proper transformation of the matrix M is thus defined as a rotation about O of 360/n degrees....

so M^0 = I --> M is an identity matrix
M^2 = I --> M is a 180 degree rotation
M^3 = I --> M is a 120 degree rotation etc
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14 years ago
#9
(Original post by BCHL85)
Hmm, can't get you.
The matrix

0 -1
1 0

represents an anti-clockwise rotatation about the origin through a right angle.

It certainly satisfies C^4 = I.

But all matrices that satisfy C^4=I aren't rotations.

Complex matrices that satisfy C^4 = I can be characterised as being similar to diagonal matrices that have entries from 1, -1, i and -i, but I'm not sure what their general form would be.

If you know the matrix is 2x2 you can say that its minimal polynomial has degree at most 2 and divides x^4-1. So the possibilities are x-1, x+1, x^2-1, x^2+1 for real matrices. The first two cases give I and -I, the third you've done and the latter you could go through the process of sticking in a general 2x2 matrix.
0
14 years ago
#10
(Original post by anchemis)
M^n = I is always a rotation where n is any integer... the proper transformation of the matrix M is thus defined as a rotation about O of 360/n degrees....

so M^0 = I --> M is an identity matrix
M^2 = I --> M is a 180 degree rotation
M^3 = I --> M is a 120 degree rotation etc
The logic doesn't go like this though. There are plenty of other matrices that satisfy M^2=I, for example every matrix representing a reflection.
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#11
Ok, so it means I rotates through 90 degree. (both anti-clockwise and clockwise)
We have 2 matrices
0 -1
1 0
and
0 1
-1 0.
But if consider just C^2 = I, that gives C = I or C = -I as the rotation of I through 180.
If C =
2 1
-3 -2
I still have C^2 = 1, and it's obviously not rotation of I through 180.
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#12
(Original post by RichE)
The logic doesn't go like this though. There are plenty of other matrices that satisfy M^2=I, for example every matrix representing a reflection.
Right, that's what Im wondering
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14 years ago
#13
To clarify my earlier comment, which I realise might not have been that helpful becuase it was quite rushed:-

If a 2x2 matrix C satisfies C^4 = I then in fact it is the case that C^2 = I or -I. This might not seem that obvious, but note that

(C^2)^2 = C^4 = I

and det(C^2) = (detC)^2 >0.

So C^2 is a 2x2 matrix which squares to I and has positive determinant. From your earlier work you know the only such matrices are I and -I.

You already know how to solve C^2=I and you can use a similar argument to solve C^2 = -I.
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#14
(Original post by RichE)
To clarify my earlier comment, which I realise might not have been that helpful becuase it was quite rushed:-

If a 2x2 matrix C satisfies C^4 = I then in fact it is the case that C^2 = I or -I. This might not seem that obvious, but note that

(C^2)^2 = C^4 = I

and det(C^2) = (detC)^2 >0.

So C^2 is a 2x2 matrix which squares to I and has positive determinant. From your earlier work you know the only such matrices are I and -I.

You already know how to solve C^2=I and you can use a similar argument to solve C^2 = -I.
Ah, right. I did work it for myself and can't get any matrix if it's not I or -I, cuz of det(C^2) can't be negative.
So I got it now, thanks
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