# Matrices Watch

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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.

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#2

(Original post by

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.

**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.

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(Original post by

What wrong with C=I? Or for variety C=-I?

**RichE**)What wrong with C=I? Or for variety C=-I?

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

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#4

(Original post by

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

**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

*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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(Original post by

I'm sorry I thought you wanted

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.

**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.

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(Original post by

its a rotation..... 90 degrees....

**anchemis**)its a rotation..... 90 degrees....

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#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

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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#9

(Original post by

Hmm, can't get you.

**BCHL85**)Hmm, can't get you.

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.

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#10

(Original post by

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

**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

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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.

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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(Original post by

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.

**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.

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#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.

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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(Original post by

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.

**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.

So I got it now, thanks

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