| 18 Apr 2008 07:28:16 pm by simonzack |
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I got some problems in linear algebra, the rest were easy, but I just couldn't solve one, maybe I need more practice...
It goes like this:
for square matrix A, if |A|=-1, transpose(A)=invert(A), then prove that |I+A|=0
Can anybody please help me?
Thanks |
| 18 Apr 2008 10:31:50 pm by alexrudd |
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I think I have it figured out for 2x2 matrices, but since my approach was analytical and not grounded in theories of matrices I can't scale it up to n x n. I'm using the calculator's formatting for matrices since it's familiar and doens't require whitespace.
Let's say A =
[a,b]
[c,d]
Given |A| = -1, you know ad-bc=-1. (1)
A-1 can be calculated since you know the determinant.
[-d,b]
[c,-a]
AT is trivial to calculate, too.
[a,c]
[b,d]
Now we can see that a=-d, and c=b. Let's use this information with our original information from the determinant (1) to rewrite it with only two variables.
-a2-b2=-1, or
a*a+b*b=1
The only way this is possible is if a=±1 and b = 0, or the other way around.
This means our matrix must be either
[±1,0]
[0,∓1] (case 1)
or
[0,±1]
[±1,0] (case 2)
Add I to case 1.
[1±1, 0]
[0, 1∓]
The determinant is then going to be (1∓1)*(1∓1) - 0*0, which will be 0
Add I to case 2
[1,±1]
[±1,1]
This determinant is (±1)2-12, which is also 0.
****
I am almost certain that I will feel rather dumb soon when someone else posts a simpler answer that's actually a proof and works with sizes greater than 2x2. What the hell, it was an interesting thinking exercise. |
| 18 Apr 2008 10:45:10 pm by simonzack |
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Yeah, thanks for the reply, I got it for 2*2 matrices as well like your method, but I don't think it's easy to be scaled up to get the general form of n*n, as for each matrix A, it's child (think that's what its called) B transpose(A) doesn't have to equal inverse(A)
Edit:
so, urr, any hints or solutions? |
| 19 Apr 2008 05:47:20 am by CoBB |
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| alexrudd wrote: |
a*a+b*b=1
The only way this is possible is if a=±1 and b = 0, or the other way around. |
That’s not true. The equation is satisfied for any x if you define a = sin(x) and b = cos(x).
I can’t really think of a simple proof at the moment. You have to realise that if AT = A-1, then A is an orthogonal matrix, and its determinant can only be 1 (rotation) or -1 (rotoinversion) without giving any more constraints. It is possible to find a matrix P (a linear transformation) that transforms A into a canonical (diagonal) form D: PAP-1=D. D has non-zero elements only in its main diagonal, either in 2x2 elementary rotation blocks or just solitary ±1s (identities or reflections). I+A is a polynomial of A, therefore it is left intact by the transformation P: I+A=P-1(I+D)P (this is easy to verify by simple substitution). Using the basic property of determinants that |A||B|=|AB| we can see that |I+A|=|I+D|, and also that |A|=|D|. Since |D|=-1, at least one of its 2x2 blocks must have a determinant of -1 or its diagonal must contain a -1 somewhere. Therefore, adding the identity matrix to it will introduce a zero factor that will make the determinant of the sum (I+D) zero, and we’re done.
If this is over the top, read up on orthogonal matrices and diagonalisation. |
| 19 Apr 2008 06:11:43 am by simonzack |
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wow, thanks for the solution
the book I was reading hasn't introduced orthogonal matrices at all, so i didn't know anything about them  , but i think its got to be somewhere in the later chapters
I've checked on them, and took some stuff in now. Thanks |
