Leading the way to the Future
 18 Apr 2008 07:28:16 pm by simonzack 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 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 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?
 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 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