Recent content by Codezion

Thank you DH! I have corrected my problem - I think all the latex syntax confused me.

Thanks for the reply! what is A dagger? Here is the actual question: http://books.google.com/books?id=bj-Lu6zjWbEC&printsec=frontcover#PPA85,M1 (look at 11.1) Thank you!

Homework Statement Prove that the ||A+|| \leq ||A1-1|| Where A+=(ATA)-1AT, ||.|| is the 2 norm and A is an mxn matrixHomework Equations A = [\stackrel{A1}{A2}] where A1 is an nxn nonsingular square matrix and A2 is any random matrix that is (m-n)xn The Attempt at a Solution All I did was...

Homework Statement If two square matrices, A and B are unitarily equivalent then A = QBQ* for some unitary Q of the same size as A and B. Prove that A and B are unitarily equivalent if and only if they have the same singular values Homework Equations The Attempt at a Solution I...

OH! I see it now :) if x = u, then x + uv*x = u + uv*u = u + -u = 0 You are right Dick, it was easy! :) Thanks a bunch!

Thanks a lot for your help guys. Dick, I think it could be the pressure:), but I still cannot see it. Here is what I did v*x is some constant beta, x + uv*x = x + beta u => x = -beta u Therefore, I concluded x is a scalar multiple of u. This is what I came up with, but it looks dodgy because...

If beta = -1 (which would imply A is singular), what would the nullspace of A be? I got only as far as writing the equation as: (I+uv*)x = 0 x + uv*x = 0... Do you have any suggestions?

I like the expression (placebo). Yup, thanks Hurkyl. Thanks so much.

Thank you! Figured it out :) (I + uv*) (I + alpha uv*)x = x (I + alpha Iuv* + Iuv* + alpha uv*uv*)x = x v*u is a scalar and let it be beta (I + alpha Iuv* + Iuv* + alpha beta uv*)x = x [I + Iuv*(alpha + 1 + alpha beta)]x = x this implies (alpha + 1 + alpha beta) = 0! alpha = -1/(1+beta)!

Wow! Thank you I missed that! I was assuming (AB)* = A*B* all along - even after I read your post! Thank you. If A is not an orthogonal matrix, then I guess I am back to square one :(

(1) - I think one of the definition of an Orthogonal matrix is a square matrix, whose columns are all linearly independent so that they make the basis for an m space. One of the properties of a nonsingular matrix is that it has the maximum rank (in this case m since it is a square matrix). These...

Wow! Thank you. I did not see that <x,x> is always positive, I was just focusing on it being nonzero! Thank you so much!

Homework Statement If A = I + uv*, where u and v are m vectors and A is known to be nonsingular, show that the inverse of A = I + \alphauv* where \alpha is a scalar value Homework Equations The Attempt at a Solution Since A is nonsingular, we know the rank of A is m. Since both...

Hmm...I played with it overnight and cannot figure out what is wrong with that statement. The only thing I can see is that x*x cannot be zero since x is a non zero vector..?

Taking the square root of each side...Sqrt(I) = S, however, the square root of an identity matrix is by no means a skew Hermitian, hence the original assumption is erroneous! Thanks Dick!