Hey math2011.
Isn't one of the definitions of a positive define matrix having positive eigenvalues?
Positive-definite matrix - Wikipedia, the free encyclopedia
QUESTION:
Let be an matrix with a factorization , where is a unit lower triangular matrix and is a diagonal matrix. Show that is positive definite if and only if , for .
ATTEMPT:
Assume for . Then is possible and
where .
Because has pivots it is nonsingular and hence if and only if . Therefore for all .
Assume is positive definite. How can I prove that has positive diagonals?
Hey math2011.
Isn't one of the definitions of a positive define matrix having positive eigenvalues?
Positive-definite matrix - Wikipedia, the free encyclopedia
Is your L matrix the matrix of eigenvectors and also is that matrix orthonormal? If so then L^T = L^(-1) and you have a eigen-decomposition of your matrix G which means that D is your matrix of eigenvalues.
I don't know if L is the matrix of eigenvectors. The problem does not state that explicitly. Are you suggesting that L actually is the matrix of eigenvectors? How can I prove it?
My suggestion is to show that LDL^t is the diagonal form by showing that L is non-singular (since it is triangular with non-zero determinant) and then use this to show that L is also orthonormal.
For the ortho-normal aspect, if you can show L*L^t = I then you're done.
By showing that L^t = L^(-1) and by using the assumption that the diagonalization produces a unique decomposition, then you have shown that this is indeed the diagonalization.
Thanks for explaining but I still cannot see how to do it. L^t is a unit upper triangular matrix and L is a unit lower triangular matrix, L^t L does not necessarily equal to the identity matrix unless the L is special. The problem does not state any other information about L that directly say L is orthogonal.