Recent content by AngrySaki

Thanks, that helps a ton.

I'm going through the book: Principles of Quantum Mechanics 2nd edition by R. Shankar 1. In the mathematical introduction to projection operators (page 22), it writes: "Consider the expansion of an arbitrary ket |V\rangle in a basis: |V\rangle=\sum_{i}^{} |i\rangle\langle i|V\rangle"...

It would come from the fact that if the matrix is singular, then column one a multiple of the other (or one row is a multiple of the other). Maybe this isn't how it's supposed to be done though, so you might be better off waiting for somebody else's ideas than going off mine :/

I'm no linear algebra expert, but I would start by writing that if the matrix is written: a b c d Then it's singular if a = x*b and c=x*d (or a=x*c and b=x*d), and work my way to the formula for the determinant.

Oh, oops. I mean to write eigenvectors. Just to be clear, does the method makes sense if I had written the word eigenvectors instead of eigenvalues?

Thanks, that helps a lot. Does this makes sense as an answer? Eigenvalues of A can be written: (\lambda I -A)X=0 So the right side of the original equation is: P(\lambda I -A)X=0 Move the P inside: (\lambda P -PA)X=0 Multiply by I: (\lambda P -PA)P^{-1}PX=0 Move P inverse...

The question is at the end of a chapter on spanning vector spaces. Homework Statement Let P denote an invertible n x n matrix. If \lambda is a number, show that E_{\lambda}(PAP^{-1}) = \left\{PX | X\;is\;in\;E_{\lambda}(A)\right\} for each n x n matrix A. [Here E_{\lambda}(A)} is...