Yes, ABv=lambda v then BA(Bv)=lambda Bv
No, the matrices change size and therefore the number of eigenvalues.
No, since you can find a counter example where AB is non - symmetric
Yes
Yes
z^TMz > 0 for any column vector z. Square and symmetric and all the eigenvalues are positive. (not necessarly symmetric)
If it has full column rank, nonzero determinante and square.
B=LL^T, where L is a lowertraingular matrix, with real and positive diagonal entries. Possible id B is possitvie semi definite. For a hermitian matrix A, A=LL*.
A=A^H, where H denotes the complex conjugate and transpose.
A=QUQ^-1, where Q is a unitary matrix such rhat the inverse of Q is also the conjugate transpose. U is an uppertriangular matrix.
A matrix is unitary if its inverse equals its conjugate transpose.
cos(theta)=(x^Ty)/(||x||||y||)
x_k = Ax_k/||Ax_k||
The largest eigenvalue, since it is the dominant term in the linear combination. lambda_1^p grows fast then the others, where p is the number of iterations.