I am applying the powermethod to get good approximations of eigenvalues of a matrix. This matrix is symmetric and got distinct (and ofcourse non complex eigenvalues).
Now i need to show: why if i start with a vector which is orthogonal to eigenvector 1, where eigenvector 1 is the eigenvector of the eigenvalue largest in absolute value, the power method will converge to labda 2, the eigenvalue second largest in absolute value.
As a hint: note that A is symmetric and write x0 as a linear combination of eignevectors.
I also need to prove the following: Put a=(labda1/(v*v^t)) (where v is the eigenvector corresponding to the largest eigenvalue). Prove that the eigenvalue of A-a(v*v^t) which is largest in absolute value equals labda 2, where labda 2 is again the eigenvalue second largest in absolute value of A.
As a hint: First show that A and A-a(v*v^t) have the same eigenvectors.
I hope someone can help me with (one or both) of the statements as i am kinda stuck in it.