S-1 is the inverse of S. thus we are asked to show that any matrix similar to A is complete.
I need to prove that if A is complete then S-1AS is complete where S is a nonsingular matrix. A complete matrix is one whose eigenvectors span Rn. Can anyone look over what i've done below. i believe it's correct but i have no confidence. i need a little reassurance. thanks!
S-1AS = B. A is complete. Every eigenvalue of A is also an eigenvalue of B. To show this we suppose Av = λv. Then (S-1A)v = λ(S-1)v. This is equivalent to (S-1ASS-1)v = (S-1AS)( S-1)v = λ(S-1)v. This shows that λ is also an eigenvalue of S-1AS and the eigenvector is S-1v. And so if v is an eigenvector of A the S-1v is an eigenvector of S-1AS. If we consider S-1V where V is the matrix whose columns are eigenvectors of A. Because by assumption A is complete V is invertible (since it's columns are linearly independent). But S-1 is invertible and so therefore is S-1V. The columns of S-1V are S-1vi for each eigenvector of A. But the S-1vi are the eigenvectors of S-1AS. Since S-1V is nonsingular it's columns are linearly independent and therefore span Rn. This is what it means for S-1AS to be complete and we are done.
Every eigenvalue of with eigenvector is also an eigenvalue of with eigenvector . This follows from
Let be the matrix whose columns are the eigenvectors of Then the above shows that has columns that are the eigenvectors of
A matrix is complete if and only if its matrix of eigenvectors is invertible. So is invertible and since is invertible, so is Hence is complete.