1. ## Eigenvector and eigenvalue. Need help please.

Let A be an n x n symmetric matrix, and let λ be an eigenvalue of A with corresponding eigenvector x.

Show that λ is also an eigenvalue of the matrix P^T AP where P is an orthogonal matrix. State the corresponding eigenvector of P^T AP.

How is the result modified if P^T AP is diagonal matrix D?

2. Could you show it using the determinate,i.e.,
$\det(P^{T}AP) = \det(P^TP)\det(A) = \det(I)\det(A) = \det(A)$.

3. I am unclear of this, could you explain further please?

4. Well, the eigenvalues of a matrix are given by $\det(\lambda I - A) = 0$, but
$P^T(\lambda I - A)P = (\lambda P^TP - P^{T}AP) = (\lambda I - P^{T}AP)$. However, we can prove that $\det(\lambda I - P^{T}AP) = \det(P^T(\lambda I - A)P) = \det(\lambda I - A)$ by the same reasoning above. Since these have the same solution, i.e., $\lambda$, then the eigenvalues of $P^{T}AP\text{ and } A$ are the same. Is that a bit clearer? It is not a formal proof, but that should be left up to you to write out.

5. Thanks, but do you have any idea of how the the result is modified if P^T AP is a diagonal matrix D?

6. The only thing I can think of is that the eigenvalues will be the diagonal.

7. I am sorry but your above explanation is unclear to me still. Can you tell me how i can write this out as a proper answer. I am very weak with matrices so find it hard to follow.

Thank you very much for your help.