This is probably a silly question, but I just confused myself, and I'd love some clarification. I know that the definition of eigenvalues does not depend on a basis, and therefore they are manifestly basis independent. However, consider the following example:
We have an invertible operator $\Lambda$ with an eigenbasis $\{\vert e_i \rangle \}$ in which $\Lambda = \text{diag} \{ \lambda_i \}$. We just rescale the eigenbasis: $$ \vert e_i \rangle \to \vert \tilde e_i \rangle = \dfrac{1}{\sqrt{\vert{\lambda_i} \vert }} \vert e_i \rangle .$$
$A$ is still diagonal in our new basis: $\langle{\tilde e_i} \vert \Lambda \vert \tilde e_j \rangle = \text{sgn}(\lambda_i) \delta_{ij}$. $\{\tilde e_i \}$ is obviously an eigenbasis and so the eigenvalues seem to have changed to $\pm 1$. What is going wrong?
I know that if I do it the usual way, nothing goes wrong: $$\Lambda \vert \tilde e_i \rangle = \frac{1}{\sqrt{\vert \lambda_i \vert}} \lambda_i \vert e_i \rangle = \lambda_i \vert \tilde e_i \rangle .$$
The two results seem to be inconsistent. This feels like a really silly question, but for whatever reason I'm really confused.
Best Answer
So apparently, the common conception that you see all your silly mistakes a few seconds after posting something on the internet is true!
The diagonals give you the eigenvalues only when your eigenbasis is orthonormal. This can be clearly seen by projecting the eigenvalue problem onto your eigenbasis.
$$ \langle \lambda_j \vert \Lambda \vert \lambda_i \rangle = \lambda_i \langle \lambda _j \vert \lambda_i \rangle .$$
After rescaling, that obviously no longer holds. Also, that's why we always normalize bases in QM haha!