Unique of eigenbasis of self-adjoint operator.

Question

Today I am reading an article: Eigenvalues and sums of Hermitian matrices there is an exercise copy from that article:

Exercise 1 Suppose that the eigenvalues $\lambda_1(A)>\cdots>\lambda_n(A)$ of an $n\times n$ Hermitian matrix are distinct. Show that the associated eigenbasis $u_1(A),\cdots,u_n(A)$ is unique up to rotating each individual eigenvector $u_j(A)$ by a complex phase $e^{i\theta_j}$. In particular, the spectral projections $P_j(A):=u_j(A)^*u_j(A)$ are unique. what happens when there is eigenvalue multiplicity?

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Assume $A$ is the matrix of self-adjoint operator $\mathcal{A}$ under some property basis of $V_{\Bbb{C}}$.
I don't know why eigenbasis $u_1(A),\cdots,u_n(A)$ is unique up to rotating. If we multiply a complex number $c_k \not=0$ respectively, then $c_1u_1(A),\cdots,c_nu_n(A)$ is also a basis of $V_{\Bbb{C}}$ consist of $\mathcal{A}$'s eigenvectors,under this basis, $\mathcal{A}$ correspond to a diagonal matrix consist of eigenvalues: $\text{diag}\{\lambda_1,\cdots,\lambda_n\}$. I think it is unique up to multiply a complex number $c_k$ respectively. what's wrong?

Answer 1

Since this is a self-adjoint operator, there is an orthonormal basis of eigenvectors, and that's what they're talking about. If you multiply basis elements by arbitrary nonzero complex numbers you still have a basis of eigenvectors, but it's no longer orthonormal.