Spectral theorem:
Let $A$ be a real symmetric $n \times n$ matrix. Then there exists a real orthogonal matrix $P$ with $B=P^{-1}AP$ diagonal.
My lecture notes only give a proof for the case where there are $n$ distinct eigenvalues.
Proof outline in this case (see here):
We know that a real symmetric matrix has only real eigenvalues, so there are $n$ distinct real eigenvalues $\lambda_1,...,\lambda_n$. Let $v_1,..,v_n$ be the corresponding eigenvalues, then we also know that $v_i \cdot v_j$ for $i \neq j$ and $v_i \cdot v_i>0$, so we can normalize them. Thus, we may assume $v_i \cdot v_i=1$ such that $v_1,...,v_n$ forms an orthonormal basis of $\mathbb{R}^n$ since the set is linearly independent. It follows that the matrix $P$ with columns $v_1,...,v_n$ is orthogonal. It is also easy so see that $B=P^{-1}AP$ is diagonal with entries $\lambda_1,...,\lambda_n$ on its diagonal.
So in this case the matrix $B$ consists of eigenvalues of $A$ and the columns of matrix $P$ are the corresponding eigenvectors.
Now I also looked up a proof for the general case which is done by induction. Basically, we just use one eigenvalue $\lambda_1$ and then build an orthonormal basis including the corresponding eigenvalue $v_1$. Then we show that $P^(-1)AP$ is symmetric and has $(\lambda_1,0,...,0)^T$ as its first column/row and note that we can break this matrix into parts and use the induction assumption to finish the proof.
Now my question is:
How do we know that the diagonal matrix still consists of the eigenvalues of $A$. Why does the orthonormal basis constructed consist of eigenvectors as in the case of $n$ distinct eigenvalues? I feel like this cannot be the case because there are not enough eigenvectors since some eigenvalues might appear more than once. Or are there multiple different unit eigenvalues that correspond to one eigenvalue if it has a multiplicity larger than $1$?
Could someone clarify this for me, please? Thanks a lot for any help and suggestions.
Edit: Removed unimportant parts and clarified what the exact question is.