I aim to compute the eigenvectors of a symmetric matrix $\breve{K}$ (here it is actually $\in \mathbb{R}^{4 \times 4}$)
$$\breve{K} \vec{u} = \lambda \vec{u}.$$
I found that the matrix $\breve{K}^2$ is in a nice form and I want to use this. After a exchanging rows and columns in a specific way (using a matrix that satisfies $\breve{P}=\breve{P}^{-1}$), I arrive at a block diagonal form of the final matrix $\breve{M}$, given by
$$ \breve{M} = \breve{P} \breve{K}^2 \breve{P}.$$
With that I compute the eigenvectors $\vec{v}$ of $\breve{M}$ corresponding to the eigenvalue $\mu$ of $\breve{M}$ (similarity transformation with $\breve{P}$ doesn't change eigenvalues such that $\text{evals}(\breve{M})=\text{evals}(\breve{K}^2)$)
$$ \breve{M} \vec{v} = \breve{P} \breve{K}^2 \breve{P} \vec{v} = \mu \vec{v} .$$
Multiplication of $\breve{M}$ with $\breve{P}^{-1}=\breve{P}$ from the left gives the eigenvector-eigenvalue relation
$$ \breve{K}^2 \breve{P} \vec{v} = \mu \underbrace{\breve{P} \vec{v}}_{=:\vec{x}} \Leftrightarrow \breve{K}^2 \vec{x} = \mu \vec{x}.$$
This gives the eigenvectors $\vec{x} = \breve{P} \vec{v}$ of $\breve{K}^2$.
Using that $\breve{K}$ is diagonalisable (since it is symmetric) it holds that $\breve{K} = \breve{Q}^{-1} \breve{D} \breve{Q}$ as well as $\breve{K}^2 = \breve{Q}^{-1} \breve{D}^2 \breve{Q}$.
If I interpret this correctly, this means that the eigenvectors of $\breve{K}$ and $\breve{K}^2$ respectively corresponding to the eigenvalues $\lambda$ and $\mu=\lambda^2$ are the same and equal to $\breve{P} \vec{v}$, where $\vec{v}$ are the eigenvectors of $\breve{M}$.
Q1: Is this so far correct/Is there anything else needed than the property of $\breve{K}$ being symmetric?
If I am correct with the above:
Q2: Do I find all eigenvectors of $\breve{K}$ this way? In theory I should, but I feel like I'm missing out on something by performing the square.
Q3: Does the above still hold for $\breve{K} \in \mathbb{C}^{4 \times 4}?$