Is it true that $A^2$ and $A$ has the same eigenvectors if $A$ is positive definite?
$Av = \lambda v$ $\implies $ $A^2 v = \lambda^2 v$ $ \, \,\, \,\, \,\, \,\, \, $what about $\iff$?
It is also known that the eigenvalues of $A^2$ are the square of eigenvalues of $A$ as in
$ A^2 v = \mu v \implies A w = \sqrt{\mu} \, w $.
How to show, if possible, that $v$ and $w$ above are the same?
Maybe (just a suggestion) using the fact that there is a unique square root of a positive definite matrix.
On
If $A$ is positive definite, then $A$ is, in particular, symmetric, and we get a basis $\{v_i\}_{1\leq i \leq n}$ of eigenvectors of $A$ with corresponding eigenvalues $\lambda_i$. Let $v=\sum_{i=1}^n a_iv_i$ be an eigenvector of $A^2$ with eigenvalue $\lambda$. Then,
$$ \sum_{i=1}^n \lambda a_i v_i=\lambda v=A^2v=A^2\sum_{i=1}^n a_i v_i=\sum_{i=1}^n \lambda_i^2 a_iv_i, $$ implying that $\lambda a_i= \lambda_i^2 a_i$ for each $i$, but then, either $\lambda=\lambda_i^2$ or $a_i=0$. This implies that $v$ must have the form $\sum_{j=1}^k a_j v_{i_j}$ for $v_{i_j}$ corresponding to the same eigenvalue. Hence, $v$ is an eigenvector of $A$ with eigenvalue some square root of $\lambda$. Since $A$ is positive definite, we get that $Av= \sqrt{\lambda} v$.
If $A$ is positive definite, the vector space has a basis $v_n$ consisting of eigenvectors of $A$. Let $\lambda_n$ be the corresponding eigenvalues. Suppose $v$ is an eigenvector of $A^2$ for eigenvalue $\mu$. We can write $$ v = \sum_n c_n v_n$$ Then $$A^2 v = \mu v = \sum_n c_n A^2 v_n = \sum_n c_n \lambda_n^2 v_n $$ But also $$\mu v = \sum_n c_n \mu v_n $$ Since $v_n$ are a basis, each vector has a unique expression as a linear combination of the $v_n$, so we must have $c_n \lambda_n^2 = c_n \mu$ for each $n$. Thus if $c_n \ne 0$, $\mu = \lambda_n^2$. Since $A$ is positive definite, $\lambda_n > 0$ so $\lambda_n = \sqrt{\mu}$. That is, $v$ is a linear combination of eigenvectors of $A$ for $\sqrt{\mu}$, and so it is itself an eigenvector of $A$ for $\sqrt{\mu}$.