eigenvectors of $A^2$ and $A$ if $A$ is positive definite. 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.
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, $\lambda_n &gt; 0$ 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 $\sqrt{\mu}$.
Suppose that $A^2v= \lambda^2v$. Then compute the norm $ ||Av-\lambda v||^2_2.$

What happens ?