If $A$ is positive definite then so is$A^2$. I am trying to prove using Eigen values that if a symmetric matrix $A$ is positive definite then so is $A^2$. 
Now let $\lambda$ be an Eigen value of $A^2$. I want to show that it is positive. Now if it is a square of an Eigen value of $A$ then I am done. 
Clearly if $\mu$ is an Eigen value of $A$ then $\mu^2$ is an Eigen value of $A^2$. 

But it is not clear to me why any Eigen value of $A^2$ is the square of an Eigen value of $A$.

Any help would be appreciated.
 A: This is a case where going with the definitions is better.

If $A$ is any matrix, then $A^TA$ is positive semidefinite, because
$$
x^T(A^TA)x=(Ax)^T(Ax)\ge0
$$
This also shows that if $A$ is square and invertible, then $A^TA$ is positive definite, because $x^T(A^TA)x=0$ implies $Ax=0$ and so $x=0$.
Since your $A$ is symmetric and invertible…

If you prefer to do it with the eigenvalues, recall that a symmetric matrix is diagonalizable (with an orthogonal matrix, but this is not needed here). Then $A=SDS^{-1}$ and $A^2=SD^2S^{-1}$.
A: There exists $P$ invertible and $J$ in the canonical Jordan form such that
$$
A = PJP ^{-1}
$$
and therefore 
$$
A^2 = PJP^{-1}PJP^{-1} = PJ^2P^{-1}
$$ 
you can square $J$ by blocks, and $J^2$ will be upper triangular with $(J^2)_{ii} = (J_{ii})^2$. Hence, the eigenvalues of $J~^2$ will be those of $J$, squared. It suffices then to note that the eigenvalues of $A^2$ are those of $J^2$. 
To prove this last claim, we can more generally assert the following: if $C$ and $D$ are conjugate matrices, they share the same eigenvalues. Since being conjugates is a symmetric relation, it will suffice to prove that if $\lambda$ is an eigenvalue of $C$, it is an eigenvalue of $D$. Let $v$ be an eigenvector of eigenvalue $\lambda$ for $C$, that is
$$
Cv = \lambda v
$$
Since $C$ and $D$ are conjugates, there exists $Q$ invertible with $C = QDQ^{-1}$, and
$$
Cv = \lambda v \iff QDQ^{-1}v = \lambda v \iff DQ^{-1}v = \lambda Q^{-1}v
$$
so $Q^{-1}v$ is an eigenvector of eigenvalue $\lambda$ for D.
A: If you go with the classical definition.
Since $A$ is symmetric, so $A^T=A$. In other words $A^2=AA^T$.
So it's enough to prove that $AA^T$ is positive definite, i.e $\langle x,A^TAx \rangle > 0$. But $\langle x,A^TAx \rangle=||Ax||^2 >0(Why?).$Therefore $A^2$ is positive semidefinite.
A: A direct proof: let $( \cdot| \cdot)$ be the usual inner product on $ \mathbb R^n$ and $|| \cdot||$ the induced norm.
If $ \mu$ is an eigenvalue of $A^2$, then there is $v \in \mathbb R^n$ sucht that $||v||=1$ and $A^2v= \mu v$. Since $A$ is invertible $A^2$ is invertible, hence $ \mu \ne 0$.
Furthermore
$$ \mu= \mu ||v||^2=\mu (v|v)=( \mu v|v)=(A^2v|v)=(Av|Av)=||Av||^2 >0.$$
A: If $A$ is symmetric, then so is $A^2$.
Let $\lambda \in \mathbb{\mathbb{R}}$ be an eigenvalue of $A^2$, and let $\sqrt{\lambda} \in \mathbb{C}$ be a square root of $\lambda$. We have$$(A-\sqrt{\lambda} I)(A-\sqrt{\lambda} I)= A^2 - \mu I$$ $A^2 - \lambda I$ is not invertible so $A-\sqrt{\lambda} I$ is also not invertible, which implies  that $\sqrt{\lambda}$ is an eigenvalue of $A$. Since $A$ is positive definite, we have $\sqrt{\lambda} > 0$ so $\lambda > 0$.
Therefore, $A^2$ is a symmetric matrix with positive eigenvalues, and hence positive definite. 
