properties of positive definite matrix If $A$ is a symmetric positive definite matrix can we conclude $A^{n}$ is positive define too? Why?
For example for $n=2$: $x^{T}(AA)x=x^{T}(AA^{T})x=(x^{T}A)^2>0$;
for $n>2$?
 A: Hint: A matrix is positive definite if and only if its eigenvalues are $>0$.
Add that the eigenvalues of $A^n$ are …
A: We have proof that for all colum vectors $\vec{v}\in\mathbb{R}^n$
$$
v^TA^nv\geq 0
$$
If $n=2m$ is even set $B=A^{m}$,
\begin{align}
v^TA^nv 
=
&
v^TBBv
\\
=
&
v^T\Big(B^T\Big)^TBv
\\
=
&
\Big(B^Tv\Big)^TBv
\\
=
&
\Big(Bv\Big)^TBv\geq 0
\\
\end{align}
If $n=2m+1$ is odd set $B=A^{m}$,
\begin{align}
v^TA^nv 
=
&
v^TBABv
\\
=
&
v^T\Big(B^T\Big)^TABv
\\
=
&
\Big(B^Tv\Big)^TA\big(Bv\big)
\\
=
&
\Big(Bv\Big)^TA\big(Bv\big)\geq 0
\\
\end{align}
Note that $B$ is no sigular and the oparator $\mathbb{R}^n\ni v\mapsto Bv\in\mathbb{R}^n$ is injetive. 
Remark Remember that, If $A^T=A$, $B_1=A$, $B_2=AA$,..., $B_{n+1}=B_nA$ then $B_1^T=A^T=A=B_1$, $B_2^T=(AA)^T=A^TA^T=AA=B_2$,..., $B_{n+1}^T=(B_nA)^T=A^TB_n^T=AB_n=B_{n+1}$.
A: A matrix is positive definite when all of its eigenvalues are positive. If we assume that all of the eigenvalues of $A$ are positive then can we show that all of the eigenvalues of $A^n$ are positive?
Assume that $A$ is positive definite. Let $\lambda > 0$ be an eigenvalue of $A$. Then $\det(I-\lambda A)=0$. Now, consider:
$$\det(I-\lambda^2A^2) = \det((I-\lambda A)(I+\lambda A)) = \det(I-\lambda A)\cdot \det(I + \lambda A) = 0.$$
It follows that $\lambda^2 > 0$ is an eigenvalue of $A^2$. Similarly, since
$$I-\lambda^nA^n = (I-\lambda A)(I+\lambda A+\lambda^2A^2 + \cdots + \lambda^{n-1}A^{n-1})$$
we see that $\lambda^n>0$ is an eigenvalue of $A^n$. We can conclude that $A^n$ is positive definite.
