Positive Definiteness Of Real Symmetric Non-singular Matrix 
For every $4 \times 4$ real symmetric non-singular matrix $A$, there exists a positive integer $p$ such that :
  A. $pI+A$ is positive definite.
  B. $A^p$ is positive definite.
  C. $A^{-p}$ is positive definite.
  D. $\mathrm exp(pA)-I$ is positive definite.  

My approach to the problem.
Since we know that a matrix $A \in \mathbb M_n(\mathbb R)$ is positive definite if it is symmetric and all its eigenvalues are strictly positive.  
For (A), I have shown $pI+A$ is symmetric as $A$ is symmetric. Also, it is clear that all eigenvalues of $pI+A$ will be of the form $p+ \lambda _i$, where $\lambda_i$ are non-zero eigenvalues of matrix $A$, as $A$ is non-singular. Therefore, I can chose $p$ in such a way that all $p+ \lambda_i \gt 0$.   
For (B), it is clear that $A^p$ is symmetric and, I can choose $p$ to be even that will show the positive definiteness for  $A^p$.  
For (C), same $p$ can be taken as in (B).   
My question is how to show that (D) is symmetric or not, and if so how to define the eigenvalues for (d).
 A: By the spectral theorem we can diagonalize $A$ as $A=Q\Lambda Q^T$. It is a property of the matrix exponential that 
$$
\exp(pQ\Lambda Q^T) = Q\exp(p\Lambda)Q^T
$$
And the matrix exponential of a diagonal matrix is just the ordinary exponential of its elements. On $(0,\infty)$ $e^x>1$. Can you see how to finish the proof?

Here's a more direct way to do it. Let $x \neq 0$ have norm $1$. Then
$$
x^T(e^{pA} - I)x = x^T\left(\sum_{k=0}^\infty \frac {p^k}{k!}A^k - I \right)x
$$
$$
= x^T\left(\sum_{k=1}^\infty \frac {p^k}{k!}A^k\right)x = \sum_{k=1}^\infty \frac {p^k}{k!} x^T A^k x.
$$
Each $A^k$ is PD and $p>0$ so we have a sum of positive things which is positive.
If you're worried about convergence, note that
$$
\lambda_{\max}(A^k) = \left(\sup_{x : ||x||=1} x^TAx\right)^k \in (0, \infty)
$$
so
$$
0 < \frac {p^k}{k!} x^T A^k x \leq \frac{(p\lambda_{\max}(A))^k}{k!}
$$
so the partial sums of $x^T(e^{pA}-I)x$ give a monotonic series bounded above by $e^{p\lambda_{\max}(A)}-1$, and therefore it converges.
