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).
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.