power of positive matrix

Question

If $0 < A $ , what are sufficient conditions for

$\forall r\in {\rm I\!R}$ , $A^r$

is also positive definite matrix.

Answer 1

for any nonsigular matrix $A\in M_n(\mathbb C)$ and for every $\alpha \in \mathbb R$, $A^{\alpha}$ is defined by $A^{\alpha}=\mathbb e^{\alpha \log(A)}$. For more details, you can see Functions of matrices: theory and computation at Chapter $11$. By this definition, and using the fact that $(\mathbb e^X)^*=\mathbb e^{X^*}$ (For proof see Lie groups, Lie algebras, and representations: An elementary introduction Proposition $2.3$) we get that $(A^{\alpha})^*=(\mathbb e^{\alpha \log(A)})^*=\mathbb e^{{(\alpha \log(A)})^*}=e^{\alpha (\log(A))^*}=e^{\alpha \log(A^*)}=(A^*)^{\alpha}$

Theorem: A symmetric square matrix $A$ is positive definite if and only if all of its eigenvalues are $> 0$.
For proof see here Theorem $5$.

Since $A$ is positive definite, So all of its eigenvalues are $> 0$. If $\lambda$ is an eigenvalue of $A$, then $\lambda^r$ is an eigenvalue of $A^r$. Since $\lambda\gt 0$, thus $\lambda^r \gt 0$ and so, all of eigenvalues of $A^r$ are $> 0$ which means $A^r$ is positive definite.