Inverse of a symmetric positive definite matrix If a matrix is symmetric and positive definite, determine if it is invertible and if its inverse matrix is symmetric and positive definite.
I know that "if a matrix is symmetric and positive definite, then its inverse matrix is also positive definite", based on a theorem. But I am not sure how to prove that the matrix even is invertible or that its inverse matrix is also symmetric.
It would really help if someone explained this a bit. Thanks
 A: If $Q$ is psd, $$y^T Q y ≥ 0 \quad ∀ y \qquad \& \qquad I = Q^{-1}Q$$because all eigenvalues are positive.
Let $x = Q^{-1}y$ for any y,
$$(Q^{-1}y)^T Q Q^{-1}y \geq 0 ⇒ y^T (Q^{-1})^T y \geq 0$$
and we know psd matrices are symmetric.
A: We have $(A^{-1})^T = (A^T)^{-1}$ for any invertible matrix. It follows from this that if $A$ is invertible and symmetric $$(A^{-1})^T = (A^T)^{-1} = A^{-1}$$ so $A^{-1}$ is also symmetric. Further, if all eigenvalues of $A$ are positive, then $A^{-1}$ exists and all eigenvalues of $A^{-1}$ are positive since they are the reciprocals of the eigenvalues of $A$. Thus $A^{-1}$ is positive definite when $A$ is positive definite.
A: If A is positive definite matrix, then its eigenvalues are $\lambda_1, \dotsc, \lambda_n >0$ so,
\begin{equation}
|A| = \prod_{i=1}^n \lambda_i > 0
\end{equation}
and A is invertible. Moreover, eigenvalues of $A^{-1}$ are $\frac{1}{\lambda_i}>0$, hence $A^{-1}$ is positive definite. To see $A^{-1}$ is symmetric consider
\begin{equation}
A^{-1} = (A^T)^{-1}=(A^{-1})^T
\end{equation}
