Show that if $A$ is a symmetric positive definite matrix, then $A^{-1}$ is a symmetric positive definite matrix We know that matrix $A$ that fulfills those conditions has to be nonsingular, thus there exists an inverse matrix.
 A: Hint: A symmetric matrix is positive definite iff all its eigenvalues are positive.
A: Big Hint: If $A$ is symmetric positive definite then by the spectral theorem you have $A = PDP^T$ for some diagonal matrix $D$ and orthogonal matrix $P$.  Think about what the diagonal elements are and what the columns of $P$ must be.  Now invert.
A: For symmetry of the inverse consider,
$$ A^{-1}A = I = I^T = (AA^{-1})^T = (A^{-1})^T A^T = (A^{-1})^T A ,  $$
from this we can conclude that,
$$ A^{-1} = (A^{-1})^T \qquad \text{(The inverse is symmetric)}.$$

Recall that a matrix is positive definite if $\langle Av,v \rangle > 0 $ for all $v$ in our vector space. Since $A$ is nonsingular we know that it represents a linear automorphism on our vector space; which means for every vector $v$ there is a unique nonzero $b$ such that $Ab = v$. This allows us to write,
$$ \langle A^{-1} v,v \rangle = \langle A^{-1} A b, Ab \rangle = \langle b,Ab\rangle = \langle Ab,b\rangle > 0 \qquad \text{(The inverse is positive definite)}.$$ 
