Suppose $X$ is symmetric and positive definite.
Then, does this imply $X$ is invertible?
Also, I read we can construct a symmetric and invertible matrix $X^{1/2}$ such that
$X^{1/2}$$X^{1/2}$=$X$.
$X^{-1/2}$$X^{-1/2}$=$X^{-1}$.
Is this an application of the Cholesky decomposition or am I confusing with some other decomposition that is applied in this case?
Thanks.
Yes, a symmetric positive definite real matrix is invertible. In addition, it is orthogonally diagonalizable so you can write $X = PDP^{-1}$ with $P$ orthogonal and $D$ diagonal. Since $X$ is positive definite, all the eigenvalues of $X$ are positive and so all the entries on the diagonal of $D$ are positive. You can then construct a matrix $\sqrt{D}$ which is diagonal and whose entries are the square roots of the diagonal entries of $D$. Clearly $\sqrt{D}$ has positive entries and $\sqrt{D}^2 = D$ but then if you define $X' = P\sqrt{D}P^{-1}$ you have
$$ X'^2 = P\sqrt{D}P^{-1}P\sqrt{D}P^{-1} = P\sqrt{D}^2P^{-1} = PDP^{-1} = X $$
so $X^{\frac{1}{2}} := X'$ is a square root of $X$. Since $P$ is orthogonal, $X'$ is symmetric. Since it has positive eigenvalues, it is positive definite and invertible.