symmetric positive definite matrices

Why must a symmetric positive definite matrix must be invertible? I'm reading a proof of the Levi-Civita theorem in differential geometry but the author states this without proof and I haven't been able to prove it.

• Because its null space is trivial. – Arin Chaudhuri Apr 25 '13 at 17:16

Suppose a positive definite matrix $P$ were not invertible. Then there would be a nonzero vector $x$ such that $Px = 0$. Multiplying both sides of this equation on the left by $x^T$ gives $x^TPx = x^T0 = 0$ for $x\neq 0$, contradicting positive definiteness.
Symmetric matrices are diagonalizable: $$S = U D U^T$$ for $U^T = U^{-1}$ and $D$ a diagonal matrix whose entries are the eigenvalues. When $S$ is positive definite, all the eigenvalues must be strictly positive. Hence if $D = \text{diag}(d_1,\dotsc,d_n)$ then $D$ is invertible with $D^{-1} = \text{diag}(d_1^{-1},\dotsc,d_n^{-1})$. Then one may check that $S^{-1} = U D^{-1} U^T$, and in particular $S$ is invertible.