Proof of Invertibility of Positive Definite Matrices

Each result cited is universally quantified over the data in its own statement.

We work in Routine Facts Used Without Citation: Arithmetic, Maps, Predicativity and Indices. By Symmetric Real Matrices and the Set $\mathcal{S}(n)$ §set and Symmetric Real Matrices and the Set $\mathcal{S}(n)$ §symmetric, $M^{\top}=M$, and by Positive Semidefinite and Positive Definite Real Matrices §pd, $x\cdot(Mx)>0$ for every $x\in\mathbb{R}^{n}$ with $x\neq0$.

Clause invertible. Let $x\in\mathbb{R}^{n}$ with $Mx=0$. Evaluating pointwise by Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations §operations and Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations §origin, the scalar multiple $0x$ is the origin $0$. Hence, by The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite §second,

$$x\cdot(Mx)=x\cdot(0x)=0\,(x\cdot x)=0.$$

If $x\neq0$, then $x\cdot(Mx)>0$, which is false; so $x=0$. Thus the only $x\in\mathbb{R}^{n}$ with $Mx=0$ is $x=0$, which is condition Criteria for Invertibility of a Square Real Matrix §trivial-kernel; by Criteria for Invertibility of a Square Real Matrix it is equivalent to condition Criteria for Invertibility of a Square Real Matrix §invertible, so $M$ is invertible.

The inverse. By the clause invertible, $M$ is invertible, so $M^{-1}$, its unique inverse by Inverse Matrix and Invertible Real Square Matrix §invertible, is in $\mathcal{M}_{n}(\mathbb{R})$ and satisfies $MM^{-1}=M^{-1}M=I_{n}$ by Inverse Matrix and Invertible Real Square Matrix §inverse.

Clause inverse-symmetric. By Rules for Matrix Inverses §transpose, $M^{\top}$ is invertible and $(M^{\top})^{-1}=(M^{-1})^{\top}$; since $M^{\top}=M$, this reads $(M^{-1})^{\top}=M^{-1}$. So $M^{-1}$ is symmetric by Symmetric Real Matrices and the Set $\mathcal{S}(n)$ §symmetric, and $M^{-1}\in\mathcal{S}(n)$ by Symmetric Real Matrices and the Set $\mathcal{S}(n)$ §set.

Clause inverse-pd. Let $y\in\mathbb{R}^{n}$ with $y\neq0$, and put $x=M^{-1}y\in\mathbb{R}^{n}$. By Rules for Matrix Inverses §solution, $x$ is the only point with $Mx=y$; in particular $Mx=y$. If $x=0$, then $y=M0=0$ by Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §matrix-vector, which is false; so $x\neq0$. By The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite §symmetric,

$$y\cdot(M^{-1}y)=(Mx)\cdot x=x\cdot(Mx)>0.$$

As $y\neq0$ is arbitrary and $M^{-1}\in\mathcal{S}(n)$ by the clause inverse-symmetric, $M^{-1}$ is positive definite by Positive Semidefinite and Positive Definite Real Matrices §pd.