Proof of Positive Semidefinite and Positive Definite Matrices by Their Eigenvalues

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.

Two preliminaries.

(E) If $x$ is an eigenvector of $A$ with eigenvalue $\mu$, then $x\cdot(Ax)=\mu\,(x\cdot x)$ and $x\cdot x>0$. Indeed $x\neq0$ and $Ax=\mu x$ by Real Eigenvalues and Eigenvectors of a Real Square Matrix §eigenvector, so $x\cdot(Ax)=x\cdot(\mu x)=\mu\,(x\cdot x)$ by The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite §second; and $x\cdot x\ge0$ by The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite §nonnegative and $x\cdot x\neq0$ by The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite §definite.

(S) By The Spectral Theorem for Real Symmetric Matrices §diagonalization, there are $Q\in\mathrm{O}(n)$, the set of orthogonal matrices, and $\lambda\in\mathbb{R}^{n}$ such that $\lambda_{j}\ge\lambda_{k}$ for all $j,k\in[n]$ with $j\le k$, and $A=(Q\operatorname{diag}(\lambda))Q^{\top}$, with the diagonal matrix $\operatorname{diag}(\lambda)$; fix them. By The Spectral Theorem for Real Symmetric Matrices §eigenvectors, for each $k\in[n]$ the vector $Qe_{k}$ is an eigenvector of $A$ with eigenvalue $\lambda_{k}$, so $\lambda_{k}$ is an eigenvalue of $A$ by Real Eigenvalues and Eigenvectors of a Real Square Matrix §eigenvalue; in particular $\lambda_{n}$ is, as $n\in[n]$. By The Spectral Theorem for Real Symmetric Matrices §rayleigh,

$$x\cdot(Ax)\ge\lambda_{n}\lVert x\rVert^{2}\qquad(x\in\mathbb{R}^{n}),\tag{1}$$

and $\lVert x\rVert^{2}=x\cdot x\ge0$ by Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square §square, with $\lVert x\rVert^{2}>0$ if $x\neq0$ by The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite §definite.

Clause psd. Let $A$ be positive semidefinite, let $\mu$ be an eigenvalue of $A$ and, by Real Eigenvalues and Eigenvectors of a Real Square Matrix §eigenvalue, let $x$ be an eigenvector of $A$ with eigenvalue $\mu$. By (E) and Positive Semidefinite and Positive Definite Real Matrices §psd, $\mu\,(x\cdot x)=x\cdot(Ax)\ge0$ with $x\cdot x>0$. If $\mu<0$, then $\mu\,(x\cdot x)<0\cdot(x\cdot x)=0$ by Rules of Arithmetic and Order in an Ordered Field §order-product, a contradiction; so $\mu\ge0$.

Conversely, let every eigenvalue of $A$ be $\ge0$. By (S), $\lambda_{n}\ge0$, so for every $x\in\mathbb{R}^{n}$, (1) and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §positive-product give $x\cdot(Ax)\ge\lambda_{n}\lVert x\rVert^{2}\ge0$. As $A\in\mathcal{S}(n)$, $A$ is positive semidefinite by Positive Semidefinite and Positive Definite Real Matrices §psd.

Clause pd. We prove: positive definite implies all eigenvalues $>0$, which implies positive definite; and positive definite implies positive semidefinite and invertible, which implies all eigenvalues $>0$.

Positive definite implies eigenvalues $>0$. Let $A$ be positive definite, $\mu$ an eigenvalue and $x$ an eigenvector with eigenvalue $\mu$. By (E) and Positive Semidefinite and Positive Definite Real Matrices §pd, as $x\neq0$, $\mu\,(x\cdot x)=x\cdot(Ax)>0$ with $x\cdot x>0$. If $\mu\le0$, then $\mu\,(x\cdot x)\le0$ by Rules of Arithmetic and Order in an Ordered Field §order-product, a contradiction; so $\mu>0$.

Eigenvalues $>0$ imply positive definite. By (S), $\lambda_{n}>0$. For $x\in\mathbb{R}^{n}$ with $x\neq0$, $\lVert x\rVert^{2}>0$, so (1) and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §positive-product give $x\cdot(Ax)\ge\lambda_{n}\lVert x\rVert^{2}$ and $\lambda_{n}\lVert x\rVert^{2}>0$, hence $x\cdot(Ax)>0$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed. As $A\in\mathcal{S}(n)$, $A$ is positive definite by Positive Semidefinite and Positive Definite Real Matrices §pd.

Positive definite implies positive semidefinite and invertible. Let $A$ be positive definite. It is positive semidefinite by Rules for Symmetric and Positive Semidefinite Matrices: Closure, Self-Adjointness, Polarization, Congruence and Cones §pd-psd. Let $x\in\mathbb{R}^{n}$ with $Ax=0$. Then $x\cdot(Ax)=x\cdot(0x)=0\,(x\cdot x)=0$ by The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite §second, as $0=0x$ (Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations §operations, Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations §origin). So $x=0$, since otherwise $x\cdot(Ax)>0$ by Positive Semidefinite and Positive Definite Real Matrices §pd. Thus condition Criteria for Invertibility of a Square Real Matrix §trivial-kernel holds, and $A$ is invertible by the equivalence of Criteria for Invertibility of a Square Real Matrix §trivial-kernel and Criteria for Invertibility of a Square Real Matrix §invertible.

Positive semidefinite and invertible imply eigenvalues $>0$. Let $A$ be positive semidefinite and invertible, let $\mu$ be an eigenvalue and $x$ an eigenvector with eigenvalue $\mu$. By clause psd, $\mu\ge0$. If $\mu=0$, then $Ax=0x=0$ (as above) with $x\neq0$, so condition Criteria for Invertibility of a Square Real Matrix §trivial-kernel fails, contradicting Criteria for Invertibility of a Square Real Matrix §invertible by the equivalence of the two. So $\mu>0$.

Clause coercive. Let $A$ be positive definite and let $c=\lambda_{n}$, with $\lambda$ as in (S). It is an eigenvalue of $A$ by (S), so $c>0$ by clause pd, and $x\cdot(Ax)\ge c\lVert x\rVert^{2}$ for every $x\in\mathbb{R}^{n}$ by (1).