Proof of Every Positive Semidefinite Matrix Has Exactly One Positive Semidefinite Square Root

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. Call a pair $(Q,\lambda)$ with $Q\in\mathrm{O}(n)$, $\lambda\in\mathbb{R}^{n}$ and $A=(Q\operatorname{diag}(\lambda))Q^{\top}$ a diagonalisation of $A$. For $M,N,P\in\mathcal{M}_{n}(\mathbb{R})$ we write $MNP$ for $(MN)P$, which equals $M(NP)$ by Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §associative; longer products are likewise unambiguous by that clause. For $Q\in\mathrm{O}(n)$, $Q^{\top}Q=I_{n}$ by Orthogonal Real Matrices §membership and $QQ^{\top}=I_{n}$ by Rules for Orthogonal Matrices: Inverse, Products, Isometry, Columns, Determinant, Orthonormal Extension and Expansion §inverse. For $s\in\mathbb{R}$, $s^{2}=s\,s$ by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §exponents and Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §product.

(C) If $M,N\in\mathcal{M}_{n}(\mathbb{R})$ and $Me_{j}=Ne_{j}$ for every $j\in[n]$, then $M=N$: for $i,j\in[n]$, $M_{ij}=(Me_{j})_{i}=(Ne_{j})_{i}=N_{ij}$ by Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §column, and $M=N$ by Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations §equality.

Step 1: the diagonal is nonnegative. Let $(Q,\lambda)$ be a diagonalisation of $A$ and $k\in[n]$. By The Spectral Theorem for Real Symmetric Matrices §eigenvectors, $Qe_{k}$ is an eigenvector of $A$ with eigenvalue $\lambda_{k}$, so $\lambda_{k}$ is an eigenvalue of $A$ (Real Eigenvalues and Eigenvectors of a Real Square Matrix §eigenvalue), and $\lambda_{k}\ge0$ by Positive Semidefinite and Positive Definite Matrices by Their Eigenvalues §psd, as $A$ is positive semidefinite. Let $\sigma=\sqrt{\lambda}\in\mathbb{R}^{n}$ be the map $k\mapsto\sqrt{\lambda_{k}}$. By The k-th Root and the Square Root of a Nonnegative Real Number §square-root and The k-th Root and the Square Root of a Nonnegative Real Number §root, $\sigma_{k}\ge0$ and $\sigma_{k}\sigma_{k}=\sigma_{k}^{2}=\lambda_{k}$ for every $k\in[n]$. Let

$$B_{Q}=Q\operatorname{diag}(\sigma)Q^{\top}.$$

Step 2: $B_{Q}$ is a positive semidefinite symmetric square root of $A$. By Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §transpose and Eigenvalues: the Determinant Criterion, Diagonal Matrices, Orthogonal Similarity and Orthogonality of Eigenvectors of Symmetric Matrices §diagonal-algebra, $B_{Q}^{\top}=(Q^{\top})^{\top}\operatorname{diag}(\sigma)^{\top}Q^{\top}=Q\operatorname{diag}(\sigma)Q^{\top}=B_{Q}$, so $B_{Q}\in\mathcal{S}(n)$ by Symmetric Real Matrices and the Set $\mathcal{S}(n)$ §set. By The Spectral Theorem for Real Symmetric Matrices §eigenvectors, applied to $B_{Q}$ with $Q$ and $\sigma$, every eigenvalue of $B_{Q}$ equals $\sigma_{k}\ge0$ for some $k\in[n]$, so $B_{Q}$ is positive semidefinite by Positive Semidefinite and Positive Definite Matrices by Their Eigenvalues §psd. Let $\sigma\sigma\in\mathbb{R}^{n}$ be the map $k\mapsto\sigma_{k}\sigma_{k}$; it equals $\lambda$ by Step 1 and Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations §equality. By Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §identity and Eigenvalues: the Determinant Criterion, Diagonal Matrices, Orthogonal Similarity and Orthogonality of Eigenvectors of Symmetric Matrices §diagonal-algebra,

$$B_{Q}B_{Q}=Q\operatorname{diag}(\sigma)(Q^{\top}Q)\operatorname{diag}(\sigma)Q^{\top}=Q\operatorname{diag}(\sigma)\operatorname{diag}(\sigma)Q^{\top}=Q\operatorname{diag}(\sigma\sigma)Q^{\top}=Q\operatorname{diag}(\lambda)Q^{\top}=A.$$

Step 3: a positive semidefinite symmetric square root acts on eigenvectors of $A$ by square roots. Let $C\in\mathcal{S}(n)$ be positive semidefinite with $CC=A$, and let $x\in\mathbb{R}^{n}$ and $\mu\in\mathbb{R}$ with $\mu\ge0$ and $Ax=\mu x$. We show $Cx=\sqrt{\mu}\,x$. By The Spectral Theorem for Real Symmetric Matrices §diagonalization applied to $C$, there are $R\in\mathrm{O}(n)$ and $\nu\in\mathbb{R}^{n}$ with $C=(R\operatorname{diag}(\nu))R^{\top}$; fix them. By The Spectral Theorem for Real Symmetric Matrices §eigenvectors and Real Eigenvalues and Eigenvectors of a Real Square Matrix §eigenvalue, each $\nu_{k}$ is an eigenvalue of $C$, so $\nu_{k}\ge0$ by Positive Semidefinite and Positive Definite Matrices by Their Eigenvalues §psd. Let $s=\sqrt{\mu}$, so $s\ge0$ and $s\,s=\mu$ by The k-th Root and the Square Root of a Nonnegative Real Number §square-root and The k-th Root and the Square Root of a Nonnegative Real Number §root, and let $y=R^{\top}x$. Since $R^{\top}C=(R^{\top}R)\operatorname{diag}(\nu)R^{\top}=\operatorname{diag}(\nu)R^{\top}$, Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §matrix-vector gives $R^{\top}(Cz)=\operatorname{diag}(\nu)(R^{\top}z)$ for $z\in\mathbb{R}^{n}$; it also gives $C(Cx)=(CC)x=Ax$; and therefore

$$\operatorname{diag}(\nu)\bigl(\operatorname{diag}(\nu)y\bigr)=R^{\top}\bigl(C(Cx)\bigr)=R^{\top}(Ax)=R^{\top}(\mu x)=\mu y .$$

Let $k\in[n]$. Taking $k$-th components with Eigenvalues: the Determinant Criterion, Diagonal Matrices, Orthogonal Similarity and Orthogonality of Eigenvectors of Symmetric Matrices §diagonal-algebra and Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations §operations, $\nu_{k}\nu_{k}y_{k}=s\,s\,y_{k}$, that is $(\nu_{k}\nu_{k}-s\,s)\,y_{k}=0$; and $\nu_{k}\nu_{k}-s\,s=\nu_{k}^{2}-s^{2}=(\nu_{k}-s)(s+\nu_{k})$ by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §factorisation with exponent $2$ (the two terms of its sum being $\nu_{k}^{0}s^{1}=s$ and $\nu_{k}^{1}s^{0}=\nu_{k}$, by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §product), so $(\nu_{k}-s)(\nu_{k}+s)\,y_{k}=0$. By Rules of Arithmetic and Order in an Ordered Field §zero, $y_{k}=0$, or $\nu_{k}=s$, or $\nu_{k}+s=0$; in the last case $0\le\nu_{k}=\nu_{k}+0\le\nu_{k}+s=0$ by Rules of Arithmetic and Order in an Ordered Field §order-sum, as $0\le s$, so $\nu_{k}=0$ and then $s=0$, so again $\nu_{k}=s$. In every case $\nu_{k}y_{k}=s\,y_{k}$. Hence $\operatorname{diag}(\nu)y=sy$, by Eigenvalues: the Determinant Criterion, Diagonal Matrices, Orthogonal Similarity and Orthogonality of Eigenvectors of Symmetric Matrices §diagonal-algebra, 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 §equality, and by Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §matrix-vector,

$$Cx=R\bigl(\operatorname{diag}(\nu)(R^{\top}x)\bigr)=R(sy)=s\,(R(R^{\top}x))=s\,\bigl((RR^{\top})x\bigr)=s\,(I_{n}x)=sx .$$

Step 4: every positive semidefinite symmetric square root equals $B_{Q}$. Let $(Q,\lambda)$ be a diagonalisation of $A$, with $\sigma$ and $B_{Q}$ as in Step 1, and let $C\in\mathcal{S}(n)$ be positive semidefinite with $CC=A$. Let $k\in[n]$. By The Spectral Theorem for Real Symmetric Matrices §eigenvectors and Real Eigenvalues and Eigenvectors of a Real Square Matrix §eigenvector, $A(Qe_{k})=\lambda_{k}(Qe_{k})$ with $\lambda_{k}\ge0$ (Step 1), so $C(Qe_{k})=\sigma_{k}(Qe_{k})$ by Step 3. By The Spectral Theorem for Real Symmetric Matrices §eigenvectors applied to $B_{Q}\in\mathcal{S}(n)$ with $Q$ and $\sigma$, also $B_{Q}(Qe_{k})=\sigma_{k}(Qe_{k})$. So $(CQ)e_{k}=(B_{Q}Q)e_{k}$ by Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §matrix-vector, for every $k\in[n]$, and $CQ=B_{Q}Q$ by (C). By Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §identity,

$$C=C(QQ^{\top})=(CQ)Q^{\top}=(B_{Q}Q)Q^{\top}=B_{Q}(QQ^{\top})=B_{Q}.\tag{1}$$

Clause square-root. By The Spectral Theorem for Real Symmetric Matrices §diagonalization, $A$ has a diagonalisation $(Q_{0},\lambda^{0})$; fix it. By Step 2, $B_{Q_{0}}\in\mathcal{S}(n)$ is positive semidefinite and $B_{Q_{0}}B_{Q_{0}}=A$. If $C,C'\in\mathcal{S}(n)$ are positive semidefinite with $CC=A$ and $C'C'=A$, then $C=B_{Q_{0}}=C'$ by (1). So there is exactly one such matrix, $B=B_{Q_{0}}$.

Clause spectral. Let $(Q,\lambda)$ be a diagonalisation of $A$. By Step 1, $\lambda_{k}\ge0$ for every $k\in[n]$. The matrix $B$ of clause square-root is positive semidefinite, lies in $\mathcal{S}(n)$ and satisfies $BB=A$, so $B=B_{Q}=(Q\operatorname{diag}(\sqrt{\lambda}))Q^{\top}$ by (1).

Clause commutes. By Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §associative, $AB=(BB)B=B(BB)=BA$.

Clause definite. Let $A$ be positive definite, and let $(Q_{0},\lambda^{0})$ and $\sigma=\sqrt{\lambda^{0}}$ be as in the proof of clause square-root, so $B=Q_{0}\operatorname{diag}(\sigma)Q_{0}^{\top}$. For $k\in[n]$, $\lambda^{0}_{k}$ is an eigenvalue of $A$ (Step 1), so $\lambda^{0}_{k}>0$ by Positive Semidefinite and Positive Definite Matrices by Their Eigenvalues §pd, and $\sigma_{k}=\sqrt{\lambda^{0}_{k}}>\sqrt{0}=0$ by Properties of k-th Roots: Inverse to Powers, Absolute Values, Products, Monotonicity, Subadditivity, Differences and Limits §monotone and Properties of k-th Roots: Inverse to Powers, Absolute Values, Products, Monotonicity, Subadditivity, Differences and Limits §power. By The Spectral Theorem for Real Symmetric Matrices §eigenvectors, applied to $B\in\mathcal{S}(n)$ with $Q_{0}$ and $\sigma$, every eigenvalue of $B$ equals some $\sigma_{k}>0$, so $B$ is positive definite by Positive Semidefinite and Positive Definite Matrices by Their Eigenvalues §pd.