Positive Square Roots of Matrices

Definition Let $\mathbb{F}$ be $\mathbb{R}$ or $\mathbb{C}$, and let $H \in \mathbb{F}^{n \times n}$ be Hermitian, which means real symmetric when $\mathbb{F} = \mathbb{R}$, and positive semidefinite, meaning $x^{*}Hx \ge 0$ for every $x \in \mathbb{F}^n$. A positive semidefinite square root of $H$ is a Hermitian positive semidefinite matrix $S \in \mathbb{F}^{n \times n}$ with

Every such $H$ has exactly one, written $H^{1/2}$. The construction and the uniqueness proof below show this. The root has the same eigenvectors as $H$, with eigenvalue $\sqrt{\lambda}$ wherever $H$ has eigenvalue $\lambda$, so $\ker H^{1/2} = \ker H$, and $H^{1/2}$ is positive definite exactly when $H$ is. For a diagonal $H = \operatorname{diag}(d_1, \dots, d_n)$ with every $d_i \ge 0$, the root is $\operatorname{diag}(\sqrt{d_1}, \dots, \sqrt{d_n})$. The same definition applies to a positive semidefinite operator on a finite-dimensional inner product space over $\mathbb{F}$.

Definition Let $H \in \mathbb{F}^{n \times n}$ be Hermitian positive semidefinite, with $\mathbb{F}$ equal to $\mathbb{R}$ or $\mathbb{C}$. The spectral theorem gives $H = Q\Lambda Q^{*}$ with $Q$ unitary, real orthogonal when $\mathbb{F} = \mathbb{R}$, and $\Lambda = \operatorname{diag}(\lambda_1, \dots, \lambda_n)$. Every $\lambda_i$ is nonnegative, because the $i$th column $q_i$ of $Q$ gives $\lambda_i = q_i^{*}Hq_i \ge 0$. Define

with nonnegative scalar roots. Then $S^{*} = S$ because the diagonal entries are real, and with $y = Q^{*}x$

so $S$ is Hermitian positive semidefinite. In the product $S^2 = Q\Lambda^{1/2}Q^{*}Q\Lambda^{1/2}Q^{*}$ the middle factor $Q^{*}Q$ is the identity, so $S^2 = Q\Lambda Q^{*} = H$. Nothing is divided by a root, so zero eigenvalues cause no trouble, and $Sx = 0$ exactly when $Hx = 0$. Grouping equal eigenvalues,

where the sums run over the distinct eigenvalues of $H$ and $P_{\lambda}$ is the orthogonal projector onto the eigenspace of $\lambda$. These projectors depend only on $H$, so $S$ does not depend on which orthonormal eigenvectors were chosen inside a repeated eigenspace.

Definition Let $H \in \mathbb{F}^{n \times n}$ be Hermitian positive semidefinite, with $\mathbb{F}$ equal to $\mathbb{R}$ or $\mathbb{C}$. If $R \in \mathbb{F}^{n \times n}$ is Hermitian positive semidefinite and $R^2 = H$, then $R$ equals the matrix $S$ built above. The proof does not assume that $R$ and $S$ share eigenvectors. First, $R$ commutes with $H$, because $RH = R^3 = HR$. So if $Hv = \lambda v$, then

and $R$ maps every eigenspace $E_{\lambda}$ of $H$ into itself. On $E_{\lambda}$, with the inherited inner product, $R$ is still Hermitian and positive semidefinite, so the spectral theorem gives an orthonormal basis of $E_{\lambda}$ made of eigenvectors of $R$. If $Rw = \mu w$ for a nonzero $w \in E_{\lambda}$, then $\mu \ge 0$ and

so $\mu^2 = \lambda$ and $\mu = \sqrt{\lambda}$, the only nonnegative number with that square. Every eigenvalue of $R$ on $E_{\lambda}$ is $\sqrt{\lambda}$, so $R$ acts on $E_{\lambda}$ as $\sqrt{\lambda}$ times the identity, exactly as $S$ does. The eigenspaces of $H$ together span $\mathbb{F}^n$, so $R = S$. The argument covers repeated eigenvalues and the eigenvalue $0$ as well.

Definition A square root of a matrix $A \in \mathbb{F}^{n \times n}$ is any $R \in \mathbb{F}^{n \times n}$ with $R^2 = A$. Without further conditions a matrix may have finitely many square roots, infinitely many, or none, and the answer depends on the field. If $R$ is a square root, so is $-R$. A Hermitian matrix $H$ can have a Hermitian square root only if it is positive semidefinite, because a Hermitian $R$ gives

for every $x$. So a Hermitian matrix with a negative eigenvalue has no Hermitian square root. Over $\mathbb{R}$ the $1 \times 1$ matrix $[-1]$ has no square root at all, while over $\mathbb{C}$ it has the two roots $[i]$ and $[-i]$, neither of them Hermitian. The matrix

has no square root over $\mathbb{R}$ or $\mathbb{C}$. A root $R$ would commute with $N$, since $RN = R^3 = NR$, and the matrices that commute with $N$ have rows $(a, b)$ and $(0, a)$. Their squares have rows $(a^2, 2ab)$ and $(0, a^2)$, which equal $N$ only if $a = 0$ and $2ab = 1$, a contradiction. At the other extreme, $I_2$ has infinitely many square roots, such as the matrix with rows $(1, a)$ and $(0, -1)$ for every scalar $a$. The positive root theorem picks out one root only in its own setting, a Hermitian positive semidefinite matrix and a Hermitian positive semidefinite root.

Definition Let $H \in \mathbb{F}^{n \times n}$ be Hermitian positive definite, with $\mathbb{F}$ equal to $\mathbb{R}$ or $\mathbb{C}$. Its Cholesky factorization is

where $R \in \mathbb{F}^{n \times n}$ is upper triangular with positive real diagonal entries. This factor $R$ exists and is unique. Many books write $H = LL^{*}$ with $L = R^{*}$ lower triangular. The positive root also gives $H = S^{*}S$ with $S = H^{1/2}$, since $S^{*} = S$, but the two factors meet different requirements. The Cholesky factor is triangular and usually not Hermitian, and the positive root is Hermitian and usually not triangular. A matrix that is both upper triangular and Hermitian is diagonal, so $R = H^{1/2}$ exactly when $H$ is diagonal. The two factors differ by a unitary matrix on the left. The matrix $W = RS^{-1}$ satisfies

so $R = WS$ with $W$ unitary. When $H$ is only positive semidefinite and singular, the positive root still exists and is unique, but a triangular factor with nonnegative diagonal can fail to be unique. For example, $\operatorname{diag}(0, 1) = R^{\mathsf{T}}R$ for every real $R$ with rows $(0, s)$ and $(0, c)$ where $s^2 + c^2 = 1$ and $c \ge 0$.