Symmetric Matrix Square Roots

Why Take a Matrix Square Root?

A scalar square root rescales a number in a way that reverses squaring. For matrices, the same idea is useful when a calculation calls for a matrix whose product with itself is a given matrix. In multivariate statistics, such transformations arise in operations involving covariance matrices, including rescaling directions that have different variances. The symmetric square root is especially useful because it treats the eigendirections of a symmetric matrix directly and does not depend on a particular triangular factorization.

This tutorial focuses on symmetric real matrices. The key tool is the spectral decomposition: for a real symmetric \(p\times p\) matrix \(A\), there is an orthogonal matrix \(Q\) and a diagonal matrix \(\Lambda\) such that \(A=Q\Lambda Q^{\mathsf T}\). The columns of \(Q\) are orthonormal eigenvectors, and the diagonal entries of \(\Lambda\) are the corresponding real eigenvalues. We use \(Q^{\mathsf T}Q=I\).

Constructing the Square Root from Eigenvalues

Suppose \(A=Q\Lambda Q^{\mathsf T}\), where \(\Lambda=\operatorname{diag}(\lambda_1,\ldots,\lambda_p)\). If every \(\lambda_i\) is nonnegative, define

The construction keeps the same orthonormal eigenvectors and replaces each eigenvalue by its nonnegative scalar square root. In particular, a zero eigenvalue remains zero. The resulting matrix is symmetric because its middle diagonal matrix is symmetric, and it is positive semidefinite because for any \(x\),

Squaring the construction recovers \(A\): the orthogonality identity \(Q^{\mathsf T}Q=I\) makes the two adjacent copies of \(Q\) cancel, and \((\sqrt{\lambda_i})^2=\lambda_i\) for each eigenvalue.

Proof. First suppose \(A\) is positive semidefinite. By the spectral theorem, \(A=Q\Lambda Q^{\mathsf T}\) with real eigenvalues \(\lambda_i\). For an eigenvector \(q_i\), positive semidefiniteness gives \(\lambda_i=q_i^{\mathsf T}Aq_i\ge 0\), since \(q_i\) is a unit vector. The construction above is therefore defined, is symmetric and positive semidefinite, and squares to \(A\).

Conversely, suppose \(A=B^2\) for a symmetric positive semidefinite \(B\). For every \(x\),

Thus \(A\) is positive semidefinite.

It remains to prove uniqueness. Let \(B\) be any symmetric positive semidefinite matrix satisfying \(B^2=A\). Then \(AB=B^2B=B^3=BB^2=BA\), so \(A\) and \(B\) commute. For each eigenvalue \(\lambda\) of \(A\), let \(E_\lambda=\{x:Ax=\lambda x\}\) be its eigenspace. If \(x\in E_\lambda\), then \(A(Bx)=B(Ax)=\lambda Bx\); hence \(B\) maps \(E_\lambda\) into itself. The restriction of \(B\) to this space is symmetric and positive semidefinite, and its square is \(\lambda I\). By the spectral theorem applied to this restriction, each of its eigenvalues \(\beta\) is nonnegative and satisfies \(\beta^2=\lambda\). Therefore every such \(\beta\) equals \(\sqrt{\lambda}\), and \(B\) acts as \(\sqrt{\lambda}I\) on \(E_\lambda\). This holds on every eigenspace of \(A\), so \(B\) must equal the spectral construction above. \(\square\)

Singular Matrices and Inverse Square Roots

The square-root construction allows zero eigenvalues, so it applies to positive semidefinite matrices that are singular as well as to positive definite ones. An ordinary inverse square root has a stricter existence condition: the matrix must be positive definite. For a positive definite \(A=Q\Lambda Q^{\mathsf T}\), all eigenvalues are strictly positive, and the inverse of its symmetric square root is

Indeed, multiplying \(A^{-1/2}\) by \(A^{1/2}\) gives \(QIQ^{\mathsf T}=I\). If a positive semidefinite matrix has a zero eigenvalue, its square root has a zero eigenvalue too and cannot have an ordinary inverse. More generally, a singular matrix itself cannot have an ordinary inverse. One can still form a matrix by taking reciprocal square roots on the positive-eigenvalue directions and leaving zeros on the zero-eigenvalue directions; that construction is not an ordinary inverse square root, since it does not invert the action on the null space.

Proof. If \(A\) is positive definite, its spectral decomposition has \(\lambda_i>0\) for every \(i\). The displayed spectral construction for \(A^{-1/2}\) is then defined, and its product with \(A^{1/2}\) is \(QIQ^{\mathsf T}=I\). Thus it is the ordinary inverse of \(A^{1/2}\). Conversely, if \(A\) is positive semidefinite but not positive definite, its spectral decomposition includes at least one zero eigenvalue. The corresponding eigenvalue of \(A^{1/2}\) is also zero, so \(A^{1/2}\) is singular and has no ordinary inverse. Therefore an ordinary inverse square root exists exactly in the positive definite case. \(\square\)

Common Mistake and Practical Checks

A common mistake is to treat every positive semidefinite matrix as invertible because its eigenvalues are nonnegative. Nonnegative is enough for a symmetric positive semidefinite square root, but not for an ordinary inverse square root: a single zero eigenvalue prevents inversion. The correction is to check whether eigenvalues are all nonnegative or all strictly positive, depending on which construction is needed.

A second mistake is to take square roots of the entries of a matrix separately. Matrix multiplication mixes entries, so entrywise square roots generally do not produce a matrix whose square is the original matrix. The construction is performed on eigenvalues in an orthonormal eigenbasis, then transformed back using \(Q\) and \(Q^{\mathsf T}\).

When checking a proposed result, verify the defining matrix equation, symmetry, and the required definiteness. For \(A^{1/2}\), check \((A^{1/2})^2=A\) and that its eigenvalues are nonnegative. For \(A^{-1/2}\), check that the eigenvalues of \(A\) are strictly positive and that \((A^{-1/2})^2=A^{-1}\), or equivalently that \(A^{-1/2}A^{1/2}=I\).