# square root of a real matrix

I want to compute the square root of a real symmetric positive definite matrix $S\in \mathcal{M}_{m,m}$ such that $S^{1/2}S^{1/2}=S$ and it's well known that this decomposition is unique.

My question is if I have $S$ a real matrix will its square root be real too or it could be a complex matrix. In case that it could be complex then $S$ could have infinitely many square roots.

• Note that the answer by user1551 shows that the statement above needs the requirement that $S^{1/2}$ is positive (semi-)definite in order for the uniqueness claim to hold. If you are asking about possible complex positive definite square roots, then you should first make precise what exactly that means (usually "positive definite" implies real symmetric, but allowing complex Hermitian matrices is a possibility). If you don't want the positive definite requirement, infinitely many real and complex solutions may exist. Apr 4, 2013 at 5:11

A real matrix $S$ can possess infinitely many real or nonreal square roots. For example, $$S=I=\begin{pmatrix}1&t\\0&-1\end{pmatrix}^2$$ for every $t\in\mathbb{C}$. Note that $S=I$ is real and positive definite.

• But.. among positive definite symmetric matrices, square root is indeed unique. Feb 25, 2013 at 2:04
• @Berci Square roots in general are not unique, but positive semidefinite matrices have unique positive semidefinite square roots. Feb 25, 2013 at 2:08
• could you please help me in this question and thank you?math.stackexchange.com/questions/2485619/… Oct 23, 2017 at 19:36
• @Student The answer over there is fine. I have nothing to add. Note that the $M$ in your question is not positive semidefinite, so it cannot possibly possess any positive semidefinite square root. Oct 24, 2017 at 10:31

If $S$ is real, symmetric and positive definite, consider its eigenvalue / eigenvector decomposition $S = X \Lambda X^T$ where $\Lambda$ is diagonal and $X$ is orthogonal. Because $S$ is positive definite, $\Lambda_{ii} > 0$ for all $i$. The unique symmetric and positive definite square root of $S$ is given by $S^{1/2} = X \Lambda^{1/2} X^T$, where $\Lambda^{1/2}$ is the diagonal matrix with the $\sqrt{\Lambda_{ii}}$ on its diagonal. Indeed, $$S^{1/2} S^{1/2} = X \Lambda^{1/2} X^T X \Lambda^{1/2} X^T = X \Lambda X^T = S,$$ because $X^T X = I$. So $S^{1/2}$ is indeed real.

• yes you're right it mostly depends on the eigenvalues. thanks Feb 25, 2013 at 2:48
• Actually if a matrix has at least one negative part eigenvalue then it won't have a real square root but for the case of positive semidefinite matrices all eigenvalues are positive hence real square root. Feb 25, 2013 at 2:51
• -1 What does this answer show at all? That the unique symmetric and positive definite square root is real. But "symmetric positive definite" implies real by definition, so this is no news (however, no news is good news). Also this fact was already stated as well known in the question. Apr 4, 2013 at 5:15

This is a very late answer, but I think that especially Dominique's answer is not very satisfying. So, if $S$ is real, symmetric and positive definite, it is trivially self-adjoint (i.e., $A^* = A$). The square root of $S$ is then usually defined to be the unique positive definite self-adjoint matrix $S^{1/2}$ which satisfies $(S^{1/2})^2 = S$. However, it does not follow immediately that $S^{1/2}$ is real. But this is in fact the case:

We have $S = XDX^{*}$ with $D$ diagonal (eigenvalues on the diagonal) and the columns of $X$ being the particular normalized eigenvectors of $S$. These can be chosen to be real, since if $x$ is an eigenvector of $S$, then also $\overline x$ (each entry being the complex conjugate of the correspoding entry in $x$) is an eigenvector (corresponding to the same eigenvalue), so $x+\overline x$ is. Thus, $X$ can be chosen to be real*. Now, $S^{1/2} = XD^{1/2}X^T$ (square roots of the (positive) eigenvalues on the diagonal) and so $S^{1/2}$ is real.

*) In case of an eigenspace $\ker(S-\lambda)$ whose dimension is greater than one, one has to work a little more. Choose an eigenvector $u_1$ and define $x_1 := u_1 + \overline{u_1}$. Then $x_1=0$ iff $u_1$ is completely imaginary. In this case, also $iu_1$ is an eigenvector. Choose this one for $x_1$. Now, consider $x_1^\perp$ in $\ker(S-\lambda)$. Choose any $z$ in this space which is not purely imaginary and put $x_2 := z+\overline z$. Now, consider $\{x_1,x_2\}^\perp$ in $\ker(S-\lambda)$ and proceed as before. In the end, and after normalization of $x_1,x_2,\ldots$ we obtain an orthonormal basis of $\ker(S-\lambda)$ with real vectors only.