Given a real symmetric positive-definite matrix $A$, it is easy to show that there is a real square root: the spectral theorem says that there will be a real eigenbasis, and that reduces the problem to taking the square root of the diagonalization. In fact, we don't need symmetry, assuming we have some other way of determining that our matrix is diagonalizable with real and positive eigenvalues.

However, there is no need for the eigenvalues to be real and positive. For example, by using the Taylor series for $\sqrt{1+x}$, one can show that a real $2\times 2$ matrix $A$ will have a real square root if $\operatorname{tr}(A)>0$ and $0<\det(A)<\operatorname{tr}(A)^2/2$.

Are there good general necessary or sufficient (or both) conditions under which a real square matrix has a real square root?

  • 1
    $\begingroup$ If the matrix is near $I$, then it has a real square root by the inverse function theorem. $\endgroup$
    – Aloizio Macedo
    Jan 17, 2016 at 2:30
  • $\begingroup$ @AloizioMacedo Yes, that is where my conditions for a $2\times 2$ matrix come from: that after scaling, subtracting the identity matrix yields a matrix whose eigenvalues are all less than $1$ in norm. $\endgroup$
    – Aaron
    Jan 17, 2016 at 2:34
  • $\begingroup$ math.stackexchange.com/questions/65227/square-root-of-a-matrix $\endgroup$ Jan 17, 2016 at 2:44
  • 2
    $\begingroup$ @mathematician I know when a matrix has a square root, assuming you're working over the complex numbers. You'll notice that the top voted answer in the link you provided was answered by me. $\endgroup$
    – Aaron
    Jan 17, 2016 at 2:46

1 Answer 1

  1. Let $A\in M_n(\mathbb{C})$; we consider the non-increasing sequence: $d_i=dim(\ker(A^i))-dim(\ker(A^{i-1}))$.

$A$ is a square IFF $A$ satisfies the following condition (P) about its iterated kernels (if $A$ is invertible, then there are no conditions!)

(P) (cf. Cross, Lancaster, Square roots of complex matrices, Linear and Multilinear Algebra): $(d_i)_i$ does not contain two successive occurrences of the same odd integer.

Note that, in Topics in Matrix Analysis, p. 472, Horn and Johnson add a condition which is useless.

  1. Let $A\in M_n(\mathbb{R})$. $A$ is a square over $\mathbb{R}$ IFF $A$ satisfies (P) and the following condition (Q) concerning the $<0$ eigenvalues of $A$.

(Q) $A$ has an even number of Jordan blocks of each size for every $<0$ eigenvalue. (Use the real Schur decomposition. cf. Functions of matrices, Higham, p.17 or Horn and Johnson -above-).

EDIT. Answer to Aaron. 1. The supplementary condition by H&J, can be rewritten: "if $dim(\ker(A))$ is odd, then $dim(\ker(A^2))<2dim(\ker(A))$". Note that $dim(\ker(A^2))\leq 2dim(\ker(A))$ is always true. Then it suffices to assume that $dim(\ker(A^2))=2dim(\ker(A))$; then $d_1=d_2$ are odd and, according to the other conditions, the square root of $A$ does not exist.

  1. The problem only stands for real $<0$ eigenvalues; indeed, if $\lambda$ is an eigenvalue of $A$ then $\bar{\lambda}$ too, and the dimensions of the Jordan blocks associated to $\lambda$ are the same as those of $\bar{\lambda}$; it is not difficult to find a square root of $diag(\lambda I+J_k,\bar{\lambda}I+J_k)$ when $\lambda\notin \mathbb{R}$ or of $\lambda I+J_k$ when $\lambda >0$ or of $diag(\lambda I+J_k,\lambda I+J_k)$ when $\lambda<0$.
  • $\begingroup$ Why do you say the condition is useless? The question is specifically about whether there is a square root in $M_{n}(\mathbb R)$, and the condition seems to address this directly. Moreover, the hint to use real Schur decomposition (which I had not seen before) seems like it might help me understand why it is true. One question: When you say $"<0"$ eigenvalue, are you referring only to the real eigenvalues, or does it mean, say, that the real part is negative? $\endgroup$
    – Aaron
    Jan 17, 2016 at 14:03
  • $\begingroup$ Re: your edit: For (1) I had misunderstood/misread, I thought you were saying that condition (Q) was useless. For (2) Yes, it is straight forward to find a square root of those Jordan matrices, but if a real matrix $A$ satisfies $A=PJP^{-1}$, with neither $P$ nor $J$ real, it is not a priori clear why $P\sqrt{J}P^{-1}$ should be a real matrix. I think that after I have read into real schur deomposition I will be able to circumvent this problem. When I have, I will likely accept this answer. $\endgroup$
    – Aaron
    Jan 17, 2016 at 14:51

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