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For $A \in M(2, \mathbb R)$ , if $A$ has a square root $B \in M(2, \mathbb R)$ , then $\det A=(\det B)^2$ is non-negative ; and if

$s:=\sqrt{\det A}$ ; then using $A^2+s^2I=(trA)A$ , evaluating $(A+sI)^2=(2s+trA)A$ and

$(A-sI)^2=(trA-2s)A$ , I can calculate the square root of $A$ . But can this method be extended

to higher order matrices ? For higher order matrices , if the matrix is diagonalizable say

$A=XDX^{-1}$ where $D$ is the diagonal matrix , then we can find a square root $S$ of $D$ taking the

real square roots of the diagonal entries of $D$ if this diagonal entries are non-negative , then

$XSX^{-1}$ is a square root of $A$ ; but what if a diagonal entry is negative ? More fundamentally ,

when can we say that a square root of $A \in M(n,\mathbb R)$ exists in $M(n,\mathbb R)$ ?

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Let $A$ be a real square matrix of size $n$. The following statements are equivalent (cf. the answers of loup blanc and mine in "Writing real invertible matrices as exponential of real matrices"):

  1. $A$ has a real logarithm, i.e. $A=e^X$ for some real matrix $X$.
  2. $A$ has a nonsingular real square root, i.e. $A=B^2$ for some nonsingular real matrix $B$.
  3. $A$ is nonsingular and in its (complex) Jordan normal form, every Jordan block corresponding to a negative eigenvalue occurs an even number of times.

So, in your example, if $D$ is a real diagonal matrix that has a simple negative eigenvalue, then it has no real matrix square root.

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