Matrices of negative determinant are not squares

Idea. The determinant is multiplicative, and a real number squared is nonnegative.

Step 1. If $A=B^2$ with $B\in M_n(\mathbb{R})$, then
$$\det A=\det(B^2)=(\det B)^2\ge0,$$
because $\det B$ is a *real* number.

Step 2: not onto. Take $A=\operatorname{diag}(-1,1,\dots,1)\in GL_n(\mathbb{R})$, with $\det A=-1<0$. By Step 1 it is not a square. So $A\mapsto A^2$ does not map $GL_n$ onto $GL_n$. More precisely, the image lies in $GL_n^+=\{\det>0\}$.

Complex squares exist. Over $\mathbb{C}$ every invertible matrix has a square root, so the obstruction is genuinely about real matrices.

Warning: $\det>0$ is not sufficient. For $n=2$ the matrix
$$A=\begin{pmatrix}-1&1\\0&-1\end{pmatrix},\qquad\det A=1>0,$$
is not the square of any real matrix. Indeed suppose $A=B^2$; the eigenvalues $\mu$ of $B$ satisfy $\mu^2=-1$, so they are $\pm i$ (for a real $2\times2$ matrix they are conjugate), distinct. Then $B$ is diagonalisable over $\mathbb{C}$, so $B^2$ is too, with both eigenvalues $-1$, which forces $B^2=-I\ne A$. Contradiction. (Whereas $-I_2=R_{\pi/2}^2$ is a square.)

Remark. So the image of squaring is a proper subset of $GL_n^+$, and *determining* it is a classical problem: a real invertible matrix is a square iff its Jordan blocks for negative eigenvalues come in pairs of equal size.