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It is known that the matrix exponential over the real matrices $\exp : M_n(\mathbb{R}) \to GL_n(\mathbb{R})$ is not surjective and that its image $S $ is the subset of all invertible matrices that are the square of a real matrix.

My question is about the "size" of that set within $GL_n(\mathbb{R})$ equipped with the Lebesgue measure of $\mathbb{R}^{n^2}$. Do we know if $S$ has full measure ? Or is it a null set ?

Phil-W
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Note that the image includes all real matrices $n\times n$ matrices that have $n$ distinct positive eigenvalues. The set of all such matrices is open and nonempty, and therefore has positive Lebesgue measure. Indeed, since this set is closed under multiplication by positive scalars, the Lebesgue measure must be infinite.

Jim Belk
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    Thank you, does that imply that the complement of S has zero measure (it's what I have in mind by "full measure") ? – Phil-W Jan 09 '21 at 00:15
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    @Phil-W No, the complement has infinite measure as well. For example, the complement includes all real $n\times n$ matrices that have $n$ distinct eigenvalues and at least one negative eigenvalue. – Jim Belk Jan 09 '21 at 00:27
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    @Phil-W (cont.) The argument is that if $\lambda$ is an eigenvalue for $A^2$, then one of the (complex) square roots of $\lambda$ must be an eigenvalue for $A$. For if $v$ is an eigenvector for $A^2$ associated to $\lambda$ and $r$ is a square root of $\lambda$, then $(A^2-r^2 I)v=0$ so $(A+r I)(A-rI)v=0$. Then either $(A-rI)v=0$ and $v$ is an eigenvector for $A$ with eigenvalue $r$, or $(A-rI)v\ne 0$ and $(A-rI)v$ is an eigenvector for $A$ with eigenvalue $-r$. If $A$ is a real matrix and $A^2$ has $n$ distinct eigenvalues, it follows that none of the eigenvalues of $A^2$ can be negative. – Jim Belk Jan 09 '21 at 00:41