It is known that every $A$ belongs to $GL(n,\mathbb C)$ equals to $\exp(B)$ for some $n \times n$ matrix $B$. How to show the following is true? Show that a matrix $M$ belonging to $GL_n(\mathbb R)$ is the exponential of a real matrix if, and only if, it is the square of another real matrix.
Image of Matrix Exponential Map
5
$\begingroup$
lie-groups
-
1The only if is easy. Suppose $M=e^{A}$. Then $M=(e^{A/2})^2$. – 2012-09-05
-
1The other direction does not seem true to me. $M$ can be singular AND a square of another real matrix. – 2012-09-05
-
1@Tunococ I believe OP is asking about invertible matrices only (the word "belongs" in the last sentence should be "belonging") – 2012-09-05
-
0If that's the case, Alex's proof works both ways :) – 2012-09-05
-
0@Tunococ: Certainly Alex's proof does _not_ show "if"; the other real matrix need not be in the image of $\exp$. – 2012-09-05
-
0@MarcvanLeeuwen I thought it was given that a non-singular real matrix is an exponential of something, and the problem asks something about square root of a matrix. – 2012-09-05
-
0@Tunococ: No the question clearly asks about a condition for being in the image if the (real) exponential. A real matrix with (for instance) negative determinant cannot be in this image. – 2012-09-05
-
0@MarcvanLeeuwen You are right. I just realized I read it wrong. So I guess the question asks for a proof that a square of an invertible real matrix is an image of $\exp$ of a real matrix. – 2012-09-05
-
0Is it valid to assume $\exp(A) = B$ implies $\exp(\bar A) = \bar B$? – 2012-09-05