Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
The only if is easy. Suppose $M=e^{A}$. Then $M=(e^{A/2})^2$. –  Alex Becker Sep 5 '12 at 2:31
1  
The other direction does not seem true to me. $M$ can be singular AND a square of another real matrix. –  Tunococ Sep 5 '12 at 2:43
1  
@Tunococ I believe OP is asking about invertible matrices only (the word "belongs" in the last sentence should be "belonging") –  user29743 Sep 5 '12 at 3:00
    
If that's the case, Alex's proof works both ways :) –  Tunococ Sep 5 '12 at 4:42
    
@Tunococ: Certainly Alex's proof does not show "if"; the other real matrix need not be in the image of $\exp$. –  Marc van Leeuwen Sep 5 '12 at 7:10

2 Answers 2

Let $A = B^2$ be an invertible real matrix, where $B$ is also real. We must show that $A$ is an exponential of a real matrix. (The converse is clear, as in the comments above.)

The first step is to conjugate $B$ by a real matrix into real Jordan form. In real Jordan form, a matrix is made up of real Jordan blocks, which have the form

$$\left(\begin{matrix} C & 1 & \\ &C & 1 & \\ & & \ldots \\ & & & C & 1 \\ & & & & C \end{matrix}\right)$$

Here $C$ can either be a 1 by 1 real scalar, or a 2 by 2 block of the form $$\left(\begin{matrix} a & b\\-b& a \end{matrix}\right).$$ (In the latter case, the 1s in the matrix must be interpreted as 2 by 2 identity matrices. The latter case corresponds to eigenvalues $a \pm bi$.)

Since both exponential and squaring commute with both conjugation and "blocking", it is sufficient to prove the result for real Jordan blocks, i.e., if $B$ is a real Jordan block, then $B^2$ is an exponential of a real matrix.

Now it is a matter of checking the various cases.

If the Jordan block $B$ is a 1 by 1 scalar matrix, then its square is a positive real number, and the result just says that positive real numbers are exponentials, which is true.

If the Jordan block $B$ is $n$ by $n$ consisting of a single nonzero real scalar on the diagonal (i.e. has $n$ equal nonzero real eigenvalues), then $A = B^2$ is an upper triangular matrix with a single positive scalar $c$ repeated on the diagonal. We must check that any such matrix is an exponential.

A candidate matrix should have form $(\log c)I + J$ where $J$ is strictly upper triangular (0's on the main diagonal). Since $\exp((\log c)I + J) = c \exp(J)$, it is sufficient to consider the case $c=1$.

Thus, we are reduced to showing that $\exp$ is surjective from the set of strictly upper triangular matrices (which I will denote $T_0$) , to upper triangular matrices with 1 on the diagonal (denoted $T_1$). This isn't too hard to do by brute force (by calculating explicitly the exponential of an element of $T_0$ ; remember that all terms of the exponential series starting with the $n$th vanish on $T_0$). Alternatively, one can use the Baker-Campbell-Hausdorff formula applied to the Lie algebra $T_0$. Because this Lie algebra is nilpotent, repeated Lie brackets eventually vanish, so the right hand side of the BCH formula becomes finite and therefore converges for all pairs of elements in $T_0$. Hence the image of $T_0$ under $\exp$ is closed under multiplication and inversion, and is hence a subgroup of $T_1$. Because $T_1$ is connected, and the image of $\exp$ contains a neighborhood of the identity (true for any Lie group), it follows that the image must be all of $T_1$.

The remaining cases are where the Jordan blocks of $B$ consist of 2 by 2 blocks, with eigenvalues $a \pm bi$. If it's a single 2 by 2 block, the result follows from the surjectivity of $\exp$ in $\mathbb{C}$ (essentially, because $(a \pm bi)^2$ has a complex logarithm). If the Jordan block is bigger, then we use a similar argument as above: first, reduce to the case where the eigenvalue is 1, and then use the Lie algebra argument as above.

share|improve this answer

This is not a complete proof, however, it may help you to come up with complete proof. I assume you have some knowledge of Lie groups and exponentials, if there is anything I mention that is not clear, say so and I'll try to clarify.

$GL_n(\mathbb{R})$ has two connected components, this fact is suggested (though not proven) by the determinant map

$\det:GL_n(\mathbb{R})\rightarrow \mathbb{R}^{\times}$

These connected components correspond to the matrices with positive and negative determinant. Call them $GL_n(\mathbb{R})^-$ and $GL_n(\mathbb{R})^+$

Then suppose that $M=A^2$ for some real matrix $A$. Then $\det(M)=\det(A)^2>0$, so $M\in GL_n(\mathbb{R})^+$

Two other important facts about lie groups and exponentials:

$(1)$ If $G$ is a lie group, then any open neighborhood of the identity element generates the connected component of the identity, usually called $G^0$

$(2)$ the exponential map is a local homeomorphism (about the origin), so there is an open neighborhood of the zero matrix in $M_n(\mathbb{R})$ that maps homeomorphically to an open neighborhood of the identity in $GL_n(\mathbb{R})$

This means that there is an open set $U$ in $GL_n(\mathbb{R})$ such that for any $A\in U$, $A=e^{B}$ for some $B$ in $M_n(\mathbb{R})$. Then $U$ generates the connected component of the identity, so any matrix in $GL_n(\mathbb{R})^+$ is a product of exponentials. Thus $M=e^{X_1}\ldots e^{X_n}$ for some real matrices $X_1,\ldots,X_n$.

This is not quite answer to the original post, because it is not the image of one matrix under the exponential. But hopefully it helps!

share|improve this answer
    
And note that $\begin{pmatrix}-1&0\\0&-2\end{pmatrix}$ is not in the image of $\exp$, in spite of it having positive determinant. –  Marc van Leeuwen Sep 5 '12 at 11:38
    
@ Marc van Leeuwen Yes exactly, and $\left(\begin{matrix} -1 & 0\\0&-2 \end{matrix}\right)$ is the product of $\left(\begin{matrix} 1 & 0\\0&2 \end{matrix}\right)$ and $\left(\begin{matrix} -1 & 0\\0&-1 \end{matrix}\right)$. The first is certainly in the image, and so is the second. In fact any matrix of form $\left(\begin{matrix} \cos{t} & -\sin{t}\\ \sin{t}& \cos{t} \end{matrix}\right)$ is in the image. An element that maps to it is $\left(\begin{matrix} 0& -t\\t&0 \end{matrix}\right)$ –  gabbering Sep 5 '12 at 11:59

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.