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.

I'm looking for a proof of this claim: "every invertible matrix can be written as the exponential of another matrix". I'm not familiar yet with logarithms of matrices, so I wonder if a proof exists, without them. I'll be happy with any proof anyways. I hope someone can help.

share|improve this question
Have you learned Jordan form of matrices? You are working with complex entries? –  Jonas Meyer Mar 18 '13 at 16:37
Yes, I'm familiar with Jordan forms and it is indeed one with complex entries :D. –  yarnamc Mar 18 '13 at 16:39
@user20327: You might like to review this en.wikipedia.org/wiki/Matrix_exponential from Jonas' comment. –  Amzoti Mar 18 '13 at 16:42
add comment

3 Answers

up vote 4 down vote accepted

I assume you are talking about complex $n\times n$ matrices. This is not true in general within real square matrices.

A simple proof goes by functional calculus. If $A$ is invertible, you can find a determination of the complex logarithm on some $\mathbb{C}\setminus e^{i\theta_0}[0,+\infty)$ which contains the spectrum of $A$. Then by holomorphic functional calculus, you can define $B:=\log A$ and it satisfies $e^B=A$.


1) There is a formula that says $\det e^B=e^{\mbox{trace}\;B}$ (easy proof by Jordan normal form, or by density of diagonalizable matrices). Therefore the range of the exponential over $M_n(\mathbb{C})$ is exactly $GL_n(\mathbb{C})$, the group of invertible matrices.

2) For diagonalizable matrices $A$, it is very easy to find a log. Take $P$ invertible such that $A=PDP^{-1}$ with $D=\mbox{diag}\{\lambda_1,\ldots,\lambda_n\}$. If $A$ is invertible, then every $\lambda_j$ is nonzero so we can find $\mu_j$ such that $\lambda_j=e^{\mu_j}$. Then the matrix $B:=P\mbox{diag}\{\mu_1,\ldots,\mu_n\}P^{-1}$ satisfies $e^B=A$.

3) If $\|A-I_n\|<1$, we can define explicitly a log with the power series of $\log (1+z)$ by setting $\log A:=\log(I_n+(A-I_n))=\sum_{k\geq 1}(-1)^{k+1}(A-I_n)^k/k.$

4) For a real matrix $B$, the formula above shows that $\det e^B>0$. So the matrices with a nonpositive determinant don't have a log. The converse is not true in general. A sufficient condition is that $A$ has no negative eigenvalue. For a necesary and sufficient condition, one needs to consider the Jordan decomposition of $A$.

5) And precisely, the Jordan decomposition of $A$ is a concrete way to get a log. Indeed, for a block $A=\lambda I+N=\lambda(I+\lambda^{-1}N)$ with $\lambda\neq 0$ and $N$ nilpotent, take $\mu$ such that $\lambda=e^\mu$ and set $B:=\mu+\log(I+\lambda^{-1}N)=\mu+\sum_{k\geq 1}(-1)^{k+1}\lambda^{-k}N^k$ and note that this series has actually finitely many nonzero terms since $N$ is nilpotent. Do this on each block, and you get your log for the Jordan form of $A$. It only remains to go back to $A$ by similarity.

6) Finally, here are two examples using the above: $$ \log\left( \matrix{5&1&0\\0&5&1\\0&0&5}\right)=\left(\matrix{\log 5&1&-\frac{1}{2}\\ 0&\log 5&1\\0&0&\log 5} \right) $$ and $$ \log\left(\matrix{-1&0\\0&1} \right)=\left(\matrix{i\pi&0\\0&0} \right) $$ are two possible choices for the log of these matrices.

share|improve this answer
Thank you very much for this post! (it'll take some time to fully comprehend it though, but that's always the case with maths :)) –  yarnamc Mar 18 '13 at 18:08
@user20327 You're welcome. Let me know if you have any question. –  1015 Mar 18 '13 at 18:23
add comment

As julien has pointed out, presumably we work over the complex field, as the statement is not true over $\mathbb{R}$ (e.g. $e^x=-1$ has no real solution). Then here is an elementary proof for the statement. Let $A=S\left(J_{k_1}(\lambda_1)\oplus\cdots\oplus J_{k_m}(\lambda_m)\right)S^{-1}$ be a Jordan decomposition of $A$, where each eigenvalue is nonzero. Then $\lambda_i=e^{z_i}$ for some $z_i\in\mathbb{C}$. Now consider the Jordan block $Y_i=J_{k_i}(z_i)$. Then $\exp(Y_i)$ is an upper triangular matrix that is always similar to $J_{k_i}(\lambda_i)$. That is, $J_{k_i}(\lambda_i)=P_i\exp(Y_i)P_i^{-1}$ for some invertible matrix $P_i$. Now define $X_i=P_iY_iP_i^{-1}$. Then $A=\exp\left(S(X_1\oplus\ldots\oplus X_m)S^{-1}\right)$.

share|improve this answer
Thank you :), that was very clear –  yarnamc Mar 18 '13 at 17:44
add comment

see this book on the net, it can be helpful Miroslav Fiedler, Special Matrices and Their Applications in Numerical Mathematics, Springer (1986)

share|improve this answer
add comment

Your Answer


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.