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I'm given the matrix exponential above and want to find $e^{-A}$. I was able to do so by simply making the negative exponents of the terms inside the matrix positive. Note how the product gives the identity matrix.

I was wondering why this works out. Does this step from some general property of the inverse of a matrix exponential or is it just a coincidence?

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    $\begingroup$ If a matrix $A$ commutes with a matrix $B$, then $e^Ae^B=e^{A+B}=e^Be^A$. Clearly enough $A$ commutes with $-A$, and $e^{A-A}=e^0=I$. en.wikipedia.org/wiki/Matrix_exponential $\endgroup$
    – Max
    Dec 14, 2017 at 2:55
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    $\begingroup$ @Max Right, I get that. I'm asking why $e^{-A}$ is equivalent to giving the terms inside the matrix positive exponents. $\endgroup$
    – Farhad
    Dec 14, 2017 at 3:03
  • $\begingroup$ @Max you should make this an answer, not just a comment. $\endgroup$ Dec 14, 2017 at 3:04
  • $\begingroup$ That's a coincidence stemming from the fact that $A$ has integer eigenvalues. Then again, what do you mean by "giving the terms positive exponents"? I guess that means changing $e^{-1}-e^{-4}$ to $e^1-e^4$, but how do you tell $e^{-1}-e^{-4}$ from $e^{-1.05107}$ (which is numerically the same)? $\endgroup$ Dec 14, 2017 at 6:12
  • $\begingroup$ Have you tried diagonalizing this matrix? $\endgroup$
    – user81375
    Dec 14, 2017 at 6:15

2 Answers 2


Let’s reverse-engineer this matrix a bit. Its eigenvalues can be found by inspection: The row sums are $e^{-4}$, so that’s an eigenvalue with eigenvector $[1,1]^T$, and it’s not too hard to discover that twice the first column added to the second results in $e^{-1}[1,2]^T$, so $[1,2]^T$ is also an eigenvector with eigenvalue $e^{-1}$. Now, a diagonalizable $2\times2$ matrix with eigenvalues $\lambda_1 \ne \lambda_2$ can be decomposed into $\lambda_1P_1+\lambda_2P_2$, where $P_1$ and $P_2$ are projections onto the respective eigenspaces of $\lambda_1$ and $\lambda_2$ such that $P_1P_2=P_2P_1=0$. A matrix and its inverse (if it exists) have the same eigenvectors, and the corresponding eigenvalues of the inverse are the inverses of its eigenvalues, so if we have $M=\lambda_1P_1+\lambda_2P_2$, then $M^{-1}=\frac1{\lambda_1}P_1+\frac1{\lambda_2}P_2$.

Fortunately, whoever constructed this exercise put the matrix in a form that makes it easy to perform this decomposition: $$e^A = e^{-1} \begin{bmatrix}-1&1\\-2&2\end{bmatrix} + e^{-4}\begin{bmatrix}2&-1\\2&-1\end{bmatrix}.$$ $e^{-A} = (e^A)^{-1}$ (see Max’s comment to your question), so to invert this matrix you just need to invert the eigenvalues and reassemble. It should be pretty clear at this point that this amounts to changing the negative exponents to positive ones in the original version of $e^A$, just as you’ve observed.

We can actually do a bit more reverse-engineering. It turns out that if you decompose the diagonalizable $2\times2$ matrix $M$ this way, then $e^{tM} = e^{\lambda_1 t}P_1+e^{\lambda_2 t}P_2$. (Verifyng this is a simple, but worthwhile exercise.) So, we can work backwards to find that $$A = -1 \begin{bmatrix}-1&1\\-2&2\end{bmatrix} -4 \begin{bmatrix}2&-1\\2&-1\end{bmatrix} = \begin{bmatrix}-7&3\\-6&2\end{bmatrix}$$ and that $$e^{tA} = e^{-t} \begin{bmatrix}-1&1\\-2&2\end{bmatrix} + e^{-4t} \begin{bmatrix}2&-1\\2&-1\end{bmatrix}.$$ We can then find both $e^A$ and $e^{-A}$ by plugging the appropriate values of $t$ into the last expression.


The eigenvalues of $e^A$ are $e^{-1},e^{-4}$; then there is $P$ invertible s.t. $e^A=Pdiag(e^{-1},e^{-4})P^{-1}$. Moreover, the entries of $e^A$ are $\mathbb{Q}$-linear in $e^{-1},e^{-4}$; then, and it's the key point, that implies that we can choose $P\in GL_n(\mathbb{Q})$. Note that the eigenvalues are not algebraic over $\mathbb{Q}$.

Consequently $e^{-A}=(e^A)^{-1}=Pdiag(e,e^4)P^{-1}$ has exactly the same form than $e^A$; it suffices to change the eigenvalues into their inverse.


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