Suppose I have a matrix $A$ with real entries such that the off-diagonal entries of $A$ are positive or zero. (The diagonal entries may be positive, negative or zero.)

From doing a few examples in Python, it looks like the following might be true of the matrix exponential $e^A$:

  • The entries of $e^A$ are all real and non-negative (both on and off the diagonal), and

  • If an entry of $A$ is non-zero, the corresponding entry of $e^A$ will be positive. (For zero entries of $A$, the corresponding entry in $e^A$ might be zero or positive.)

Are these things indeed the case? How can this be shown? Is there a result that will allow me to predict which elements of $e^A$ will be positive, depending on which elements of $A$ are non-zero?


Your first conjecture is true.

Lemma. If the square matrix $B$ has real non-negative entries, then $e^B$ has real non-negative entries.

Proof. Clearly every power of $B$ has real non-negative entries, and the result follows from the definition $$e^B=I+B+{\textstyle\frac{1}{2!}}B^2+{\textstyle\frac{1}{3!}}B^3+\cdots\ .$$

Theorem. If $A$ is a real square matrix for which the off-diagonal entries are non-negative, then all entries of $e^A$ are real and non-negative.

Proof. As all off-diagonal entries of $A$ are non-negative, adding a suitable scalar multiple of $I$ to $A$ will give a matrix $$B=A+kI$$ in which all entries are non-negative. By the lemma, $e^B$ has real non-negative entries, and so $$e^A=e^{B-kI}=e^{-kI}e^B=e^{-k}e^B$$ also has real non-negative entries. Note that we have used the fact that $e^{X+Y}=e^Xe^Y$ whenever the matrices $X$ and $Y$ commute.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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