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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?

N. Virgo
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1 Answers1

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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.

David
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