# Is there a basis-independent proof of Abel's identity?

Abel's identity states that if $X(t)$ and $A(t)$ are $n\times n$ matrix-valued functions such that $X'(t)=A(t)X(t)$, then $\frac{d}{dt}(\det X(t)) = \mathrm{tr}\,A(t) \cdot \det X(t)$.

The question is whether there's a nice high-brow basis-independent way to see this. Given that you can state the problem without ever referring to matrix entries, I want prove it without ever referring to matrix entries. I'd expect the identity $\det e^A = e^{\mathrm{tr}(A)}$ to play a central role in the proof, but I have yet to come up with such an argument.

You can prove this without too much trouble in a bare-hands way, but I don't see how to turn this into a basis-free proof. Suppose $X_i(t)$ is the matrix you get from $X(t)$ by taking the derivative of every entry in the $i$-th row. Then $\frac{d}{dt}(\det X(t)) = \sum_{i=1}^n \det(X_i(t))$. Using the relation $X'(t)=A(t)X(t)$ you can express the derivative of the $i$-th row of $X'(t)$ in terms of the entries of $A$ and $X$. When you do this, you find that $\det X_i(t)$ is simply $\det X(t)$ multiplied by the $(i,i)$-th entry of $A(t)$.

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Please forgive this argument that plays rather fast and loose with infinitesimals. If you ignore second-order terms, an infinitesimal change $\delta t$ in $t$ takes $X(t)$ to $\big(I + \delta t A(t)\big) X(t)$. Since the determinant distributes over multiplication, all you need to show is that $\det\big(I + \delta t A(t)\big)$ is $1 + \delta t \operatorname{tr}A$ for infinitesimal $\delta t$. More formally, you need $\frac{d}{d\tau} \det(I + \tau A) = \operatorname{tr}A$. Since $I + \tau A$ agrees with $e^{\tau A}$ to first order (according to the Taylor series), you can differentiate the identity $\det e^{\tau A} = e^{\operatorname{tr} \tau A}$ with respect to $\tau$ and get the desired result.
Update: For a high-brow view, I think it makes much more sense to cast this in a geometric light, as it has much to do with the geometric interpretation of the trace. Consider how the vector ODE $x'(t) = A(t) x(t)$ acts on Euclidean $n$-space. The identity $\det e^A = e^{\operatorname{tr}A}$ is a statement of the fact that for constant $A$, the volume of any region grows (or shrinks) exponentially at the rate $\operatorname{tr}A$ under this ODE. (This is fine for varying $A(t)$ too as long as we don't care about second-order effects.) Now think of $X(t)$ as defining a parallelepiped in $n$-space as it flows around under the action of the ODE. The volume of the parallelepiped is $\det X(t)$, and its rate of change is therefore $\operatorname{tr}A$ times that. I think this is the answer I should have given in the first place.