The derivative of a determinant

Harald Hanche-Olsen

                       hanche@math.ntnu.no

Abstract? No, not really. This is a classical result.

Background and a simple result

Let Φ(t) be an n × n matrix depending on a parameter t. If Φ is a differentiable function of t — that is, each of its components is differentiable with respect to t — then so is det Φ(t), since we know that the determinant is a polynomial in the components of Φ. To get from this to an actual computation of the derivative of det Φ(t) is a different matter, though.

What we shall need is the fact that the determinant is a multilinear function of its rows: If we write the rows of Φ as φ_{1}, …, φ_{n} and think of the determinant as a function of the rows

det Φ = d(φ_{1}, …, φ_{n})

then d is a linear function of each of its arguments as long as we keep each of the remaining rows constant. We then get

\frac{d}{dt} det Φ(t) = d(φ̇_{1}, φ_{2}, …, φ_{n}) + d(φ_{1}, φ̇_{2}, …, φ_{n}) + ⋯

+ d(φ_{1}, φ_{2}, …, φ̇_{n}) (1)

I outline the proof of this only for n = 3, to keep the notation simple. It should be clear how to generalize the proof to arbitrary n. If h ≠ 0 then

h^{−1}(Φ(t + h) − Φ(t)) = d(h^{−1}(φ_{1}(t + h) − φ_{1}(t)), φ_{2}(t + h), φ_{3}(t + h))
+ d(φ_{1}(t), h^{−1}(φ_{2}(t + h) − φ_{2}(t)), φ_{3}(t + h))
+ d(φ_{1}(t), φ_{2}(t), h^{−1}(φ_{3}(t + h) − φ_{3}(t)))

which has the stated limit as h → 0. (We must use the continuity of d for this argument to work.)

A better result

Equation (1) requires the computation of n determinants for the computation of a single derivative. We can do much better than this! For example, if Φ(t) is the identity matrix then a moment’s contemplation of the righthand side of (1) shows it is the trace of Φ̇. Indeed, the first term will be

|\begin{matrix}φ̇_{11} & φ̇_{12} & ⋯ & φ̇_{1n} \\ 0 & 1 & ⋯ & 0 \\ ⋮ & ... & ... & ⋮ \\ 0 & 0 & ⋯ & 1 \end{matrix}| = φ̇_{11}

and so forth, with the sum φ̇_{11} + φ̇_{22} + ⋯ + φ̇_{nn} = tr Φ̇. The resulting formula

\frac{d}{dt} det Φ(t) = tr Φ̇(t) when Φ(t) = I

may seem like a rather useless special case, but appearances deceive! For, let A be a constant, invertible matrix and apply the above result to the function det(AΦ(t)) = det A det Φ(t). Now, the above formula states that

det A \frac{d}{dt} det Φ(t) = tr(AΦ̇(t)) when AΦ(t) = I

Whenever Φ(t) is invertible we can apply this result with A = Φ(t)^{−1} and rearrange to get the result

\frac{d}{dt} det Φ(t) = det Φ(t) tr (Φ(t)^{−1}Φ̇(t))

This result can also be written in the following useful form:

\frac{d}{dt} ln det Φ(t) = tr (Φ(t)^{−1}Φ̇(t)).