Let \(F(t)\) be an \(n\times n\) matrix whose entries are continuously differentiable real functions of \(t\), and define \[ u(t)=\operatorname{tr}(F(t)^3). \] Show that \(u\) is differentiable and that \[ u'(t)=3\operatorname{tr}(F(t)^2F'(t)). \]
Solution
1. The matrix-valued function \(F^3\) is differentiable and\[ (F^3)' = F'F^2+FF'F+F^2F'. \]
Each entry of a product of two matrix-valued functions is a finite sum of products of scalar functions. Hence the ordinary product rule applied entrywise gives \[ (AB)'=A'B+AB' \] whenever \(A\) and \(B\) are differentiable matrix-valued functions. Applying this twice, \[ \begin{aligned} (F^3)' &= (F^2)'F+F^2F'\\ &= (F'F+FF')F+F^2F'\\ &= F'F^2+FF'F+F^2F'. \end{aligned} \]
2. The function \(u\) is differentiable and\[ \boxed{ u'(t)=3\operatorname{tr}\bigl(F(t)^2F'(t)\bigr) }. \]
The trace is a finite sum of diagonal entries, so step 1 implies \[ \begin{aligned} u' &= \operatorname{tr}\bigl((F^3)'\bigr)\\ &= \operatorname{tr}(F'F^2) +\operatorname{tr}(FF'F) +\operatorname{tr}(F^2F'). \end{aligned} \] Using the cyclic identity \[ \operatorname{tr}(AB)=\operatorname{tr}(BA), \] one obtains \[ \operatorname{tr}(F'F^2) = \operatorname{tr}(F^2F') \] and \[ \operatorname{tr}(FF'F) = \operatorname{tr}(F^2F'). \] Thus all three terms are equal, giving the claimed formula.
3. Q.E.D.
Step 2 proves both the differentiability of \(u\) and the required derivative formula.