$\det A(t)=1$, $A(0)=E$, show $\operatorname{tr} A'(0)=0$.

For $n\times n$ continuously differentiable matrix $A(t)$.

$\det A(t)=1$, $A(0)=E$, show $\operatorname{tr} A'(0)=0$.

I have just know something like $(e^{at})'=ae^{at}$.

Solution 1:

Consider $A'(t)=(A'(t)A^{-1}(t))A(t)$. Thus $A$ can be seen as a fundamental matrix of the linear ode system $X'(t)=C(t)X(t)$ with $C(t)=A'(t)A^{-1}(t)$. For each such fundamental matrix the differential equation for the Wronski determinant is valid: $\frac{d}{dt} \det(X(t)) = \operatorname{tr}(C(t))\det(X(t))$. Thus $$ 0=\frac{d}{dt} \det(A(t)) = \operatorname{tr}(A'(t)A^{-1}(t))\det(A(t)) = \operatorname{tr}(A'(t)A^{-1}(t)). $$ Inserting $t=0$ gives $0=\operatorname{tr}(A'(0)E)=\operatorname{tr}(A'(0))$.

Solution 2:

Hint: Jacobi's formula tells you that

$$ \frac{d}{dt} \det A(t) = \left(\operatorname{det}A(t) \right) \cdot \operatorname{tr} \left (A(t)^{-1} \cdot \, \frac{dA(t)}{dt}\right )$$ Now plugging in your information about $A(t)$ gives the result.