Derivative of the Wronskian

Question

Consider a non-autonomous linear system of ode's: $X'=A(t)X$, $X:\mathbb{R}\rightarrow\mathbb{R}^n$. Let $B(t)$ be a fundamental matrix solution $\dot{B}=A(t)B$ of the system and $W(t):=\det B(t)$ the Wronskian. Show that $$\dot{W}=tr(A(t))W.$$ My idea: I started by induction. For $n=2$ we have $B=\left(\begin{array}{cc} x_{1} & y_{1}\\ x_{2} & y_{2} \end{array}\right)$ and $$W =\det B =x_{1}y_{2}-y_{1}x_{2}.$$ Therefore, $$ \begin{eqnarray} \dot{W} &=&\dot{x_{1}}y_{2}+x_{1}\dot{y_{2}}-\dot{y_{1}}x_{2}-y_{1}\dot{x_{2}}\\ &=&\dot{x_{1}}y_{2}-\dot{y_{1}}x_{2}+x_{1}\dot{y_{2}}-y_{1}\dot{x_{2}}\\ &=&\det\left(\begin{array}{cc} \dot{x_{1}} & x_{2}\\ \dot{y_{1}} & y_{2} \end{array}\right)+\det\left(\begin{array}{cc} x_{1} & \dot{x_{2}}\\ y_{1} & \dot{y_{2}} \end{array}\right)\\ &=&\det\left(\begin{array}{cc} a_{11}x_{1}+a_{12}x_{2} & x_{2}\\ a_{11}y_{1}+a_{12}y_{2} & y_{2} \end{array}\right)+\det\left(\begin{array}{cc} x_{1} & a_{21}x_{1}+a_{22}x_{2}\\ y_{1} & a_{21}y_{1}+a_{22}y_{2} \end{array}\right)\\ &=&\det\left(\begin{array}{cc} a_{11}x_{1} & x_{2}\\ a_{11}y_{1} & y_{2} \end{array}\right)+\det\left(\begin{array}{cc} a_{12}x_{2} & x_{2}\\ a_{12}y_{2} & y_{2} \end{array}\right)+\det\left(\begin{array}{cc} x_{1} & a_{21}x_{1}\\ y_{1} & a_{21}y_{1} \end{array}\right)+\det\left(\begin{array}{cc} x_{1} & a_{22}x_{2}\\ y_{1} & a_{22}y_{2} \end{array}\right)\\ &=&\det\left(\begin{array}{cc} a_{11}x_{1} & x_{2}\\ a_{11}y_{1} & y_{2} \end{array}\right)+\det\left(\begin{array}{cc} x_{1} & a_{22}x_{2}\\ y_{1} & a_{22}y_{2} \end{array}\right)\\ &=&a_{11}\det\left(\begin{array}{cc} x_{1} & x_{2}\\ y_{1} & y_{2} \end{array}\right)+a_{22}\det\left(\begin{array}{cc} x_{1} & x_{2}\\ y_{1} & y_{2} \end{array}\right)\\ &=&(a_{11}+a_{22})\det\left(\begin{array}{cc} x_{1} & x_{2}\\ y_{1} & y_{2} \end{array}\right)\\ &=&tr(A)W. \end{eqnarray}$$ Now I suppose that it is valid for $n>2$ and consider taking $n+1$ and the formula for the determinant $$W=\sum_{k=1}^{n+1}(-1)^{i+k}b_{ik}\det B_{ik}$$ where $B_{ik}$ is the $ik$-minor of $B$. Up to here I haven't been able to advance. Any suggestions?

Answer 1

The link in the comments contains interesting responses towards solving the question, but here I would like to add a derivation based on the properties of the Levi-Civita symbol, which is very concise in my opinion, at the cost of handling some indices and tensor calculus conventions.

The determinant of the $n\times n$ matrix can be written

$$n!\det B=\epsilon_{i_1...i_n}\epsilon_{j_1...j_n}B_{i_1j_1}...B_{i_nj_n}$$

where all the indices here are summed over their range. Since $\dot{B}_{i_kj_k}=A_{j_k \ell}B_{i_k\ell}$ we can take the derivative of this relation:

$$n!\dot{W}=\sum_k\epsilon_{i_1...i_n}\epsilon_{j_1...j_n}A_{j_k \ell}B_{i_k\ell}\prod _{m\neq k}B_{i_mj_m}$$

Now notice that $\epsilon_{i_1...i_n}B_{i_1j_1}...B_{i_k\ell}...B_{i_nj_n}=\epsilon_{j_1...\ell...j_n}W $

which leads us, after reordering indices appropriately and taking account of the minus signs generated, to the relation

$$n!\dot{W}=W\sum_k\epsilon_{\ell j_1...j_n}\epsilon_{j_k j_1...j_n}A_{j_k \ell}$$

Here, note that $\epsilon_{\ell j_1...j_n}\epsilon_{j_k j_1...j_n}=(n-1)!\delta_{\ell j_k}$ which allows us to write

$$n!\dot{W}=(n-1)!W\sum_k\delta_{ \ell j_k }A_{j_k\ell}=(n-1)!\sum_{k=1}^n(\text{tr} A) W\Rightarrow\dot{W}=(\text{tr}A(t))W $$