# Ngô Quốc Anh

## March 16, 2011

### Cofactor matrix has divergence-free rows

Filed under: PDEs — Ngô Quốc Anh @ 15:45

In this entry, we prove the following interesting result

Let $\mathbf{u} : \mathbb R^n \to \mathbb R^n$ be a smooth function. Then

$\displaystyle \sum_{i=1}^n (\mbox{cof}D\mathbf{u})_{i,x_i}^k=0$

for each $k=\overline{1,n}$ fixed.

For simplicity, let us write $\mathbf{u}=(u^1,...,u^n) \in \mathbb R^n$. Then

$\displaystyle D\mathbf{u} = \left( {\begin{array}{*{20}{c}} {u_{{x_1}}^1}&{u_{{x_2}}^1}& \cdots &{u_{{x_2}}^1} \\ {u_{{x_1}}^2}&{u_{{x_2}}^2}& \cdots &{u_{{x_2}}^2} \\ \vdots & \vdots & \ddots & \vdots \\ {u_{{x_1}}^n}&{u_{{x_2}}^n}& \cdots &{u_{{x_n}}^n} \end{array}} \right).$

Proof. We recall from this entry that for each matrix $A=(a_{ij})$,

$\displaystyle \frac{\partial}{\partial a_{ij}}\mbox{det}(A)=(\mbox{cof}A)_i^j.$

This and the fact that

$\displaystyle A^{-1}=\frac{1}{\mbox{det}(A)}\mbox{cof}A$

or equivalently

$\displaystyle \mbox{det}(A)I=A\mbox{cof}A$

i.e.,

$\displaystyle \mbox{det}(A)\delta_{ij}=\sum_{k=1}^n a_{ik}\mbox{cof}A_j^k.$

give us after differentiating with respect to $x_j$

$\displaystyle\frac{d}{{d{x_j}}}{\text{det}}({\text{cof}}D{\mathbf{u}}){\delta _{ij}} = \frac{d}{{d{x_j}}}\sum\limits_{l = 1}^n {u_{{x_l}}^i} ({\text{cof}}D{\mathbf{u}})_j^l = \sum\limits_{l = 1}^n {\left[ {u_{{x_l}{x_j}}^i({\text{cof}}D{\mathbf{u}})_j^l + u_{{x_l}}^i{{{\text{(}}({\text{cof}}D{\mathbf{u}})_j^l)}_{{x_j}}}} \right]} .$

Summing with respect to $j$ we get

$\displaystyle\sum\limits_{j = 1}^n {\frac{d}{{d{x_j}}}{\text{det}}({\text{cof}}D{\mathbf{u}}){\delta _{ij}}} = \sum\limits_{j = 1}^n {\sum\limits_{l = 1}^n {\left[ {u_{{x_l}{x_j}}^i({\text{cof}}D{\mathbf{u}})_j^l + u_{{x_l}}^i{{{\text{(}}({\text{cof}}D{\mathbf{u}})_j^l)}_{{x_j}}}} \right]} } .$

Again, it is not hard to see that

$\displaystyle\sum\limits_{j = 1}^n {\frac{d}{{d{x_j}}}{\text{det}}({\text{cof}}D{\mathbf{u}}){\delta _{ij}}} = \sum\limits_{j = 1}^n {\sum\limits_{l = 1}^n {u_{{x_l}{x_j}}^i({\text{cof}}D{\mathbf{u}})_j^l} }.$

Thus

$\displaystyle\sum\limits_{j = 1}^n {\sum\limits_{l = 1}^n {u_{{x_l}}^i{{{\text{(}}({\text{cof}}D{\mathbf{u}})_j^l)}_{{x_j}}}} } = \sum\limits_{l = 1}^n {u_{{x_l}}^i\sum\limits_{j = 1}^n {{\text{(cof}}D{\mathbf{u}})_{j,{x_j}}^l} } = 0$

which completes the proof.

Source: L. Evans, Partial differential equations, AMS, 1998.