Past exam of the solid mechanics course of the University of Oxford / 2011 / 1 / b

Define the invariants by the characteristic polynomialBecause $B=F{F}^T$ has positive eigenvalues ${\lambda }_j^2$, its invariants are the elementary symmetric functions in those eigenvalues:The Frobenius norm gives $|F{|}^2=\mathrm{tr}(F^{T}F)={\sum }_j{\lambda }_j^2$. Write $J=\det F>0$. Since $\mathrm{cof}F=J{F}^{-T}$, its singular values are $J/{\lambda }_j$, soFinally $\det B=\det F\det F^T=J^2$. ThereforeThese identities also give $I_B=\mathrm{tr}B$ and $I{I}_B=((\mathrm{tr}B{)}^2-\mathrm{tr}(B^2))/2$.