The isotropic elastic strain energy is $\mu \mathrm{\mid }{e}_0{\mathrm{\mid }}_F^2+(\lambda \mathrm{/}2+\mu \mathrm{/}3)(\mathrm{tr}e{)}^2$, hence positive definite under $\mu >0$, $\lambda +2\mu \mathrm{/}3>0$

Let $e\in \mathrm{Sym}(3)$ with deviatoric part $e_0=e-\frac{\mathrm{tr}e}{3}{I}_3$ and Lamé parameters $\lambda ,\mu \in \mathbb{R}$ with $\mu >0$ and $\lambda +\frac{2\mu }{3}>0$ (The small-strain isotropic linear elastic constitutive law and its strain energy). Then

$W(e)=\mu \mathrm{\mid }{e}_0{\mathrm{\mid }}_F^2+(\frac{\lambda }{2}+\frac{\mu }{3})(\mathrm{tr}e{)}^2\ge 0,$

and $W(e)=0$ if and only if $e=0$. Here $\mathrm{\mid }A{\mathrm{\mid }}_F^2=\mathrm{tr}(A^{\mathsf{T}}A)$ is the squared Frobenius norm (The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$, The trace $\mathrm{tr}(A)$ as the sum of the diagonal entries).

• The decomposition is the standard deviatoric/volumetric split: $W$ is the sum of a shear term controlled by $\mu$ and a volumetric term controlled by the bulk modulus $K=\lambda +\frac{2\mu }{3}$. The hypotheses are exactly the inequalities needed for both coefficients to be positive; they are strictly weaker than requiring $\lambda >0$.

• No stress-strain inversion, no bounds on the stress matrix, and no constitutive interpretation beyond the stated model are asserted. The lemma is a statement about symmetric matrices.