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For symmetric and positive semidefinite matrices, we can compute $$(Ax,y):=y^\top Ax =y^\top A^{1/2} A^{1/2} x = (A^{1/2} y)^\top A^{1/2} x= (A^{1/2} x, A^{1/2} y)$$ where $(A^{1/2})^\top=A^{1/2}$. Alternatively, we could have directly used that $A^{1/2}$ is self-adjoint.

Now, I am in the situation of the inner product $$(CA,B)=CA:B$$ where $C$ is a symmetric and positive definite fourth-order elasticity tensor, that is, it maps $\mathbb{R}^{n \times n}$ matrices to $\mathbb{R}^{n \times n}$ matrices. Further, $A$ and $B$ are symmetric $\mathbb{R}^{n \times n}$ matrices. Here, the inner product is defined via the double dot product. In Gurtin's "Continuum Mechanics" book it is stated that a symmetric and positive definite tensor has a unique square root in the sense that there is a unique (positive definite and symmetric) tensor $D$ such that $D^2=C$ and we call $D=\sqrt{C}$. In that sense $\sqrt{C}$ again maps matrices to matrices.

From a functional analysis point of view (using that everything acts on the Hilbert space $\mathbb{R}^{n \times n}$ and the operations are linear), I should be able to use that $\sqrt{C}$ is self-adjoint (as it is symmetric and hermitian) $$(CA,B)=(C^{1/2} C^{1/2}A,B)=(C^{1/2} A, C^{1/2} B).$$

Is this correct? If not, is there a class of fourth-order tensors so that it works like adding properties? It surely works for scaled identifity matrices like $C=\eta I$ for $\eta>0$.

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  • $\begingroup$ You keep referring to $C$ as a tensor rather than as a matrix, but you treat $C$ as though it's a matrix. If $C$ is a tensor (presumably a (2,0) tensor), then what exactly is $CA$? What exactly does $C^{1/2}C^{1/2}$ mean if $C^{1/2}$ is a similar tensor? $\endgroup$ Commented Apr 30 at 18:37
  • $\begingroup$ For a positive definite matrix $C$, it is indeed true that $(CA,B) = (C^{1/2}A, C^{1/2}B)$. To see this, note that $A:B = \operatorname{tr}(A^\top B)$. $\endgroup$ Commented Apr 30 at 18:39
  • $\begingroup$ @BenGrossmann Ah, indeed, I should have been more precise. $C$ is a fourth order tensor, like the elasticity tensor. Then $C^{1/2}$ should exits according to the square-root theorem, whatever it looks like. At least it maps again $(n \times n)$ matrices to $(n \times n)$ matrices, just like $C$, so everything should be well-defined. $\endgroup$
    – Cahn
    Commented May 3 at 12:11
  • $\begingroup$ What does it mean for a fourth order tensor to be “symmetric” or “positive definite”? $\endgroup$ Commented May 3 at 16:31
  • $\begingroup$ @BenGrossmann If we write $C^{ijkl}$ for its components, then symmetric means $C^{ijkl}=C^{klij}=C^{jikl}$ and positive definite means $C^{ijkl} A_{kl} A_{ij} \geq c \sum_{i,j=1}^3 A_{ij}^2$ for any symmetric $A$ with $A_{ij}=A_{ji}$ (and $c$ is an independent constant). $\endgroup$
    – Cahn
    Commented May 3 at 17:17

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Yes, it is completely legitimate. Just apply the spectral theorem on the space of real symmetric matrices. You have a dot product given by: $$ \langle A,B\rangle = \text{Tr}(AB) $$ which is positive definite. The elasticity tensor is symmetric, since it derives from the quadratic form. You can interpret the form as energy, taking as an input the strain tensor $\epsilon$: $$ E = \frac12 C_{ijkl}\epsilon_{ij}\epsilon_{kl} $$ By construction, $C$ is therefore a real symmetric operator on real symmetric operators. In terms of indices, since it acts on symmetric matrices: $$ C_{ijkl} = C_{jikl} = C_{ijlk} $$ And because it is derived from a quadratic form: $$ C_{ijkl} = C_{klij} $$ You can apply the spectral theorem. Since energy is positive for physical reasons, you can define the positive square root unambiguously as another positive tensor.

A useful example of this is for homogeneous isotropic materials where $C$ depends only on the two Lamé coefficients: $$ C_{ijkl} = 2\mu \delta_{ik}\delta_{jl}+\lambda\delta_{ij}\delta_{kl} \\ \langle \epsilon,C\epsilon\rangle = 2\mu\text{Tr}(\epsilon^2)+\lambda\text{Tr}(\epsilon)^2 $$ Recall that $C$ is positive iff $2\mu\geq 0$ and $2\mu+n\lambda\geq 0$, with $n$ the spatial dimension (in terms of the more familiar Young modulus $E$ and Poisson ratio $\nu$, it is equivalent to $E>0$ and $\nu\in[-1,1/2]$). You can calculate that: $$ (\sqrt C)_{ijkl} = 2\alpha\delta_{ik}\delta_{jl}+\beta\delta_{ij}\delta_{kl} $$ (the square root will also be isotropic) with: $$ \begin{align} \alpha &= \frac{\sqrt{2\mu}}2 & \beta &= \frac{\sqrt{2\mu+n\lambda}-2\mu}n \end{align} $$ with $n$ the spatial dimension. You can check the consistency as the existence of $\alpha,\beta$ gives: $$ 2\mu = (2\alpha)^2 \geq 0 \\ 2\mu+n\lambda = (2\alpha+n\beta)^2\geq 0 $$ so $C$ is automatically positive. Furthermore, the square root is well defined precisely because $C$ is positive.

In fact you can generalise the discussion to arbitrary square matrices, not necessarily symmetric ones with the usual Frobenius product: $$ \langle A,B\rangle = \text{Tr}(A^TB) $$ The analogue of the elasticity tensor is still symmetric (the second index identity only, not the first one), and you can similarly compute square roots. In the simple isotropic case, you now have three parameters: $$ C_{ijkl} = \mu_1 \delta_{ik}\delta_{jl}+\mu_2 \delta_{il}\delta_{jk}+\lambda\delta_{ij}\delta_{kl} \\ \langle M,CM\rangle = \mu_1\text{Tr}(M^TM)+\mu_2\text{Tr}(M^2)+\lambda\text{Tr}(M)^2 $$ It is positive iff $\mu_1+\mu_2\geq 0$, $\mu_1-\mu_2\geq 0$, and $\mu_1+\mu_2+n\lambda\geq 0$. The square root is: $$ (\sqrt C)_{ijkl} = \alpha_1 \delta_{ik}\delta_{jl}+\alpha_2 \delta_{il}\delta_{jk}+\beta\delta_{ij}\delta_{kl} $$ with: $$ \begin{align} \alpha_1 &= \frac{\sqrt{\mu_1+\mu_2}+\sqrt{\mu_1-\mu_2}}2 & \alpha_2 &= \frac{\sqrt{\mu_1+\mu_2}-\sqrt{\mu_1-\mu_2}}2 & \beta &= \frac{\sqrt{\mu_1+\mu_2+n\lambda}-\sqrt{\mu_1+\mu_2}}n \end{align} $$ which are well defined because $C$ is positive and conversely, the existence of $\sqrt C$ implies the positivity of $C$: $$ \begin{align} \mu_1+\mu_2 &= (\alpha_1+\alpha_2)^2 & \mu_1-\mu_2 &= (\alpha_1-\alpha_2)^2 & \mu_1+\mu_2+n\lambda &= (\alpha_1+\alpha_2+n\beta)^2 \end{align} $$

Hope this helps.

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