Suppose that I have a symmetric matrix $\textbf{A}$, which represents the strain (or deformation) matrix. Then, what is the physical meaning behind the sum of the Hadamard product of $\textbf{A}$ with itself, i.e.

$\sum \textbf{A} \circ \textbf{A}$,

where the sum is taken over all elements of the product matrix?

  • 1
    $\begingroup$ Of what is the sum being taken here? $\endgroup$
    – klunkean
    Jun 28, 2019 at 7:02
  • $\begingroup$ @klunkean The sum is taken over all elements of the product matrix. $\endgroup$
    – KratosMath
    Jun 28, 2019 at 8:02

1 Answer 1


The expression in the question $$\sum \boldsymbol A \circ \boldsymbol A$$ is the double contraction of a tensor $\boldsymbol A$ with itself commonly written as $$\boldsymbol A:\boldsymbol A$$ It is the square of the Frobenius-Norm of the Tensor.

The only application I can think of off the top of my head is the elastic strain energy in the isotropic case. It is given by $$\psi=\frac12\lambda\operatorname{tr}^2\boldsymbol E+\mu\boldsymbol E:\boldsymbol E$$ where $\lambda, \mu$ are Lamé constants and $\boldsymbol E$ is the (symmetric) linear strain tensor. Since $\mu$ is the shear modulus you could interpret the expression as shearing contribution to the elastic energy. This, however, is not completely true since the full decomposition into bulk and shearing parts involves the deviatoric part $\boldsymbol E'=\boldsymbol E-\frac13\operatorname{tr}(\boldsymbol E)\boldsymbol 1$: $$\psi=3K\operatorname{tr}^2\boldsymbol E+\mu \boldsymbol E':\boldsymbol E'$$

Another occurence is found in classical plasticity theory:

The widely used scalar Mises equivalent stress is given by $$\sigma_{\text{vm}}=\sqrt{\frac 32\boldsymbol T':\boldsymbol T'}$$ with Cauchy stress deviator $\boldsymbol T'$. To achieve dissipation power equivalence in 3D and 1D, i.e. $$\boldsymbol T:\dot{\boldsymbol E}=\sigma_{\text{vm}}\dot\varepsilon_{\text{eq}}$$ you can calculate that $$\dot\varepsilon_{\text{eq}}=\sqrt{\frac23\dot{\boldsymbol E}:\dot{\boldsymbol E}}$$ The equivalent strain is then $$ \varepsilon_\text{eq}=\int\dot\varepsilon_\text{eq}\;\text dt $$

  • $\begingroup$ Thanks for your response. In the literature, the expression $\sum \textbf{A} \circ \textbf{A}$ multiply by $2/3$ under the square root sign $\sqrt{}$ is defined as the equivalent strain. I was wondering if you can elaborate more on this. $\endgroup$
    – KratosMath
    Jun 28, 2019 at 10:14

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