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In continuum mechanics we use finite deformation tensors to exprime deformations in a point. The 9 components of the tensor (in reality 6 because of its symmetry) are defined as $$ \sigma_{ik}(\vec{r}) = \frac{dF_k^{(i)}}{dS_i} $$ where $k$ is the component of the force that acts over the surface perpendicular to axe $i$. We put this components on a "table" and we use matrix calculus to manipulate the tensor.

In linear algebra we saw only the definition of a tensor. It is a multilinear map $$ T: V\times \cdots \times V \times V^* \times \cdots \times V^* \to K. $$

In fact I really can't see that this definition correspond to my finite deformation tensors. Cans someone show me the relations between physics and linear algebra in this case? Why I can use a tensor as a matrix?

I read that if we use orthogonal axis we can avoid tensor analysis. Why?

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This case is a very simple specific case of the general tensor you reference, to wit: $$ T: V \rightarrow U $$ i.e. a linear mapping $T$ from a vector space $V$, whose elements define the possible surfaces, to another vector space $U$ giving the force acting on a particular surface (if I've understood your description correctly).

A 3x3 matrix represents this mapping. (See the related questions for details, but simply put such a matrix also provides a linear mapping from an "input" vector space to an "output" vector space.)

Finally, it's a property of real symmetric matrices (like yours) that you can find a set of orthogonal basis vectors $$\boldsymbol{e_1,e_2,e_3}$$ (in general different from the starting x-y-z coordinate basis), in which the matrix is diagonal (off-diagonal elements all zero). If you work in that basis, you're still doing matrix math, but the calculations are very simple, because the various element equations are decoupled from each other. [For example, if the input $\boldsymbol{v_1 = e_1}$, the output $\boldsymbol{u_1} = \lambda_1 \boldsymbol{e_1}$, with $\lambda_1$ the (1,1) element of the matrix.]

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