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Does a “cubic” matrix exist?

There is some theory about spatial matrices ?, for example matrices of order $3\times 3\times 3 ,A=(a_{ijk}),i,j,k=1,2,3$ where $A$ would be a cube in space with entries inside the cube.

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For example if $f:\mathbb{R}^3\longrightarrow \mathbb{R}$ , $f\in C^3$

$f(a+h)=f(a)+f'(a).h+\frac{1}{2}f''(a).h^2+\frac{1}{6}f'''(a).h^3+r(h)$ , $\lim_{h \to 0}{\frac{r(h)}{||h||^3}}=0$ where $f'''(a)$ would be a spatial matrix of order $3\times3$ (application symmetric trilinear).

$f'''(a)=(\frac{\partial^3 f}{x_ix_jx_k}),i,j,k=1,2,3$ $f'''(a):\mathbb{R}^3 \times \mathbb{R}^3 \times \mathbb{R}^3 \longrightarrow \mathbb{R}$ $f'''(a)(x,y,z)=\langle (x^t.f'''(a)).y,z\rangle$ Where $x^t.f'''(a)$ would be the multiplication of each plane matrix of the spatial matrix by the vector $x$ resulting in a plane matrix.

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marked as duplicate by J. M., tomasz, t.b., sdcvvc, William Aug 30 '12 at 17:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Please take a look at this meta question on the reasons for accepting answers. – joriki Jul 29 '12 at 19:08
This thread might be of interest to you. – t.b. Jul 29 '12 at 19:43

These triply-indexed objects are not considered as "matrices" but rather tensors.

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