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Is it possible to generate an inverse of an order 3 tensor? If so, how? I have been searching for a couple days, and cannot seem to find anything online to help with this.

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What do you mean by inverse here? – Qiaochu Yuan Jun 17 '12 at 7:31
With a matrix M, the inverse of the matrix M multiplied by the matrix M is the Identity Matrix. Much like x * 1/x = 1 for scalar values. – BryanLemon Jun 19 '12 at 23:35
That is a definition of the inverse of a matrix. What do you mean by the inverse of an order 3 tensor? – Qiaochu Yuan Jun 20 '12 at 1:11

Think about how to regard an order 3 tensor as a linear map of spaces of the same dimension?

$$V^3 \rightarrow \mathbb R$$

$$V\rightarrow L(V,V)$$

$$L(V,V)\rightarrow V$$


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I am trying to figure out what to make of this answer. Would I be correct in assuming that it is possible to take the inverse of an order 3 tensor? – BryanLemon Jun 19 '12 at 23:42
@Bryan: no. Blah is trying to clue you in on the fact that an order $3$ tensor cannot be naturally realized as a map from a vector space to itself; it can only be realized as maps between vector spaces of different dimension (if $\dim V > 1$), and no such map can have an inverse. – Qiaochu Yuan Jun 20 '12 at 1:11
I was afraid of that. Thanks. – BryanLemon Jun 20 '12 at 4:40
@QiaochuYuan I would be grateful if you could give me a reference or helpful link to a proof/further exposition on the statement that "no such map can have an inverse". – Drone Scientist Mar 11 '15 at 12:40

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