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[ Disclaimer: I am only starting to get my head around concepts of tensor calculus, so I apologize in advance for the lack of clarity or asking something trivial. I came across it trying to apply it somewhere]

I understand that a tensor can be seen a multilinear map. Is there a formal study of the kernel(s) of this map, i.e. a null-space of the tensor? I am guessing such a definition would have to take into account the tensor-rank of the tensor $T$ and the tensor-rank of tensor $X$ such that $X_IT^I = \mathbf{0} $, so we would have multiple null-spaces for one tensor $T$ each with a different tensor-rank.

If so could you provide with a reference? Is there a way to compute it as well?

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A case:

In a finite dimensional vector space $V$ over the reals, if $Q:V\times V\to\mathbb R$ is a bilinear map then this could be set by $(v,w)\mapsto v^{\top}Qw$ where $Q$ is a square matrix.

If this matrix has eigenvectors $u_0$ with eigenvalue equal to zero then $v^{\top}Qu_0=0$ and if $Q^{\top}$ has also $u_1$ as an eigenvector of eigenvalue equal to zero then $(u_1^{\top}Qw)^{\top}=w^{\top}Q^{\top}u_1=0$.

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  • $\begingroup$ I was thinking about a $Q: V \times V \times V \times \cdots \times V \rightarrow \mathbb{R}$, let's say $k$-linear, defining a kernel of dimension $k-1$ with regards to $Q$. I have found this link to be answering part of my question $\endgroup$
    – shnnnms
    Jun 2 at 11:27
  • $\begingroup$ let's talk about rank 3: you will have $Q^{ijk}$ and $Q(u,v,w)=Q^{ijk}u_iu_ju_k$ for the value assigned, the by fixing an index, let say $i_0$ then $Q^0=Q^{i_0 jk}$ gonna give you a rank 2 tensor for which you can apply the above idea: that will give that the eigenvectors $u_0$ of $Q^0$ with egenvalue zero: $Q(u_0,v,w)=0$. Same with the other indexes $j\ ,\ k$. $\endgroup$
    – janmarqz
    Jun 2 at 15:50
  • $\begingroup$ interessant the link you mention $\endgroup$
    – janmarqz
    Jun 2 at 15:54

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