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Suppose we have a matrix $H$ and I want to find out whether there exists a vector $r\ne 0$ such that $H^{ij}r_i r_j=0$. To this end we could simply compute if the determinant $\det(H)$ is zero, which is computationally inexpensive and explicit (we don't have to find such $r$ explicitly).

Is there an analogue of such a test to see if there exists a vector $r\ne0$ such that $G^{ijk}r_i r_j r_k=0$ for a $3$-tensor $G$?

I didn't specify the number of dimensions of the vector space, but if it's relevant $r\in \mathbb R^3$.

This question could be related to Determinant of a tensor.

Community
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pwl
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2 Answers2

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There's always such a vector. To see this, let $$ f(r) = G^{ijk}r_i r_j r_k, \qquad r \neq 0 . $$ If $f$ is identically zero, any $r \neq 0$ will do. Otherwise, $f$ assumes both positive and negative values, since $f(-r)=-f(r)$. Then, by continuity of $f$ and connectedness of the space $\mathbb{R}^3\setminus \{ 0 \}$, we see that $f$ must also take the value zero for some $r \neq 0$.

Hans Lundmark
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  • This will work for odd dimensions only. – mvw Oct 13 '16 at 13:05
  • @mvw: It works for tensors of odd valence (like $G^{ijk}$ of valence three), since it's the number of factors $r_i$ which determines if the sign changes or not. The dimension of the space is irrelevant; it works in $\mathbb{R}^n$ for any $n>1$. – Hans Lundmark Oct 13 '16 at 13:08
  • Ah, I see now. That's a beautiful counter example! – pwl Oct 13 '16 at 13:21
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Example: $$ H = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} $$ Then $\det H = -1 \ne 0$, but for $r = (0,1,0)^\top \ne 0$ we have $$ H^{ij} r_i r_j = r_1 r_1 + r_2 r_3 + r_3 r_2 = 0 $$ So your criterion is not good enough for you, it leaves out the cases $x^\top H x$ where $x \perp Hx$, covers only $Hx = 0$.

mvw
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  • Thanks! This is a valid point. Now I see that $\det(H)=0$ implies the existence of such $r$ but, by your example, $\det(H)\ne0$ does not exclude such $r$. – pwl Oct 13 '16 at 13:20