All Questions
Tagged with tensor ag.algebraic-geometry
14 questions
0
votes
1
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157
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Can the derivative of eigenvectors with respect to its components be taken as zero if all eigenvalues are equal?
I want to ask a couple of follow up questions to the question answered on the thread "Derivative of eigenvectors of a matrix with respect to its components".
I noticed that in the accepted ...
1
vote
0
answers
124
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Space of all orthogonal partially complex $2\times2\times2$ tensors
I am trying to understand the space of all orthogonal tensors, I asked a more general version of this question here but with no solution yet found. I want to look at the simplest case first, namely a $...
5
votes
1
answer
241
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Is the asymptotic rank of a tensor bounded by (naive) border rank?
Write $R(T)$ for the rank of an order-$3$ tensor $T \in \mathbb C^{m \times n \times p}$ over the complex numbers. If $T' \in \mathbb {C}^{m' \times n' \times p'}$ is another such tensor then let $T \...
6
votes
1
answer
510
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Waring rank of monomials, and how it depends on the ground field
The Waring rank of a degree-$d$ homogeneous polynomial $p$ is the least integer $r$ such that you can write $p$ as a linear combination of $r$ $d$-th powers of linear forms $\{\ell_k\}$:
$$
p = \sum_{...
8
votes
0
answers
267
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Generalization of a standard algebraic group theory result for a tensor problem
$\DeclareMathOperator\GL{GL}$Let $X$, $Y$, $Z$ be $\mathbb{C}$-vector spaces, and let $A\subseteq X$ and $B\subseteq Y$ and $C\subseteq Z$ be linear subspaces. Let $V=X \otimes Y \otimes Z$, acted on ...
3
votes
1
answer
227
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A different notion of a decomposable symmetric tensor (besides Veronese)
$\DeclareMathOperator{\complex}{\mathbb{C}}$
Let $\bigvee^m(\complex^n)\subseteq (\complex^n)^{\otimes m}$ denote the space of symmetric tensors, i.e. the set of $x \in (\complex^n)^{\otimes m}$ that ...
0
votes
1
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234
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Rank of matrices and secant varieties
Consider the Segre embedding $\mathbb{P}^n\times \mathbb{P}^n\rightarrow \mathbb{P}^N$, and let $S\subset\mathbb{P}^N$ be its image.
Then $rank(Z)\leq k$ implies that $Z\in Sec_k(S)$. Moreover if $Z\...
4
votes
1
answer
980
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Symmetric tensor decomposition
Let $T$ be an order-$k$, rank-$m$ symmetric tensor, that is, $T=\sum_{j=1}^m v_j\otimes v_j \otimes \cdots \otimes v_j$, where the Segre outer product is taken $k$ times, with $v_j\in\mathbb{R}^d$ for ...
2
votes
0
answers
67
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Bounds on the tensor and border rank ratios of tensor unfoldings?
Consider $T_1 \in \mathbb{C}^{d_1 \times \cdots \times d_k}$ as an order-$k$ tensor of the format $(d_1, \cdots , d_k)$. Now, let $T_2$ be an unfolding of $T_1$ that is obtained by merging $\mathbb{C}^...
5
votes
1
answer
610
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Is a flattening rank a lower bound for the border rank?
Suppose $T \in V_1 \otimes \cdots \otimes V_k$ is a tensor, where each $V_i$ is a finite dimensional complex vector space. A $1$-flattening (or a flattening) is a realization of $T$ as a matrix in the ...
2
votes
0
answers
69
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Doubt on the best low rank approximation of a symmetric tensor
I have a matrix $M\in\mathbb{R}^{n\times k}$, with $k<n$ whose columns $m_i$ are linearly independent.
So we have $M := [m_1|..|m_k]$.
From the columns of $M$ I can define the following matrix
$$
...
2
votes
0
answers
211
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Smoothness of a (given) global section of a vector bundle over G(2,6)
Let $G=Gr(2,6)$ the Grassmannian of two planes in $V=\mathbb C^6$, and let $\mathcal Q(1)$ the rank four quotient bundle on it twisted with $\mathcal O_G(1) \cong $ det$(S^*)$, $S$ being the ...
11
votes
2
answers
10k
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Derivative of eigenvectors of a matrix with respect to its components
Suppose that $B$ is a real, positive-definitive symmetric ($3\times3$) matrix (more accurately, $B$ is a tensor) with distinct eigenvalues, and that we can write it as
$$
B= \sum_{i=1}^3 \lambda_{i}(...
18
votes
2
answers
2k
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What is the largest tensor rank of $n \times n \times n$ tensor?
The tensor rank of a three dimensional array $M[i,j,k], i,j,k\in [1,\ldots,n]$ is the minimal number of vectors $x_i,y_i,z_i$, such that $M=\sum_{i=1}^d x_i\otimes y_i\otimes z_i$.
From dimension ...