Questions tagged [tensor-rank]

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Rank 1 tensors, how to describe them? (specific case)

I want to undestand a specific case. I consider two $\mathbb{C}$-vectorial spaces, $\mathbb{C}^2$ both. Then, I want to work with $\mathbb{C}^2\otimes \mathbb{C}^2$. Now, I consider basis for each ...
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Dimension of the variety of rank 1 decompositions of a matrix

Let $A\in\mathrm{GL}_n(\mathbb{C})$, and let's consider its decompositions into a sum of rank 1 matrices $$A=\sum_{i=1}^t A_i,\ \text{rank}(A_i)=1,$$ where $t\geqslant n$. When $t=n$, Wikipedia claims ...
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What is the transpose of a 3rd rank tensor?

If I have a 3rd rank tensor $\stackrel{\leftrightarrow}{A}$ or in index notation $A_{ijk}$, how to make the transpose of this in index notation? Does the concept of transpose of third rank tensor even ...
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Rank of Matrix Multiplication tensor $\langle1,n,1\rangle$

$\newcommand{\rank}{\operatorname{rank}}$I am reading through Belzer's Fast Matrix Multiplication, available here. I want to prove tjat $$\rank(t)=\rank(\langle1,n,1\rangle)=n$$ In "usual tensor ...
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Trying to effect permutating a tensor on its rank

I am reading through Fast Matrix Multiplication by Markus Blaser. I am trying to prove Lemma 5.3 from page 19. It states the following: For any tensor $T\in \mathbb{F}^{n\times m\times t}$, and any ...
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Do orthogonal transformations preserve the symmetry of the tensors?

I have the following doubt: Do orthogonal transformations preserve the symmetry of the 2 rank mixed tensors? It seems logical to me since the symmetry of the tensor needs to be preserved if we change ...
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Is the determinant a tensor?

I was reading Schutz's book on General Relativity. In it, he says that a(n) $M \choose N$ tensor is a linear function of $M$ one-forms and $N$ vectors into the real numbers. So does that mean the ...
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Tensor contraction and notational problems

I am going through a chapter in my book on tensors, and it gives a basic understanding of tensors. The question posed is simple: "Show that the contracted tensor $T_{ijk}V_k$ is a rank-2 tensor.&...
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Diagonalize matrix multiplication

Let $V$ be the space of $2\times 2$ matrices with complex coefficients. Let $A \in V$ and let $L_A:V \to V$, defined by $L_A(X)=A\cdot X$. I am trying to solve the exercise (10) from this book: find a ...
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Uniqueness of the matrix representation of tensors

Note that both maps below satisfy the universal property of the tensor product. \begin{align*} \mathbb{R}^2 \times \mathbb{R}^2 &\rightarrow \mathbb{R}^{2 \times 2} &\mathbb{R}^2 \times \...
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Tensor rank decomposition for some vectors

I have a problem with solving of following problem. We have a vector space $V$ over field $k$ with basis $v_1, v_2$ Rank of a vector $v$ of $V \otimes V \otimes V$ is minimum length of decomposition ...
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Intuitive understanding of 2-forms, (1,1)-tensors, and other fundamental objects of exterior algebra or tensor algebra

My background consists mostly of a good level in linear algebra, abstract algebra, undergrad calculus, topology & probability, and some working knowledge of geometric algebra and category theory. ...
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Is higher rank tensor always the product of lower rank tensor?

I remember that I saw the definition of tensor somewhere as tensor is an object in $E\otimes F$ for some vector space $E$ and $F$. (here I used an example for rank two.) But most of the time in ...
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Tensor confusion - Are coordinates a type of tensor?

(All of this question uses $V$ as the real vector space where tensors are being built from). I am currently trying to learn about tensor notation, and I am running into a road block with ...
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Understanding a discrepancy in tensor multiplication

I have seen in texts about quantum computing people take two vectors and do tensor multiplication. Now, what confuses me is that vectors are (1,0)-tensors. This means that when I multiply two of them, ...
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Why a vector is a (1,0) tensor?

I am looking for some familiar examples of Tensors and I am wondering why a vector is a (1,0) tensor type? That is it takes some covector and gives and scalar!! How?
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Abelian group Rank ιs equal to the dimension of the tensor product [closed]

If $G$ is a finitely generated abelian group then why it's rank is ιs equal to the dimension of $G\otimes_\mathbb{Z} \mathbb{Q}$ as a vector space over $\mathbb{Q}$? Thank you in advance.