# Questions tagged [multilinear-algebra]

For questions about the extension of linear algebra to multilinear transformations of vector spaces.

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### I want to find $m,l,n$ and $k$ and $(A_1,B_1),(A_2,B_2)\in V\times W$ such that $A_1B_1+A_2B_2\neq A_3B_3$ for any $(A_3,B_3)\in V\times W$.

I am reading "Tensor Algebra" by Takeo Yokonuma (in Japanese). Problem 2 (on p. 329) Let $V,W$ and $U$ be vector spaces over $k$, where $k$ is a field. Let $\mathcal{L}(V,W;U)$ be the set ...
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### What allows a bilinear form (output: field element) to also work as a linear map (output: vector)?

A bilinear form $B: V × V → K$, when the inputs are 2 vectors, has 1 element of their field as output. A linear map $L: V → W$, when the input is 1 vector, has 1 vector as output. I have seen cases ...
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### Do all bilinear forms have a signature? [closed]

If yes: starting from any arbitrary bilinear form, what is the algorithm to calculate its signature? If no: what are the conditions necessary for a bilinear form in order to have a signature? Is there ...
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### Matrix-rank nonincreasing unitary tensor operations

I have a multidimensional array $A_{ijkl}$ $\in\mathbb{C}^{m\times n\times o \times p}$ indexed by four integers $i,j,k,l$. I will call $i$ and $j$ the "left" indices, $j$ and $k$ the "...
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### Generalization of "symmetric positive definite" to higher dimensions?

Suppose I have a positive-definite function $f:\mathbb{R}^n\to \mathbb{R}^1$. Then the Hessian $\nabla^2 f$ is symmetric positive definite and I can write $\nabla^2 f=QQ^T$ for some real-valued matrix ...
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Prove: $$\begin{array}{|cccccccccc|} 1 & 0 & 0 & \cdots & 0 & 1 & 0 & 0 & \cdots & 0 \\ x & x & x & \cdots & x & y & y & y & \cdots ... 1 vote 1 answer 68 views ### Embedding from exterior product to tensor product space I came across the following: For a basis \phi_1,\dots,\phi_n of V there is a natural embedding V^{\wedge n}\hookrightarrow V^{\otimes n} defined as$$(\phi_1\wedge \cdots \wedge \phi_n) \... 34 views

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### Proving that $\dim(\bigwedge^k(V^*)) = \binom{n}{k}$ without constructing an explicit basis

Text: Discussion: I find this argument kind of hard to follow because an explicit basis is never constructed; the argument seems kind of indirect. I'm relatively comfortable with the first sentence ...
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### Show that $v_1\wedge\dots\wedge v_k = x_1\wedge\dots\wedge x_k \implies \text{span}\{v_1,\dots, v_k\} = \text{span}\{x_1,\dots, x_k\}$
Let $V$ be an $n$-dimensional space and $v_1,\dots, v_k \in V$ are linearly independent. It is clear that if $x_1,\dots, x_k \in V$ have the same span as $v_1\dots v_k \in V$ then there is a scalar $t$...