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Questions tagged [multilinear-algebra]

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

A multilinear function is a function from $V_1\times V_2\times\dots \times V_n\to V$ where the $V_i$ and $V$ are all vector spaces, and the function is a linear function into $V$ when restricted to each of the $V_i$. This includes the special case of bilinear functions.

For example, the ordinary dot product in $\Bbb R^n=V$ is a multilinear function from $V\times V\to \Bbb R$.

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From linear transformation to alternating linear transformation

I'm reading Kenneth Hoffman's Linear Algebra, Ed2. In $\S5.6$ "Multilinear Functions" it talks about generating an alternating linear transformation from a linear transformation. The collection of all multilinear functions on $V$ will be denoted…
athos
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expression of the 4-tensor $f \otimes g$ in given basis

Let $f$ and $g$ be bilinear functions on $\mathbb{R}^n$ with matrices $a = \{a_{ij}\}$ and $B = \{b_{ij}\}$, respectively. How would I go about finding the expression of the $4$-tensor $f \otimes g$ in the basis$$x_{i_1} \otimes x_{i_2} \otimes…
user198839
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What is $(V^*)^{\otimes l} \stackrel{f}{\rightarrow} S^l( (V^*)^{\oplus l} )\cong S^l( (V^{\oplus l})^* ) \stackrel{g}{\rightarrow} S^l(V^*)$?

Let's work over an algebraically closed field and let $V$ be a finite dimensional vector space. Then is there a natural chain of maps $$ (V^*)^{\otimes l} \stackrel{f}{\rightarrow} S^l( (V^*)^{\oplus l} )\cong S^l( (V^{\oplus l})^* …
user85336
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Are alternating and anticommutative forms equivalent?

A form is defined to be alternating iff having two equal arguments means it is equal to 0. It is defined to be anticommutative iff permuting its arguments means multiplying by the sign of the permutation. Unless my definitions are wrong, I'm pretty…
David Cian
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Properties of k-linear form

I would like to know if this is correct: Let be $f:V\times V\times\text{ }...\times V\rightarrow W $ a symetric k-linear form. If $f(v,..,v)=f\cdot v^k=0,\forall v\in V$, then $f=0$. I proved this for k=2, but I don't know if it's worth it in…
Felipe Gonçalves
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$x_{\alpha}\in E$ and $y_{\beta} \in F$ are linearly independent then so are the vectors $\varphi(x_{\alpha},y_{\beta})$ iff $\otimes_2$ holds

Let $\varphi: E \times F \to G$ be a bilinear mapping. Show that the following property is equivalent to $\otimes_2$: Whenever the vectors $x_{\alpha}\in E$ and $y_{\beta} \in F$ are linearly independent then so are the vectors…
Curious
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Equivalence between the product of a skew symmetric matrix and the product of a bivector and a vector

I stumbled upon the following statement on the wikipedia page (https://en.wikipedia.org/wiki/Cross_product#cite_note-lounesto2001-14) about the cross product: The vector cross product also can be expressed as the product of a skew-symmetric matrix…
mathology
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Form of a 2-form if and only if wedge is zero.

Let $\{\alpha^1,...,\alpha^p\}$ be a collection of linearly independent 1-forms. Show a 2-form $\beta$ is of the form \begin{align*} \beta=\gamma^1\wedge\alpha^1+\cdots+\gamma^p\wedge\alpha^p \end{align*} for 1-forms $\gamma^i$ if and only if…
Walt
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What is $S^{(1^{nd})}(V)$?

Given a vector space $V$ of dimension $n$, I can form the module $$ S^{(1^{nd})}(V)$$ where $(1^{nd})$ denotes the partition $$ (1,1,\dots,1)\vdash nd$$ Now for $d=1$, I know that this module is in fact $\bigwedge^n(V)$. However I don't understand…
Levent
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Prove that the image of the bilinear application is not a vector space

Prove that the image of the bilinear application $$ \mathbb{R}^2\times\mathbb{R}^2 \ni \big((x_1,x_2),(y_1,y_2)\big) \mapsto \varphi \big((x_1,x_2),(y_1,y_2)\big) = (x_1y_1,x_1y_2,x_2y_1,x_2y_2)\in\mathbb{R}^4 $$ is not a vector space. My…
Elias Costa
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Bilinear extension of a map defined only on pairs of independent vectors

Let $V={\mathbb R}^d$ and $$ A=\bigg\lbrace (v_1,v_2) \in V \times V \bigg| \ v_1 \ \text{and} \ v_2 \ \text{are linearly independent} \bigg\rbrace $$ Consider the maps $f:A \to {\mathbb R}$ satisfying the following three properties :…
Ewan Delanoy
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Dimension of "self-dual" and "anti-self-dual"

Let $V$ a $2k$-dimensional nondegenerate quadratic space and assume that the index is such that $**=\text{Id}: \text{Alt}^kV \rightarrow \text{Alt}^kV$ (Hodge star operator). Then $V$ is the direct sum of the subspaces "self-dual" ($*\omega=\omega$)…
Lotte
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Analogy of covectors

Given that the bases of a vector and their duals satisfy: $${\epsilon}^ie_j={\delta}^i_j$$ in which $\epsilon$ and $e$ are the dual bases and bases respectively, and ${\delta}^i_j$ is the Kronecker symbol, and that the definition of the dual vector…
user518704
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Relation between symmetric powers and G-linear morphisms

My multilinear algebra is pretty bad, so I just wanted to check if I my intuition is correct: $$ Hom_{S_n}(V^{\otimes n}, V^{\otimes n}) \cong Sym^n(End(V)) $$ where V is a finite dimensional vector space and $S_d$ is the group of permutations of…
AnonAnon
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Dimension of symmetric $n$-forms on vector space of dimension $n$

Let $E$ be a vector space of dimension $n$ over a field $k=\mathbb R$ or $\mathbb C$, and consider $S^n E^*$ to be the space of $n$-linear symmetric forms. What is the dimension of $S^n E^*$? I would say 1, because I can't think of anything other…
Joe
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