Mathematics Stack Exchange
Mathematics Stack Exchange

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 $V\times V\to \Bbb R$.

1390 questions
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Does a "cubic" matrix exist?

Well, I've heard that a "cubic" matrix would exist and I thought: would it be like a magic cube? And more: does it even have a determinant - and other properties? I'm a young student, so... please don't get mad at me if I'm talking something…
Ian Mateus
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Does the $32$-inator exist?

Background It is common popular-math knowledge that as we extend the real numbers to complex numbers, quaternions, octonions, sedenions, $32$-nions, etc. using the Cayley-Dickson construction, we lose algebraic properties at each step such as…
pregunton
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understanding of the "tensor product of vector spaces"

In Gowers's article "How to lose your fear of tensor products", he uses two ways to construct the tensor product of two vector spaces $V$ and $W$. The following are the two ways I understand: $V\otimes W:=\operatorname{span}\{[v,w]\mid v\in V,w\in…
user9464
44
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4 answers

Cross product in higher dimensions

Suppose we have a vector $(a,b)$ in $2$-space. Then the vector $(-b,a)$ is orthogonal to the one we started with. Furthermore, the function $$(a,b) \mapsto (-b,a)$$ is linear. Suppose instead we have two vectors $x$ and $y$ in $3$-space. Then the…
goblin GONE
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Symmetric and wedge product in algebra and differential geometry

Which is the correct identity? $dx \, dy = dx \otimes dy + dy \otimes dx$ $~~~$or$~~~$ $dx \, dy = \dfrac{dx \otimes dy + dy \otimes dx}{2}~$? $dx \wedge dy=dx \otimes dy - dy \otimes dx$ $~~~$or$~~~$ $dx \wedge dy=\dfrac{dx \otimes dy - dy \otimes…
Seub
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What is the kernel of the tensor product of two maps?

Assume that $f_1\colon V_1\to W_1, f_2\colon V_2\to W_2$ are $k$-linear maps between $k$-vector spaces (over the same field $k$, but the dimension may be infinity). Then the tensor product $f_1\otimes f_2\colon V_1\otimes V_2\to W_1\otimes W_2$ is…
Lao-tzu
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Geometric interpretation of the determinant of a complex matrix

A complex $n$-dimensional vector space $V$ can be thought of as a real $2n$-dimensional vector space equipped with a map $J:V \to V$ with $J^2 = -I$. Complex-linear maps are then linear maps $V \to V$ which commute with $J$. One can think of $J$ as…
Ash Malyshev
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Exterior power "commutes" with direct sum

I know that for vector spaces $V, W$ over a field $K$, we have the following identity : $$ \bigoplus_{k=0}^n \left[ \Lambda^k(V) \otimes_K \Lambda^{n-k}(W) \right] \simeq \Lambda^n(V \oplus W) $$ which holds in a canonical way : for $0 \le k \le…
Patrick Da Silva
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Why the whole exterior algebra?

So, I've been reading up on multilinear algebra a bit. In particular, I've been reading up on the construction of of the exterior algebra of a finite dimensional vector space $X$, say over $\mathbb{R}$. $$ \Lambda(X) = \bigoplus_{n \geq 0}…
Mike F
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Polarization: etymology question

The polarization identity expresses a symmetric bilinear form on a vector space in terms of its associated quadratic form: $$ \langle v,w\rangle = \frac{1}{2}(Q(v+w) - Q(v) - Q(w)), $$ where $Q(v) = \langle v,v\rangle$. More generally (over…
KCd
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3 answers

A user's guide to Penrose graphical notation?

Penrose graphical notation seems to be a convenient way to do calculations involving tensors/ multilinear functions. However the wiki page does not actually tell us how to use the notation. The several references, especially ones with Penrose as…
Hui Yu
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What's the difference between geometric, exterior and multilinear algebra?

I've studied what I think is geometric algebra, but can't seem to understand the difference between it and exterior and multilinear algebra. And is it linked to Clifford and Grassmann algebras in any way?
Anivarya Kumar
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Tensors: Acting on Vectors vs Multilinear Maps

I have the feeling like there are two very different definitions for what a tensor product is. I was reading Spivak and some other calculus-like texts, where the tensor product is defined as $(S \otimes T)(v_1,...v_n,v_{n+1},...,v_{n+m})=…
Squirtle
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Proving that the coefficients of the characteristic polynomial are the traces of the exterior powers

Let $T$ be an endomorphism of a finite-dimensional vector space $V$. Let $$f(x)=x^n+c_1x^{n-1}+ \dots + c_n$$ be the characteristic polynomial of $T$. It is well known that $c_m=(-1)^m\text{tr}(\bigwedge^mT)$. If the base field is $\mathbf{C}$,…
Bruno Joyal
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5 answers

Why is the tensor product constructed in this way?

I've already asked about the definition of tensor product here and now I understand the steps of the construction. I'm just in doubt about the motivation to construct it in that way. Well, if all that we want is to have tuples of vectors that behave…
Gold
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