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

1172 questions
151
votes
7 answers

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
  • 7,071
  • 3
  • 29
  • 57
45
votes
6 answers

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
36
votes
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…
35
votes
3 answers

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…
30
votes
2 answers

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
  • 39,395
  • 5
  • 76
  • 126
29
votes
1 answer

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…
28
votes
4 answers

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}…
27
votes
2 answers

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…
23
votes
2 answers

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
  • 52,320
  • 6
  • 122
  • 220
21
votes
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
  • 14,131
  • 4
  • 33
  • 97
21
votes
3 answers

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
  • 6,348
  • 28
  • 57
20
votes
2 answers

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…
19
votes
3 answers

Basis for tensor products

Suppose $V_1$ and $V_2$ are $k$-vector spaces with bases $(e_i)_{i \in I}$ and $(f_j)_{j \in J}$, respectively. I've seen the claim that the collection of elements of the form $e_i \otimes f_j$ (with $\left(i,j\right) \in I \times J$) forms a basis…
19
votes
3 answers

Multiplying 3D matrix

I was wondering if it is possible to multiply a 3D matrix (say a cube $n\times n\times n$) to a matrix of dimension $n\times 1$? If yes, then how. Maybe you can suggest some resources which I can read to do this. Thanks!
Sahil Chaudhary
  • 343
  • 1
  • 3
  • 10
19
votes
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…
1
2 3
78 79