# Tag Info

## Hot answers tagged multilinear-algebra

3

A linear operator $f$ on $R^3$ has a real-valued eigenvalue $x$, because if you represent $f$ by a matrix with respect to any basis for $R^3$, the equation $\det (f-xI)=0$ is a real cubic. Now if $f_1$ is invertible, let $x$ be an eigenvalue of the operator $g=f_1^{-1}(f_2+f_3)$ and let $g(y)=xy$ with $y \ne 0$. Then $(-x)f_1(y)+f_2(y)+f_3(y)=0$. But if ...

3

It seems as though you have the right idea, but you're getting caught up on the notation. Here's a suitable generalization of your proof: \begin{align} \phi(Av_1,\cdots,Av_n) &= \phi\left(\sum_{j_1=1}^n a_{1j_1} v_{j_1}, \sum_{j_2=1}^n a_{2j_2} v_{j_2}, \dots, \sum_{j_n=1}^n a_{nj_n} v_{j_n}\right) \\ & = \sum_{j_1=1}^n \sum_{j_2=1}^n \cdots ... 3 The map V^n\to K, (v_1,\ldots, v_n)\mapsto \phi(Av_1,\ldots, Av_n) is readily checked to be alternating multilinear. As the space of alternating n-forms is onedimensional, it must be a multiple c(A)\phi of \phi. To determine c(A) it suffices to evaluate \phi(e_1,\ldots, e_n), i.e., \phi applied to the columns of A. By alternating ... 3 I think you are encountering an issue with your calculation due to a misreading of the problem. The way I understand it, the question is, for any 2 form \omega, does there exist a basis \sigma_i such that\omega(u, v) = \sigma_1\wedge\sigma_2(u, v)+\dots +\sigma_{2r-1}\wedge\sigma_{2r}(u, v)$$In your example, it is fine to start with a basis ... 2 There is nice characterisation of tensors which Alt(T)=0. Thm. Let V be a vector space over \mathbb{R} (without any additional assumptions). Consider subspace N^n(V) of \otimes^nV generated by elements v_1\otimes\dots\otimes v_n such that v_i=v_j for at least one pair i\neq j. We have that$$\ker(Alt)=N^n(V).$$Equivalently. For every ... 2 For the commutative case, we know that$$ \dim k^i [x_1, \cdots, x_n] = \sum_{e_1 + \cdots + e_n = i} 1. This gives \begin{align*} P_{\Bbb{C}[x_1, \cdots, x_n]}(t) &= \sum_{i=0}^{\infty} \sum_{e_1 + \cdots + e_n = i} t^i \\ &= \sum_{i=0}^{\infty} \sum_{e_1 + \cdots + e_n = i} t^{e_1} \cdots t^{e_n} \\ &= \sum_{e_1, \cdots, e_n} t^{e_1} ... 2 The algebra of polynomials on a vector space W (of dimension N, say) is just a coordinate-free way of saying "polynomials in N variables". So, if you like, a polynomial on W is just a function q:W\to \mathbb{R} such that if you pick a linear isomorphism f:\mathbb{R}^N\to W, the composition qf:\mathbb{R}^N\to\mathbb{R} is a polynomial function ... 1 The function g defined by g(A) = \det(A)f(\mathrm{Id}) is alternating and multilinear. Sinceg(\mathrm{Id}) = \det(\mathrm{Id}) f(\mathrm{Id}) = f(\mathrm{Id}),$$the statement you quoted implies that g = f. 1 You might want take a look at adjoint matrix to get the idea. 1 Let me reformulate the question. You are essentially asking if the isomorphism \Lambda^2(V^*) \rightarrow (\Lambda^2 V)^* where \varphi^* \wedge \psi^* \mapsto [v \wedge w \mapsto \varphi(v)\psi(w) - \psi(v)\varphi(w)] is "canonically" induced by the natural isomorphism$$V^* \otimes V^* \rightarrow (V \otimes V)^* where $\varphi \otimes \psi \mapsto ... 1 Say the quiver$Q$has adjacency matrix$A$. Then the$ij$entry of$A^n$counts the number of paths of length$n$from$i$to$j$. So$\dim k^{(n)}Q$is the sum of the entries of$A^n$. Therefore your generating function is the sum of the entries of the matrix$\sum_{n\ge0} A^nt^n=({\rm Id}-At)^{-1}\$. Note one can write the entries of an inverse matrix ...

1

A nice textbook which introduces higher order Fréchet derivatives is "Differential Calculus" by Cartan.

Only top voted, non community-wiki answers of a minimum length are eligible