Tag Info

Hot answers tagged

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. Since $$g(\mathrm{Id}) = \det(\mathrm{Id}) f(\mathrm{Id}) = f(\mathrm{Id}),$$ 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 Fr├ęchet derivatives is "Differential Calculus" by Cartan.



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