New answers tagged

1

Maybe try this paper? Fast Matrix Multiplication: Limitations of the Laser Method It goes into enough detail for me to think I understand what's going on. I think the laser method is figuring out answers for N values with less rounding errors than what occurs using Strassen's method using some differential equation or numerical method. In recent years ...


1

If $\frac{\partial^2}{\partial x \partial \lambda}$ is denoting an outer product of $\frac{\partial}{\partial x}$ with $\frac{\partial}{\partial \lambda}$ we have $$ \begin{align} \partial_{x^m} \partial_{\lambda^n} (A^{ij} \lambda^i x^j + B^{ij} x^i \lambda^j) &= A^{ij} \delta^{in} \delta^{jm} + B^{ij} \delta^{im} \delta^{jn} \\ &=A^{nm} + B^{mn} ...


0

I think this is what you're after: If $R$ is a ring, then the map $f : R \times R \to R$ given by $f(x, y) = x y$ is $R$-bilinear, so it induces a linear map $\tilde{f}:R\otimes_R R \to R$ given by $\tilde{f}(x \otimes y) = f(x, y) = x y$. In general, tensor products of modules (or, more concretely, vector spaces) turn bilinear maps $f : A \times B \to C$ ...


0

It is the first definition: Given a basis $\mathcal F = (f_i)$ of a finite-dimensional vector space $\Bbb V$, the dual basis of the dual space, $\Bbb V^*$, is the unique basis $(\hat f{}^i)$ of $\Bbb V^*$ such that $$\hat f{}^i(f_j) = \left\{\begin{array}{rl}1,& i = j \\ 0,& i \neq j\end{array}\right.$$ In particular, if $v = ...


0

If you are having trouble with tensors, try this: Imagine that you want to locate a point in 3-dimensional Cartesian plane. You need (x,y,z) coordinates to locate the point. Now imagine that each of these points has a numerical value. Then this three dimensional plane with these numerical value will constitute a matrix. If the length of x, y and z axes, each ...


4

Tensors in general are multilinear coordinate-free objects that can be represented with respect to some basis by a multi-dimensional arrays indexed appropriately. Just like you can represent a bilinear form $B \colon V \times V \rightarrow \mathbb{F}$ when $V$ is an $n$-dimensional vector space (or a linear map $T \colon V \rightarrow V$) by a $n \times n$ ...


1

I'ts a tensor. But so are normal numbers, vectors, and matrices which can be considered 1X1 tensor, 1Xn tensor, mXn Tensor. All data structures of this form (mXnX...)are what are called Tensors.


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You can think of tensors as multidimensional arrays.


3

You can think of a rank three tensor as a three dimensional array. A matrix is a rank two tensor, or a two dimensional array. A vector then is a rank one tensor and scalar a rank zero. This is a simplification of the subject of tensors, but it's useful to think of them as a generalization of matrices.


0

In physics (and in mathematics, in differential geometry) you are working with so called manifolds $M$, which you can think of as smooth $k-$ dimensional subsets of some $\mathbb{R}^n$. A vector (field) along $M$ can be thought of as an assignment of a vector $X(p)$ in each point $p$ of $M$ which is tangent to $M$. A differential form is an assignment of a ...


1

how exactly to represent specific linear transformations in tensor notation. Indeed every linear map $A:V\rightarrow W$ has a representation as a tensor $T\in V^{*}\otimes W$ such that $\forall x\in V:Ax=T(x)\in W$. In chosen bases, $$ T=\sum a_{ij}\;v_j^*\otimes w_{i} $$ where $a_{ij}$ is the matrix of the linear map and $\{v_j^*\}$ is the basis in ...


1

$$ u=\{\text{u1},\text{u2}\}, v=\{\text{v1},\text{v2}\} $$ $$ \text{uv}=u\otimes v=\left( \begin{array}{cc} \text{u1} \text{v1} & \text{u1} \text{v2} \\ \text{u2} \text{v1} & \text{u2} \text{v2} \\ \end{array} \right) $$ $$ \left| \text{uv}\right|=0 $$ The determinant of the matrix that represents the tensor product of any 2 vectors will always be ...


0

Let $E:=\mathbb{R}^n$ and $E^*$ its dual. Here is a "concrete" way of seeing the equivalence of a linear transform $L:E \rightarrow E$ with a linear application from $L:E \otimes E^* \rightarrow \mathbb{R}$: It is nothing more that the decomposition of matrix $M=[m_{i,j}]$ of linear mapping $L$ (with respect to canonical bases of $E$ and $E^*$) under the ...


0

One can use the matrix for $g$ within the construction of the reciprocal basis $$\partial^k=\sum_sg^{ks}\partial_s,$$ or simply $\partial^k=g^{ks}\partial_s$. It happens that $\partial_k\mapsto\partial^k$ is a change of basis in the tangent space.


1

Ok, lets formalise all that have been said in the comments: Let $(v_1,\dots,v_n)$ and $(v^1,\dots,v^n)$ be basis of $V_p$ and $V^*_p$, respectively. Take a tensor $\tau \in \mathcal{T}^r_s(V_p)$, and pick indexes $k \leq r, l \leq s$. Then we define the contraction $C^k_l\tau \in \mathcal{T}^{r-1}_{s-1}(V_p)$ as ...


2

An element of the domain of $T$ is of the form $(f_1,\dots, f_k, v_1, \dots, v_l),\, f_1,\dots,f_k\in V^*, v_1,\dots,v_l \in V$, so the first $k$ elements of $T$ should be able to take elements of $V^*$ and the next $l$ terms should be able to take elements of $V$, hence $v_{\mu_1} \otimes \cdots \otimes v_{\mu_k}(f_1,\dots, f_k)$ first, and $v^{{\nu_1}^*} ...


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For finite dimensional vector spaces, we have an isomorphism $V \cong V^{**}$, so you are really using a basis of $V^{**}$


0

I'm going to write $\varepsilon^i$ instead of $e^i$ to make the notation clearer: we have $\varepsilon^i(e_j) = \delta_{ij}$. We want to write $B$ as some linear combination of the $\varepsilon^i \otimes \varepsilon^j$. Note that $(\varepsilon^i \otimes \varepsilon^j)(e_p \otimes e_q) = \delta_{ip} \delta_{jq}$ just as for the 1-dimensional case. This means ...


1

The second line is wrong. At first act with the transformation of the covariant derivative $Y^{j'}_{,p}$. Then you will get $$\frac{\partial x'^p}{\partial x^i}(Y^{j'})=\frac{\partial x'^p}{\partial x^i}(\frac{\partial x^{j'}}{\partial x^q} Y^p_q).$$ So, you will get a term that will cancel the last term in last line in your solution.


3

The definition of the tensor product is $(f \otimes g)(u,v) = f(u)g(v)$. You should be able to check that whenever $f,g \in T^*$ (i.e. are linear), their product $f \otimes g$ is bilinear; and the product $f \otimes f$ is symmetric and weakly positive definite. Thus the action of $dx \otimes dx$ on a pair $u,v$ of tangent vectors is $$(dx \otimes dx)(u,v) = ...


3

If $$\omega = \sum_{i=1}^n\frac{(-1)^{i-1}x_i}{\|x\|^n}\,dx_1 \wedge\cdots \wedge \widehat{dx_i}\wedge \cdots \wedge dx_n,$$then: $$d\omega = \sum_{i=1}^n\sum_{j=1}^n\frac{\partial}{\partial x_j}\left(\frac{(-1)^{i-1}x_i}{\|x\|^n}\right) dx_j \wedge dx_1 \wedge\cdots \wedge \widehat{dx_i}\wedge \cdots \wedge dx_n.$$Now, the only surviving term is when $j = ...



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