hazbijav6

2022-10-08

If ${\alpha }_{ij}{A}^{i}{B}^{j}=0$ and ${A}^{i}$ and ${B}^{j}$ are arbitrary vectors, then prove that ${\alpha }_{ij}=0$
This problem appeared in my Differential Geometry class, the professor explained the problem by, first taking an arbitrary vector and demonstrating that ${\alpha }_{ii}=0$. and then proceeded to demonstrate that, ${A}^{l}={B}^{m}=1,\left(1\leqq l\le n,1\leqq m\leqq n,l\ne m\right)$. I get the proof somewhat. Can any of you elucidate it or give an alternative proof?

Johnathon Mcmillan

${A}^{i}{B}^{j}$ are the component of the tensor product of two vector $\mathbf{A}\otimes \mathbf{B}$. Among all these tensors there are also the tensors ${\mathbf{e}}_{i}\otimes {\mathbf{e}}_{j},$ where $B=\left\{{\mathbf{e}}_{1},\dots ,{\mathbf{e}}_{n}\right\}$ is a base of the vector space V,and is a base of the space of rank 2 tensors over tensors over V.
${\alpha }_{i,j}{A}^{i}{B}^{j}$ is the inner product of the tensor $\mathbit{\alpha }$ and the tensor $\mathbf{A}\otimes \mathbf{B}$
${\alpha }_{i,j}{A}^{i}{B}^{j}=\mathbit{\alpha }:\left(\mathbf{A}\otimes \mathbf{B}\right)$
You know that if the inner product of a vector for each element of a base vanish, then the vector is the null vector. This is true also for the inner product vector space of tensors.

rialsv

Fix, h,k, then if you take
$\begin{array}{r}{A}^{i}=\left\{\begin{array}{lll}1,& & i=h,\\ 0,& & i\ne h,\end{array}\end{array}\phantom{\rule{2em}{0ex}}\begin{array}{r}{B}^{j}=\left\{\begin{array}{lll}1,& & j=k,\\ 0,& & j\ne k\end{array}\end{array}$
then
${\alpha }_{ij}{A}^{i}{B}^{j}=0\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{\alpha }_{hk}=0,$
for the arbitrariness of h,k, this is true for all h,k.

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