hazbijav6

Answered

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, ${A}^{l}={B}^{m}=1,(1\leqq l\le n,1\leqq m\leqq n,l\ne m)$. I get the proof somewhat. Can any of you elucidate it or give an alternative proof?

Answer & Explanation

Johnathon Mcmillan

Expert

2022-10-09Added 7 answers

${A}^{i}{B}^{j}$ 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 $B=\{{\mathbf{e}}_{1},\dots ,{\mathbf{e}}_{n}\}$ is a base of the vector space V,and ${B}^{\prime}=\{{\mathbf{e}}_{i}\otimes {\mathbf{e}}_{j},\text{}i,j=1,\dots ,n\}$ 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 $\mathit{\alpha}$ and the tensor $\mathbf{A}\otimes \mathbf{B}$

${\alpha}_{i,j}{A}^{i}{B}^{j}=\mathit{\alpha}:(\mathbf{A}\otimes \mathbf{B})$

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

Expert

2022-10-10Added 3 answers

Fix, h,k, then if you take

$\begin{array}{r}{A}^{i}=\{\begin{array}{lll}1,& & i=h,\\ 0,& & i\ne h,\end{array}\end{array}\phantom{\rule{2em}{0ex}}\begin{array}{r}{B}^{j}=\{\begin{array}{lll}1,& & j=k,\\ 0,& & j\ne k\end{array}\end{array}$

then

${\alpha}_{ij}{A}^{i}{B}^{j}=0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\alpha}_{hk}=0,$

for the arbitrariness of h,k, this is true for all h,k.

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