let $f$ : ${M}\times{M}$ $\mapsto$ $F$ be bilinear map
if $T$ = {$u_{1}$ , $u_{2}$ , .... , $u_{n}$} is a subset of ${M}$
with $f(u_{i} , u_{j}$) = $0$ if i $\neq$ j and otherwise $f(u_{i} , u_{j}$) $\neq$ $0$
show that $T$ is linearly independent set .
Edit : to show that $T$ is linearly independent set let $a_{1}$ , $a_{2}$ , ... $a_{n}$ $\in$ $F$ and if $a_{1}$$u_{1}$ + $a_{2}$$u_{2}$ + .... + $a_{n}$$u_{n}$ = $0$ then we must show that $a_{1}$ = $a_{2}$ ... = $a_{n}$ = $0$ but i dont know how i can complete any help please ,
Let, $\sum_{i=1}^{n} c_i u_i=0$ , where $c_i\in F$
Claim :$c_i=0 \space$ for all $i\in\Bbb{N_n}$.
\begin{align}0=f(\sum_{i=1}^{n} c_i u_i, u_j) &=\sum_{i=1}^{n} f(c_i u_i, u_j)\\&=\sum_{i=1}^{n} c_if( u_i, u_j)\\&=c_j\end{align}
Hence, $c_j=0 \space \space, \forall j\in \Bbb{N_n}$
Hence, the set $ T=\{u_1,u_2,...,u_n\}$ is linearly independent.