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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 ,

Liyla morad
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    Hint: assume that one of the $a_i$, say $a_n$, is nonzero. What is $f(a_1u_1 + \dots + a_{n - 1}u_{n - 1}, u_n)$, and why is this a contradiction? – CJ Dowd Jan 12 '22 at 07:54

1 Answers1

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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.

SoG
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