Hi,
how can one shows that an orthonomal set in an inner product space is automatically an independent set?
Let $\displaystyle \{v_1,\ldots,v_n\}$ be n orthonormal vectors , and suppose $\displaystyle \sum^n_{k=1}a_kv_k=0\,,\,\,a_i\in\mathbb{F}=$ the linear space's definition field. If we denote by $\displaystyle <,>$ the inner product, we get:
$\displaystyle \forall j=1,2,\ldots,n\,,\,\,0=<0,v_j>=<\sum^n_{k=1}a_kv_k ,vj>=\sum^n_{k=1}a_k<v_k,vj>=$ $\displaystyle \sum^n_{k=1}a_i\delta_{k,j}=a_j$ ...and we're done.
Tonio
Well, any such summation is zero when all the coefficients are zero......what you wanted to prove, and we achieved, is that only then the sum is zero, or in other words: if the sum is zero then all the coefficients equal zero. This, according to definition, means the vectors are lin. independient .
Tonio