Hi,

how can one shows that an orthonomal set in an inner product space is automatically an independent set?

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- May 4th 2010, 11:50 AMmtmathorthonomal set
Hi,

how can one shows that an orthonomal set in an inner product space is automatically an independent set? - May 4th 2010, 12:58 PMtonio

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. (Wink)

Tonio - May 5th 2010, 02:13 AMmtmath
- May 5th 2010, 02:29 AMHallsofIvy
Yes, and since it is "not linearly dependent", it is linearly independent.

- May 5th 2010, 03:25 AMtonio

Well,such summation is zero when all the coefficients are zero...(Worried)...what you wanted to prove, and we achieved, is that__any__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.__only__.__independient__

Tonio