# orthonomal set

• May 4th 2010, 11:50 AM
mtmath
orthonomal 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 PM
tonio
Quote:

Originally Posted by mtmath
Hi,
how can one shows that an orthonomal set in an inner product space is automatically an independent set?

Let $\{v_1,\ldots,v_n\}$ be n orthonormal vectors , and suppose $\sum^n_{k=1}a_kv_k=0\,,\,\,a_i\in\mathbb{F}=$ the linear space's definition field. If we denote by $<,>$ the inner product, we get:

$\forall j=1,2,\ldots,n\,,\,\,0=<0,v_j>=<\sum^n_{k=1}a_kv_k ,vj>=\sum^n_{k=1}a_k=$ $\sum^n_{k=1}a_i\delta_{k,j}=a_j$ ...and we're done. (Wink)

Tonio
• May 5th 2010, 02:13 AM
mtmath
Quote:

Originally Posted by tonio
Let $\{v_1,\ldots,v_n\}$ be n orthonormal vectors , and suppose $\sum^n_{k=1}a_kv_k=0\,,\,\,a_i\in\mathbb{F}=$ the linear space's definition field. If we denote by $<,>$ the inner product, we get:

$\forall j=1,2,\ldots,n\,,\,\,0=<0,v_j>=<\sum^n_{k=1}a_kv_k ,vj>=\sum^n_{k=1}a_k=$ $\sum^n_{k=1}a_i\delta_{k,j}=a_j$ ...and we're done. (Wink)

Tonio

thanks a lot Tonio,

so this simply tells that the summation equals zero when all a's are zeros, hence the so mention set is not linearly dependant?
• May 5th 2010, 02:29 AM
HallsofIvy
Yes, and since it is "not linearly dependent", it is linearly independent.
• May 5th 2010, 03:25 AM
tonio
Quote:

Originally Posted by mtmath
thanks a lot Tonio,

so this simply tells that the summation equals zero when all a's are zeros, hence the so mention set is not linearly dependant?

Well, any such summation is zero when all the coefficients are zero...(Worried)...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