In general, for an inner product space over ( or ) the equality holds iff and are linearly dependent.
when does equality hold in the scwharz inequality. When ai, bi are real numbers, then there is equality when ai=kbi, for i =1,...,n. But when they are complex sequences I tried setting ai=kb*i, (where b* is b conjugate) for i=1,...,n and it didn't work.
Thanks. I am not showing my work because I have not got the time to do latex atm.
The complex sequences are also vectors so, for any inner vector space whose elements are complex sequences, you only have to translate the condition of linear dependence. For example, translate it to the vector space of all finite non null sequences with . What do you obtain?
Right in that case, a few questions.
1) How can a vector space be a set of sequences (I thought it was a set of vectors)?
2) In that sum, should it be k instead of infinity since we are dealing with finite sequences?
3) So you are saying this is how we define the dot product for vectors with complex components. So in that case I can set ai=kb*i and get equality?
Thanks, you have been very helpful.
Define and . With these operations, is a vector space over . For that reason we call vector to any sequence .
It is irrelevant, the sum is finite:2) In that sum, should it be k instead of infinity since we are dealing with finite sequences?
We get the equality iff and are linearly dependent, i.e. iff there exists such that for all or for all .3) So you are saying this is how we define the dot product for vectors with complex components. So in that case I can set ai=kb*i and get equality?