We are given that <s, s>≥ <v, s>+ <s, v> not that they are equal so really saying that
<v, s>+ <s, v> =0 ( something we know will be true seeing as we are proving v is in the orthogonal complement of S) only shows that <s,s> can be 0 or anything greater than zero, not defiantly 0
I did for get to state that this was for all s in S.
Could you explain a little more why <s,s>=0
would you be referring to the fact that we may be able to pick a case where v=s and therefor
2<s,s> ≤ <s,s>
and this would only happen when <s,s> = 0 because <x,x> ≥ 0?
Homework Statement
Let V be a complex inner product space and let S be a subset of V. Suppose that v in V is a vector for which
<s,v > + <v,s> \leq <s,s>
Prove that v is in the orthogonal set S\bot
Homework Equations
We have the three inner product relations:
1) conjugate symmetry...
Homework Statement
The Riesz representation theorem gives us that forall f in V* there exists a unique R_f in V such that f(x) = <x, R_f >. (<,> is my attempt to type inner product angle brackets) Verify that R_f is unique.
Homework Equations
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If two vector spaces, say V and W, have equal cardinality |V|=|W| ... do they then have the same dimension? That is dim(V)=dim(W)?
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