6. Let S = {V₁,..., Vn} be a set of orthonormal vectors such that Span(S) = V, and let x € V. Then there is a unique set of coefficients C₁,..., Cò such that x=0V1+tên Vn
6. Let S = {V₁,..., Vn} be a set of orthonormal vectors such that Span(S) = V, and let x € V. Then there is a unique set of coefficients C₁,..., Cò such that x=0V1+tên Vn
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Please answer the questions 6,7,8.

Transcribed Image Text:1. Triangle inequality for inner products: For all a, b, c € V, (a, c) ≤ (a, b) + (b, c).
=
2. Transitivity of orthogonality: For all a, b, c € V, if (a, b) = 0 and (b, c) = 0 then (a, c) = 0.
3. Orthogonality closed under addition: Suppose S {V₁,..., Vn} V is a set of vectors, and x is
orthogonal to all of them (that is, for all i = 1,2,...n, (x, v₁) = 0). Then x is orthogonal to any
y € Span(S).
4. Let S = {V₁, V2,..., Vn} V be an orthonormal set of vectors in V. Then for all non-zero
x € V, if for all 1 ≤ i ≤ n we have (x, vi) = 0 then x Span(S).
5. Let S =
{V₁, V2, ..., Vn} ≤ V be a set of vectors in V (no assumption of orthonormality). Then for
all non-zero x € V, if for all 1 ≤ i ≤ n we have (x, V₁) = 0 then x Span(S).
6. Let S = {V₁,..., Vn} be a set of orthonormal vectors such that Span(S) = V, and let x € V.
Then there is a unique set of coefficients c₁,..., Cn such that
X=CV1+tCnvn
7. Let S = {V₁,..., Vn} be a set of vectors (no assumption of orthonormality) such that Span(S) = V,
and let x € V. Then there is a unique set of coefficients c₁,..., Cn such that
X=CV1+…+Cnvn
8. Let S = {V₁, V2,...)
, Vn} CV be a set of vectors. If all the vectors are pairwise linearly independent
(i.e., for any 1 ≤ i ‡ j ≤n, then only solution to ciV₁ +CjVj = 0 is the trivial solution c¿ = c; = 0.)
then the set S is linearly independent.
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