THEOREM 1.7 (Gram-Schmidt process). Let X be an inner product space and let {₁,..} be a linearly independent set of vectors. Then there exists an orthonormal set of vectors {x₁, X,...} such that, for any n, **** [{₁, 2, yn}] = [{x₁,x₂,. x₁}], 29 ****
THEOREM 1.7 (Gram-Schmidt process). Let X be an inner product space and let {₁,..} be a linearly independent set of vectors. Then there exists an orthonormal set of vectors {x₁, X,...} such that, for any n, **** [{₁, 2, yn}] = [{x₁,x₂,. x₁}], 29 ****
Advanced Engineering Mathematics
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Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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![THEOREM 1.7 (Gram-Schmidt process). Let X be an inner product space and
let {₁,.} be a linearly independent set of vectors. Then there exists an
orthonormal set of vectors {x₁, ..., X, ...} such that, for any n,
[{₁, 2, Yn}] = [{X₁, X2₂, ..., *}],
where the square brackets indicate the space spanned by the set of vectors enclosed;
for example, if x, y = X, then [{x, y}] = {ax +By|a, B = F}.
Proof. The proof will be by induction. First we verify that the theorem is true in
the case when n = 1. Since y₁ = 0, it makes sense to consider x₁ = y₁/|y₁|l, which,
clearly, has norm 1. Further, the spaces that x₁ and y₁ span must be identical,
because the two vectors are linearly dependent. Now that this is done, assume
orthonormal vectors {x₁, x₂, ..., xn-1} have been constructed such that
[{x₁, x₂,...,x₁}] = [{₁, 2, ..., y}]
for any between 1 and (n − 1) inclusive. Our job now is to construct the nth vector
with the same properties. To this end, consider the vector
and take, for 1 ≤j≤n-1,
n-1
w=y-Έch, xxi,
¡= 1
(w, xj) = (xn, xj) —
n-1
On, X1) (X₁, ×z)
= (yn, xj) - (yn, x) = 0.
Hence w 1 x; for i = 1, 2, ..., n - 1. Suppose it were possible for w to be zero.
This would imply that
Yn = Σ(yn, X₂)X₁
or, in words, that y, could be written as a linear combination of X₁, X₂, ..., X₁-19
which, by the induction hypothesis, would imply that y, could be written as a linear
combination of y₁, 2,-1, which is contrary to our assumption that the
{₁, 2, ...} was a linearly independent set. Thus w cannot be zero. It now
remains only to choose x, = w/||w, with the fact that the spaces spanned by
{x₁, x₂,...,x) and {₁,2,...} are the same being left as a simple exercise
for the reader.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe15b7304-cc73-4505-92c3-23aa2fda4f71%2Fa2165a9b-a346-47c2-aed1-4c6eb494bc12%2Fa8sjnkw_processed.png&w=3840&q=75)
Transcribed Image Text:THEOREM 1.7 (Gram-Schmidt process). Let X be an inner product space and
let {₁,.} be a linearly independent set of vectors. Then there exists an
orthonormal set of vectors {x₁, ..., X, ...} such that, for any n,
[{₁, 2, Yn}] = [{X₁, X2₂, ..., *}],
where the square brackets indicate the space spanned by the set of vectors enclosed;
for example, if x, y = X, then [{x, y}] = {ax +By|a, B = F}.
Proof. The proof will be by induction. First we verify that the theorem is true in
the case when n = 1. Since y₁ = 0, it makes sense to consider x₁ = y₁/|y₁|l, which,
clearly, has norm 1. Further, the spaces that x₁ and y₁ span must be identical,
because the two vectors are linearly dependent. Now that this is done, assume
orthonormal vectors {x₁, x₂, ..., xn-1} have been constructed such that
[{x₁, x₂,...,x₁}] = [{₁, 2, ..., y}]
for any between 1 and (n − 1) inclusive. Our job now is to construct the nth vector
with the same properties. To this end, consider the vector
and take, for 1 ≤j≤n-1,
n-1
w=y-Έch, xxi,
¡= 1
(w, xj) = (xn, xj) —
n-1
On, X1) (X₁, ×z)
= (yn, xj) - (yn, x) = 0.
Hence w 1 x; for i = 1, 2, ..., n - 1. Suppose it were possible for w to be zero.
This would imply that
Yn = Σ(yn, X₂)X₁
or, in words, that y, could be written as a linear combination of X₁, X₂, ..., X₁-19
which, by the induction hypothesis, would imply that y, could be written as a linear
combination of y₁, 2,-1, which is contrary to our assumption that the
{₁, 2, ...} was a linearly independent set. Thus w cannot be zero. It now
remains only to choose x, = w/||w, with the fact that the spaces spanned by
{x₁, x₂,...,x) and {₁,2,...} are the same being left as a simple exercise
for the reader.
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