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Letting x, be any one of the basis vectors in M, we have (y,x₁) = (z, x;) - (z, xj) = 0 (j = 1, 2, ..., n) or that y is orthogonal to each of the basis vectors in M. Hence it is orthogonal to every vector in M, which implies that y e M¹. Therefore, we have the representation, for any vector z e X, = Σ (z, x₁)x₁ + y₂ Z=

Letting x, be any one of the basis vectors in M, we have (y,x₁) = (z, x;) - (z, xj) = 0 (j = 1, 2, ..., n) or that y is orthogonal to each of the basis vectors in M. Hence it is orthogonal to every vector in M, which implies that y e M¹. Therefore, we have the representation, for any vector z e X, = Σ (z, x₁)x₁ + y₂ Z=

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter6: Vector Spaces
Section6.4: Linear Transformations
Problem 24EQ
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THEOREM 1.8. Let X be an inner product space and let M be a subspace of X such that dim M<∞o. Then we have the direct sum decomposition X = MOM¹. [The notation M N = X means that any vector z e X can be written uniquely as z = x + y, where x = M and y = N, or, equivalently, that every z e X can be written as z = x + y, where x = M and ye N and M N = {0}.] Proof. First it will be shown that any vector in the space can be written as the sum of an element from M and an element from M¹. Since dim M <∞, we can choose an orthonormal basis for M : X₁, X₂, Now it is clear that (z, x₁)x₁ € M (1.10) i=1 for any vector z e X, because expression (1.10) is just a linear combination of the basis vectors in M. Now consider the vector y = z - Σ (z, x₁)x₁. i=1 Letting x, be any one of the basis vectors in M, we have ... Xn. (y, xj) = (z, x;) - (z, x;) = 0 (j = 1, 2, ..., n) or that y is orthogonal to each of the basis vectors in M. Hence it is orthogonal to every vector in M, which implies that ye Mt. Therefore, we have the representation, for any vector z € X, z = [ (z, x₁)x; + y₂ i=1 where the first term on the right is in M and the second is in M+, and have completed the first part of the proof.
Transcribed Image Text:THEOREM 1.8. Let X be an inner product space and let M be a subspace of X such that dim M<∞o. Then we have the direct sum decomposition X = MOM¹. [The notation M N = X means that any vector z e X can be written uniquely as z = x + y, where x = M and y = N, or, equivalently, that every z e X can be written as z = x + y, where x = M and ye N and M N = {0}.] Proof. First it will be shown that any vector in the space can be written as the sum of an element from M and an element from M¹. Since dim M <∞, we can choose an orthonormal basis for M : X₁, X₂, Now it is clear that (z, x₁)x₁ € M (1.10) i=1 for any vector z e X, because expression (1.10) is just a linear combination of the basis vectors in M. Now consider the vector y = z - Σ (z, x₁)x₁. i=1 Letting x, be any one of the basis vectors in M, we have ... Xn. (y, xj) = (z, x;) - (z, x;) = 0 (j = 1, 2, ..., n) or that y is orthogonal to each of the basis vectors in M. Hence it is orthogonal to every vector in M, which implies that ye Mt. Therefore, we have the representation, for any vector z € X, z = [ (z, x₁)x; + y₂ i=1 where the first term on the right is in M and the second is in M+, and have completed the first part of the proof.
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