Section 18 β Linear Algebra (Axler)
π Todayβs Lesson: Section 18 From: Linear Algebra (Axler) β Section 18
Chapter 6.C: Orthogonal Complements and Minimization
The orthogonal complement of a subspace and the orthogonal projection onto it are among the most powerful tools in applied mathematics. They solve the fundamental minimization problem: find the point in a subspace closest to a given vector. This section develops these ideas and connects them to the geometry of inner product spaces.
Orthogonal Complements
π Definition β Orthogonal Complement
If U is a subset of an inner product space V, the orthogonal complement of U is
String.rawU^β₯ = {v β V : β¨ v, u β© = 0 for every u β U}.
In words, U^β₯ consists of all vectors orthogonal to every vector in U.
π Theorem β Properties of the Orthogonal Complement
That U^β₯ is a subspace follows from linearity of the inner product in the first slot. For U β© U^β₯ = {0}: if v β U β© U^β₯, then β¨ v, v β© = 0, so v = 0 by definiteness.
Let U be a subspace of V. Then:
β’ U^β₯ is a subspace of V. β’ {0}^β₯ = V and V^β₯ = {0}. β’ U β© U^β₯ = {0}. β’ If Uβ β Uβ, then Uβ^β₯ β Uβ^β₯.
Orthogonal Decomposition
π Theorem β Direct Sum with Orthogonal Complement
Let eβ, β¦, eβ be an orthonormal basis of U (obtained via Gram-Schmidt). For any v β V, write
String.rawv = β¨ v, eβ β© eβ + β― + β¨ v, eβ β© eβ_β_ _U + v - β¨ v, eβ β© eβ - β― - β¨ v, eβ β© eβ_β_ _U_^_β₯.
The second term is in U^β₯ because for each eβ:
String.rawβ¨ v - Ξ£β±Ό β¨ v, eβ±Ό β© eβ±Ό, eβ β© = β¨ v, eβ β© - β¨ v, eβ β© = 0.
This shows V = U + U^β₯. The sum is direct because U β© U^β₯ = {0}. The dimension formula V = U + U^β₯ follows from the direct sum.
Suppose U is a finite-dimensional subspace of V. Then
String.rawV = U β U^β₯.
In particular, V = U + U^β₯.
Double complement: An immediate consequence is that (U^β₯)^β₯ = U when V is finite-dimensional. The orthogonal complement is an involution on the lattice of subspaces.
Orthogonal Projection
π Definition β Orthogonal Projection
Suppose U is a finite-dimensional subspace of V. The orthogonal projection of V onto U, denoted P_U, is the operator defined by: for v = u + w with u β U and w β U^β₯,
String.rawP_U v = u.
π Theorem β Properties of Orthogonal Projection
If eβ, β¦, eβ is an orthonormal basis for U, then from the decomposition in the previous theorem:
String.rawP_U v = β¨ v, eβ β© eβ + β― + β¨ v, eβ β© eβ.
This shows range P_U = U and null P_U = U^β₯. Also P_UΒ² = P_U because projecting a vector already in U gives itself. Finally, v - P_U v β U^β₯ by construction.
Let U be a finite-dimensional subspace of V. Then:
β’ P_U β L(V) and P_UΒ² = P_U. β’ range P_U = U and null P_U = U^β₯. β’ v - P_U v β U^β₯ for every v β V. β’ βP_U vβ βvβ for every v β V.
βοΈ Example β Projection in R^3
Let U = span{(1,0,0), (0,1,0)} (the xy-plane in RΒ³). For v = (3, 4, 5):
String.rawP_U v = β¨ v, eβ β© eβ + β¨ v, eβ β© eβ = 3(1,0,0) + 4(0,1,0) = (3,4,0).
The residual v - P_U v = (0, 0, 5) β U^β₯.
Minimization: Closest Point in a Subspace
π Theorem β Minimization Principle
Let u β U. Then
String.rawβv - uβΒ² = βv - P_U v + P_U v - uβΒ².
Since v - P_U v β U^β₯ and P_U v - u β U, the Pythagorean theorem gives
String.rawβv - uβΒ² = βv - P_U vβΒ² + βP_U v - uβΒ² βv - P_U vβΒ².
Equality holds iff βP_U v - uβ = 0, i.e., u = P_U v.
Suppose U is a finite-dimensional subspace of V and v β V. Then
String.rawβv - P_U vβ βv - uβ for every u β U.
Furthermore, equality holds if and only if u = P_U v.
The geometry: The orthogonal projection P_U v is the unique point in U closest to v. The βerrorβ vector v - P_U v is perpendicular to the subspace. This is the geometric essence of least-squares approximation.
βοΈ Example β Least-Squares Approximation
Find the point in U = span{(1,1,0), (0,1,1)} closest to v = (1, 0, 0). First apply Gram-Schmidt to get an orthonormal basis of U:
String.raweβ = 1/β(2)(1,1,0), eβ = 1/β(6)(-1, 1, 2).
Then compute the projection:
String.rawP_U v = β¨ v, eβ β© eβ + β¨ v, eβ β© eβ = 1/β(2) Β· 1/β(2)(1,1,0) + (-1)/β(6) Β· 1/β(6)(-1,1,2).
String.raw= 1/2(1,1,0) + 1/6(1,-1,-2) = (2/3, 1/3, -1/3).
One can verify: v - P_U v = (1/3, -1/3, 1/3) is orthogonal to both (1,1,0) and (0,1,1).
Key Takeaways
β’ 1.
U^β₯ is the set of all vectors orthogonal to every vector in U. It is always a subspace. β’ 2.
Orthogonal decomposition: V = U β U^β₯. Every vector splits uniquely into a component in U and a component perpendicular to U. β’ 3.
The orthogonal projection P_U extracts the U-component. It satisfies P_UΒ² = P_U. β’ 4.
Minimization: P_U v is the unique closest point in U to v. The error is perpendicular to the subspace. β’ 5.
This is the mathematical foundation of least-squares methods used throughout science, engineering, and data science.
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