Answer :
Certainly! Let's address the two parts of the problem.
### Part 1: True or False Statement
Statement: If [tex]\( E \)[/tex] is a subspace of [tex]\( V \)[/tex], then [tex]\(\operatorname{dim} E + \operatorname{dim}(E^{\perp}) = \operatorname{dim} V\)[/tex].
Analysis:
The statement is true. Here's why:
When you have a subspace [tex]\( E \)[/tex] of a finite-dimensional inner product space [tex]\( V \)[/tex], the orthogonal complement [tex]\( E^{\perp} \)[/tex] consists of all vectors in [tex]\( V \)[/tex] that are orthogonal to every vector in [tex]\( E \)[/tex]. One important property is that [tex]\( V \)[/tex] can be expressed as the direct sum of [tex]\( E \)[/tex] and [tex]\( E^{\perp} \)[/tex] (i.e., [tex]\( V = E \oplus E^{\perp} \)[/tex]).
According to this property, the dimensions relate by:
[tex]\[
\operatorname{dim} V = \operatorname{dim} E + \operatorname{dim} E^{\perp}
\][/tex]
This is because each vector in [tex]\( V \)[/tex] can be uniquely written as the sum of a vector in [tex]\( E \)[/tex] and a vector in [tex]\( E^\perp\)[/tex]. Thus, the statement is true.
### Part 2: Orthogonal Projection
We are given:
- [tex]\( P \)[/tex] is the orthogonal projection onto a subspace [tex]\( E \)[/tex] of an inner product space [tex]\( V \)[/tex].
- [tex]\(\operatorname{dim} V = n\)[/tex], [tex]\(\operatorname{dim} E = r\)[/tex].
Eigenvalues and Eigenspaces of [tex]\( P \)[/tex]:
1. Eigenvalue:
- The eigenvalues of orthogonal projection [tex]\( P \)[/tex] onto subspace [tex]\( E \)[/tex] are 1 and 0. This is because projection maps vectors in [tex]\( E \)[/tex] to themselves and vectors in [tex]\( E^{\perp} \)[/tex] to zero.
2. Eigenspaces:
- Eigenvalue 1: The eigenspace corresponding to eigenvalue 1 is the subspace [tex]\( E \)[/tex]. This includes all vectors that remain unchanged by the projection.
- Eigenvalue 0: The eigenspace corresponding to eigenvalue 0 is the orthogonal complement [tex]\( E^{\perp} \)[/tex]. This includes all vectors that are mapped to zero by the projection.
3. Algebraic and Geometric Multiplicities:
- Eigenvalue 1:
- Algebraic Multiplicity: [tex]\( r \)[/tex] (dimension of [tex]\( E \)[/tex]).
- Geometric Multiplicity: [tex]\( r \)[/tex] (also the dimension of [tex]\( E \)[/tex]).
- Eigenvalue 0:
- Algebraic Multiplicity: [tex]\( n - r \)[/tex] (dimension of [tex]\( E^{\perp} \)[/tex]).
- Geometric Multiplicity: [tex]\( n - r \)[/tex] (also the dimension of [tex]\( E^{\perp} \)[/tex]).
These foundational concepts for projections in linear algebra allow us to make these conclusions regarding the eigenvalues and eigenspaces of the orthogonal projection operator [tex]\( P \)[/tex].
### Part 1: True or False Statement
Statement: If [tex]\( E \)[/tex] is a subspace of [tex]\( V \)[/tex], then [tex]\(\operatorname{dim} E + \operatorname{dim}(E^{\perp}) = \operatorname{dim} V\)[/tex].
Analysis:
The statement is true. Here's why:
When you have a subspace [tex]\( E \)[/tex] of a finite-dimensional inner product space [tex]\( V \)[/tex], the orthogonal complement [tex]\( E^{\perp} \)[/tex] consists of all vectors in [tex]\( V \)[/tex] that are orthogonal to every vector in [tex]\( E \)[/tex]. One important property is that [tex]\( V \)[/tex] can be expressed as the direct sum of [tex]\( E \)[/tex] and [tex]\( E^{\perp} \)[/tex] (i.e., [tex]\( V = E \oplus E^{\perp} \)[/tex]).
According to this property, the dimensions relate by:
[tex]\[
\operatorname{dim} V = \operatorname{dim} E + \operatorname{dim} E^{\perp}
\][/tex]
This is because each vector in [tex]\( V \)[/tex] can be uniquely written as the sum of a vector in [tex]\( E \)[/tex] and a vector in [tex]\( E^\perp\)[/tex]. Thus, the statement is true.
### Part 2: Orthogonal Projection
We are given:
- [tex]\( P \)[/tex] is the orthogonal projection onto a subspace [tex]\( E \)[/tex] of an inner product space [tex]\( V \)[/tex].
- [tex]\(\operatorname{dim} V = n\)[/tex], [tex]\(\operatorname{dim} E = r\)[/tex].
Eigenvalues and Eigenspaces of [tex]\( P \)[/tex]:
1. Eigenvalue:
- The eigenvalues of orthogonal projection [tex]\( P \)[/tex] onto subspace [tex]\( E \)[/tex] are 1 and 0. This is because projection maps vectors in [tex]\( E \)[/tex] to themselves and vectors in [tex]\( E^{\perp} \)[/tex] to zero.
2. Eigenspaces:
- Eigenvalue 1: The eigenspace corresponding to eigenvalue 1 is the subspace [tex]\( E \)[/tex]. This includes all vectors that remain unchanged by the projection.
- Eigenvalue 0: The eigenspace corresponding to eigenvalue 0 is the orthogonal complement [tex]\( E^{\perp} \)[/tex]. This includes all vectors that are mapped to zero by the projection.
3. Algebraic and Geometric Multiplicities:
- Eigenvalue 1:
- Algebraic Multiplicity: [tex]\( r \)[/tex] (dimension of [tex]\( E \)[/tex]).
- Geometric Multiplicity: [tex]\( r \)[/tex] (also the dimension of [tex]\( E \)[/tex]).
- Eigenvalue 0:
- Algebraic Multiplicity: [tex]\( n - r \)[/tex] (dimension of [tex]\( E^{\perp} \)[/tex]).
- Geometric Multiplicity: [tex]\( n - r \)[/tex] (also the dimension of [tex]\( E^{\perp} \)[/tex]).
These foundational concepts for projections in linear algebra allow us to make these conclusions regarding the eigenvalues and eigenspaces of the orthogonal projection operator [tex]\( P \)[/tex].
