Question

Think of \(\mathbb{R}^{n}\) as consisting of rows. a. Let \(E\) be an \(n \times n\) matrix, and let \(U=\left\\{\mathbf{x} E \mid \mathbf{x}\right.\) in \(\left.\mathbb{R}^{n}\right\\} .\) Show that the following are equivalent. i. \(E^{2}=E=E^{T}(E\) is a projection matrix). ii. \((\mathbf{x}-\mathbf{x} E) \cdot(\mathbf{y} E)=0\) for all \(\mathbf{x}\) and \(\mathbf{y}\) in \(\mathbb{R}^{n}\) iii. \(\operatorname{proj}_{U} \mathbf{x}=\mathbf{x} E\) for all \(\mathbf{x}\) in \(\mathbb{R}^{n}\). [Hint: For (ii) implies (iii): Write \(\mathbf{x}=\mathbf{x} E+\) \((\mathbf{x}-\mathbf{x} E)\) and use the uniqueness argument preceding the definition of proj \(_{U} \mathbf{x} .\) For (iii) implies (ii): \(\mathbf{x}-\mathbf{x} E\) is in \(U^{\perp}\) for all \(\mathbf{x}\) in \(\left.\mathbb{R}^{n} .\right]\) b. If \(E\) is a projection matrix, show that \(I-E\) is also a projection matrix. c. If \(E F=0=F E\) and \(E\) and \(F\) are projection matrices, show that \(E+F\) is also a projection matrix. d. If \(A\) is \(m \times n\) and \(A A^{T}\) is invertible, show that \(E=A^{T}\left(A A^{T}\right)^{-1} A\) is a projection matrix.

Short answer

The equivalencies establish that E is a projection matrix based on properties such as orthogonality and symmetry. Both \(I-E\) and \(E+F\) form projection matrices under given conditions, and the defined \(E\) in part d is valid by its idempotency and symmetry.

Step-by-step solution

Understanding Projection Matrix (Part a i)

A projection matrix is defined by the properties that it is idempotent and symmetric. Therefore, for a matrix \(E\) to be a projection matrix, it must satisfy two conditions: \(E^2 = E\) and \(E^T = E\). Verify these properties given the matrix \(E\) and apply these conditions.

Prove (i) Implies (ii) (Part a ii)

Consider that \(E\) is a projection matrix, meaning \(E^2 = E\). Now, \((\mathbf{x} - \mathbf{x}E) \cdot (\mathbf{y}E) = 0\) for all \(\mathbf{x}, \mathbf{y} \in \mathbb{R}^n\). Use the property that a projection matrix projects vectors orthogonally, which implies that the difference \((\mathbf{x} - \mathbf{x}E)\) is orthogonal to the image of any vector projected by \(E\), including \(\mathbf{y}E\).

Prove (ii) Implies (iii) (Part a iii)

Given \((\mathbf{x} - \mathbf{x}E) \cdot (\mathbf{y}E) = 0\) for any \(\mathbf{x}, \mathbf{y} \in \mathbb{R}^n\), decompose \(\mathbf{x}\) as \(\mathbf{x} = \mathbf{x}E + (\mathbf{x} - \mathbf{x}E)\). The conditions ensure that \((\mathbf{x} - \mathbf{x}E)\) is in \(U^\perp\), thus \(\mathbf{x}E\) is the unique projection of \(\mathbf{x}\) in \(U\), making \(\operatorname{proj}_U \mathbf{x} = \mathbf{x} E\).

Prove (iii) Implies (i)

Assume \(\operatorname{proj}_U \mathbf{x} = \mathbf{x}E\) for all \(\mathbf{x}\). This states that \(\mathbf{x}\) can be split as \(\mathbf{x} = \mathbf{x}E + (\mathbf{x} - \mathbf{x}E)\) with \((\mathbf{x} - \mathbf{x}E) \cdot (\mathbf{y}E) = 0\) assuring orthogonality. Therefore, \(E\) must satisfy both \(E^2 = E\) (idempotency) and \(E^T = E\) (symmetry), proving it as a projection matrix.

Prove Part b: \(I - E\) is a Projection Matrix

Since \(E\) is a projection matrix with properties \(E^2 = E\) and \(E^T = E\), consider \(I - E\). The matrix \(I - E\) satisfies \((I - E)^2 = I - 2E + E^2 = I - E\) due to \(E^2 = E\), and \((I - E)^T = I - E^T = I - E\), indicating it is also a projection matrix.

Prove Part c: \(E + F\) is a Projection Matrix

Given \(EF = 0 = FE\) and \(E\) and \(F\) are projection matrices, note that \(E^2 = E\) and \(F^2 = F\). For \(E + F\), compute \((E + F)^2 = E^2 + EF + FE + F^2 = E + 0 + 0 + F = E + F\). Therefore, \((E + F)^2 = E+F\) and since \(E\) and \(F\) are symmetric, \((E+F)^T = E^T + F^T = E + F\). Hence \(E + F\) is also a projection matrix.

Prove Part d: Define \(E\) and Show It Is a Projection Matrix

For an \(m \times n\) matrix \(A\) where \(A A^T\) is invertible, define \(E = A^T(A A^T)^{-1}A\). Show \(E^2 = E\) by calculating \(E^2 = A^T(A A^T)^{-1}A A^T(A A^T)^{-1}A = A^T(A A^T)^{-1}A = E\) using properties of inverses. Also, \(E^T = (AA^T)^{-1}AA^T = E\), thus proving \(E\) is a projection matrix.

Key concepts

Linear Transformations

Linear transformations are fundamental concepts in linear algebra. They describe how vectors are transformed through a linear map. In the context of matrices, these transformations can be seen as operations that convert input vectors into new output vectors. A linear transformation can be represented using a matrix. This allows us to use matrix operations to analyze and compute transformations.
For example, if we have a linear map represented by a matrix \(A\), applying this to a vector \(\mathbf{x}\) results in a new vector \(\mathbf{y}\). This process can be written as \(\mathbf{y} = A\mathbf{x}\).
Key characteristics of a linear transformation include:

• Additivity: \(T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})\) for any vectors \(\mathbf{u}\) and \(\mathbf{v}\).
• Scalar multiplication: \(T(c\mathbf{u}) = cT(\mathbf{u})\) for any vector \(\mathbf{u}\) and scalar \(c\).

These properties make linear transformations predictable and easy to work with, especially when using matrices, as they simplify complex vector operations.

Matrix Algebra

Matrix algebra forms the backbone of many complex computations in mathematics, physics, and engineering. It involves operations such as addition, subtraction, multiplication, and finding inverses of matrices. When dealing with transformations or systems of equations, understanding matrix operations is critical.
One key operation is matrix multiplication, which allows us to combine transformations. For instance, multiplying two matrices \(A\) and \(B\) results in a new matrix \(C\), which represents a transformation combining those of \(A\) and \(B\).
Key points about matrix algebra include:

• Associative Property: \((AB)C = A(BC)\).
• Distributive Property: \(A(B + C) = AB + AC\).
• Non-commutative Nature: Generally, \(AB eq BA\).

These properties help us manipulate and solve systems linearly represented by matrices. They ensure we can systematically approach solutions by applying known rules and simplifications.

Orthogonality

Orthogonality is a concept that describes the perpendicular relationships between vectors in a vector space. In matrix terms, orthogonality refers to the property where lines or planes are positioned at right angles to each other. This concept is crucial in simplifying computations involving projections, which are often dealt with when using projection matrices.
For vectors \(\mathbf{u}\) and \(\mathbf{v}\), being orthogonal means their dot product is zero: \(\mathbf{u} \cdot \mathbf{v} = 0\). In spaces involving matrices, projecting a vector onto a subspace can be thought of as creating an orthogonal component.
Key aspects of orthogonality include:

• Simplifies computations: Orthogonality reduces complexity in calculations, particularly in projections and least squares methods.
• Basis simplification: Orthogonal vectors can form bases that are easier to work with.

Understanding orthogonality is key to mastering projections and transformations in linear algebra, making computations more intuitive and manageable.

Algebraic Properties of Matrices

The algebraic properties of matrices determine how matrices behave under various operations, like addition or multiplication. These properties frame our understanding of more complex mathematical concepts like vector spaces and transformations.
Some basic properties are:

• Idempotency: A matrix \(E\) is idempotent if \(E^2 = E\). This occurs in many matrices, including projection matrices, where repeated applications of the transformation yield the same result.
• Symmetric Matrices: A matrix \(A\) is symmetric if \(A = A^T\). Symmetric matrices often have special properties, including real eigenvalues and orthogonal eigenvectors.

These algebraic properties are crucial when determining the role and effects of matrices in transformations, systems solutions, and optimizations. They provide a structured and predictable framework, ensuring matrices can be precisely manipulated for desired outcomes.