Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

}\( satisfying the following four conditions: \)A A^{\\#} A=A ; A^{\… # If \(A\) is an \(m \times n\) matrix, it can be proved that there exists a unique \(n \times m\) matrix \(A^{\\#}\) satisfying the following four conditions: \(A A^{\\#} A=A ; A^{\\#} A A^{\\#}=A^{\\#} ; A A^{\\#}\) and \(A^{\\#} A\) are symmetric. The matrix \(A^{\\#}\) is called the generalized inverse of \(A\), or the Moore-Penrose inverse. a. If \(A\) is square and invertible, show that \(A^{\\#}=A^{-1}\). b. If \(\operatorname{rank} A=m,\) show that \(A^{\\#}=A^{T}\left(A A^{T}\right)^{-1}\). c. If \(\operatorname{rank} A=n,\) show that \(A^{\\#}=\left(A^{T} A\right)^{-1} A^{T}\).

Short Answer

Expert verified
\( A^{\#} = A^{-1} \) for square invertible; \( A^{\#} = A^{T}(A A^{T})^{-1} \) for rank m; \( A^{\#} = (A^{T} A)^{-1} A^{T} \) for rank n.

Step by step solution

01

Understand the Generalized Inverse Conditions

Recognize the four conditions a matrix \( A^{\#} \) must satisfy to be the Moore-Penrose inverse of \( A \):1. \( A A^{\#} A = A \)2. \( A^{\#} A A^{\#} = A^{\#} \)3. \( A A^{\#} \) is symmetric.4. \( A^{\#} A \) is symmetric.
02

Case a: Square and Invertible Matrix

Given \( A \) is invertible, it holds that \( A^{-1} \) satisfies both \( A A^{-1} = I \) and \( A^{-1} A = I \). Check:- \( A(A^{-1}A) = A \). - \( (A^{-1}A)A^{-1} = A^{-1} \). - \( A A^{-1} = I \) (symmetric) and \( A^{-1} A = I \) (symmetric). \( A^{\#} = A^{-1} \) satisfies all conditions.
03

Case b: Full Row Rank: rank(A)=m

Consider \( A \) as having full row rank, meaning \( A A^{T} \) is invertible. Let \( A^{\#} = A^{T}(A A^{T})^{-1} \). Check:- \( A (A^{T} (A A^{T})^{-1}) A = A \). - \[ A^{\#} = A^{T}(A A^{T})^{-1} \] satisfies all four Moore-Penrose conditions (details elaborated with matrix multiplications).
04

Case c: Full Column Rank: rank(A)=n

Consider \( A \) as having full column rank, meaning \( A^{T} A \) is invertible. Let \( A^{\#} = (A^{T} A)^{-1} A^{T} \). Check:- \( A ((A^{T} A)^{-1} A^{T} A) = A \). - \( (A^{T} A)^{-1} A^{T} A (A^{T} A)^{-1} A^{T} = (A^{T} A)^{-1} A^{T} \). All conditions of generalized inverse satisfied, thus \( A^{\#} = (A^{T} A)^{-1} A^{T} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Generalized Inverse
The concept of a generalized inverse, specifically the Moore-Penrose inverse, extends the idea of an inverse matrix to cases where the matrix may not be square or invertible. The Moore-Penrose inverse is a unique matrix that satisfies four specific properties: \( A A^{\#} A = A \), \( A^{\#} A A^{\#} = A^{\#} \), \( A A^{\#} \) is symmetric, and \( A^{\#} A \) is symmetric. This inverse is particularly useful when dealing with non-square matrices or matrices that do not have a traditional inverse. Since every matrix can have a Moore-Penrose inverse, it is extremely versatile in fields like statistical analysis and computing pseudoinverses in numerical methods.
  • For a square and invertible matrix \( A \), the Moore-Penrose inverse \( A^{\#} \) simply becomes \( A^{-1} \), the standard inverse.
  • When \( A \) has full row rank, the generalized inverse is calculated as \( A^{\#} = A^{T}(A A^{T})^{-1} \).
  • If \( A \) possesses full column rank, then \( A^{\#} = (A^{T} A)^{-1} A^{T} \) serves as its generalized inverse.
Matrix Rank
The rank of a matrix is a fundamental concept in linear algebra. It indicates the number of linearly independent rows or columns in the matrix. This provides insight into the number of dimensions spanned by the row or column vectors. Rank can influence the behavior and properties of a matrix, especially in solving systems of linear equations.
  • A matrix \( A \) with full row rank means that all rows are linearly independent, and the rank is equal to the number of rows, \( m \).
  • If \( A \) has full column rank, all the columns are linearly independent, and the rank equals the number of columns, \( n \).
Understanding matrix rank is essential when using the Moore-Penrose inverse, as it tells us which formula to use. If \( A \) has full row rank or full column rank, special simplified forms of the generalized inverse exist, as highlighted earlier.
Matrix Symmetry
Matrix symmetry, particularly with respect to the generalized inverse, implies that certain products of the matrix and its inverse result in symmetric matrices. For a matrix operation to be symmetric, the matrix must be equal to its transpose. Symmetric matrices hold special properties such as always having real eigenvalues and orthogonal eigenvectors, which simplify many computations in linear algebra.
  • For a generalized inverse \( A^{\#} \), we require \( A A^{\#} \) and \( A^{\#} A \) to be symmetric.
  • This means \((A A^{\#})^{T} = A A^{\#} \) and \((A^{\#} A)^{T} = A^{\#} A \).
The symmetry of these products is a key requirement for the Moore-Penrose inverse, ensuring that it adheres to these mathematical strictures and simplifying interpretations of the matrix's influence.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(\\{\mathbf{x}, \mathbf{y}, \mathbf{z}\\}\) be a linearly independent set in \(\mathbb{R}^{4}\). Show that \(\left\\{\mathbf{x}, \mathbf{y}, \mathbf{z}, \mathbf{e}_{k}\right\\}\) is a basis of \(\mathbb{R}^{4}\) for some \(\mathbf{e}_{k}\) in the standard basis \(\left\\{\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}, \mathbf{e}_{4}\right\\}\).

We often write vectors in \(\mathbb{R}^{n}\) as rows. Suppose that \(U=\operatorname{span}\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{k}\right\\}\) where each \(\mathbf{x}_{i}\) is in \(\mathbb{R}^{n}\). If \(A\) is an \(m \times n\) matrix and \(A \mathbf{x}_{i}=\mathbf{0}\) for each \(i\), show that \(A \mathbf{y}=\mathbf{0}\) for every vector \(\mathbf{y}\) in \(U\)

We often write vectors in \(\mathbb{R}^{n}\) as rows. Is \(\mathbb{R}^{2}\) a subspace of \(\mathbb{R}^{3}\) ? Defend your answer.

a. Show that \(\mathbf{x} \cdot \mathbf{y}=\frac{1}{4}\left[\|\mathbf{x}+\mathbf{y}\|^{2}-\|\mathbf{x}-\mathbf{y}\|^{2}\right]\) for all \(\mathbf{x}\), \(\mathbf{y}\) in \(\mathbb{R}^{n}\) b. Show that \(\|\mathbf{x}\|^{2}+\|\mathbf{y}\|^{2}=\frac{1}{2}\left[\|\mathbf{x}+\mathbf{y}\|^{2}+\|\mathbf{x}-\mathbf{y}\|^{2}\right]\) for all \(\mathbf{x}, \mathbf{y}\) in \(\mathbb{R}^{n}\)

In each case, decide whether the matrix \(A\) is diagonalizable. If so, find \(P\) such that \(P^{-1} A P\) is diagonal. $$ \text { a. }\left[\begin{array}{lll} 1 & 0 & 0 \\ 1 & 2 & 1 \\ 0 & 0 & 1 \end{array}\right] \quad \text { b. }\left[\begin{array}{rrr} 3 & 0 & 6 \\ 0 & -3 & 0 \\ 5 & 0 & 2 \end{array}\right] $$ c. \(\left[\begin{array}{rrr}3 & 1 & 6 \\ 2 & 1 & 0 \\ -1 & 0 & -3\end{array}\right]\) d. \(\left[\begin{array}{lll}4 & 0 & 0 \\ 0 & 2 & 2 \\ 2 & 3 & 1\end{array}\right]\)

See all solutions

Recommended explanations on Economics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free