Question

Use the definition of the pseudo-inverse of a matrix \(A\) in terms of its singular values and singular vectors, as given in the discussion on solving linear least squares problems via the SVD, to show that the following relations hold: (a) \(A A^{\dagger} A=A\). (b) \(A^{\dagger} A A^{\dagger}=A^{\dagger}\). (c) \(\left(A A^{\dagger}\right)^{T}=A A^{\dagger}\). (d) \(\left(A^{\dagger} A\right)^{T}=A^{\dagger} A\).

Short answer

Using the definition of the pseudo-inverse and properties of the Singular Value Decomposition, we proved that for a matrix \(A\) and its pseudo-inverse \(A^{\dagger}\), the following relations hold: (a) \(AA^{\dagger}A=A\) (b) \(A^{\dagger}AA^{\dagger}=A^{\dagger}\) (c) \(\left(AA^{\dagger}\right)^{T}=AA^{\dagger}\) (d) \(\left(A^{\dagger}A\right)^{T}=A^{\dagger}A\) These properties are important because they are used to define the properties of a pseudo-inverse matrix and help in solving linear least squares problems using the SVD.

Step-by-step solution

Recall the SVD and Pseudo-Inverse

The SVD of matrix \(A\) is given by \(A=U\Sigma V^T\), where \(U\) and \(V\) are orthogonal matrices and \(\Sigma\) is a diagonal matrix with singular values of \(A\). The pseudo-inverse of a matrix \(A\), denoted by \(A^{\dagger}\), is given by \(A^{\dagger}=V\Sigma^{\dagger}U^T\). Here, \(\Sigma^{\dagger}\) is formed by taking the reciprocal of the non-zero singular values of \(A\).

Prove (a)

To prove (a), we calculate \(AA^{\dagger}A\): \(AA^{\dagger}A=(U\Sigma V^T)(V\Sigma^{\dagger}U^T)(U\Sigma V^T)\) Since \(V\) is an orthogonal matrix, \(V^TV=I\) (Identity matrix). Thus, the equation becomes: \((U\Sigma \cancel{V^T})(\cancel{V}\Sigma^{\dagger}U^T)(U\Sigma V^T) = (U\Sigma\Sigma^{\dagger}U^T)(U\Sigma V^T)\) Now, since \(U\) is also orthogonal, \(U^TU=I\). The equation becomes: \((U\Sigma\Sigma^{\dagger}\cancel{U^T})(\cancel{U}\Sigma V^T) = U\Sigma\Sigma^{\dagger}\Sigma V^T\) Notice that \((\Sigma\Sigma^{\dagger}\Sigma = \Sigma)\) since non-zero singular values get multiplied and then divided by themselves when we take the product of \(\Sigma\) and \(\Sigma^{\dagger}\). Thus, \(AA^{\dagger}A = U\Sigma V^T = A\)

Prove (b)

To prove (b), we calculate \(A^{\dagger}AA^{\dagger}\): \(A^{\dagger}AA^{\dagger}=(V\Sigma^{\dagger}U^T)(U\Sigma V^T)(V\Sigma^{\dagger}U^T)\) We use the properties of orthogonal matrices again, finding: \((V\Sigma^{\dagger}\cancel{U^T})(\cancel{U}\Sigma V^T)(V\Sigma^{\dagger}U^T)=(V\Sigma^{\dagger}\Sigma V^T)(V\Sigma^{\dagger}U^T)\) Now, \(V^TV=I\): \((V\Sigma^{\dagger}\Sigma\cancel{V^T})(\cancel{V}\Sigma^{\dagger}U^T)=V\Sigma^{\dagger}\Sigma\Sigma^{\dagger}U^T\) When we take the product \(\Sigma^{\dagger}\Sigma\), all non-zero singular values cancel each other, and we are left with: \(A^{\dagger}AA^{\dagger} = V\Sigma^{\dagger}U^T = A^{\dagger}\)

Prove (c)

To prove (c), we take the transpose of \(AA^{\dagger}\): \(\left(AA^{\dagger}\right)^{T} = \left(U\Sigma V^T V\Sigma^{\dagger}U^T\right)^T\) Transpose reverses the order of multiplication: \(\left(U\Sigma V^T V\Sigma^{\dagger}U^T\right)^T = (U^T)^T(\Sigma^{\dagger})^T (V^T)^T V^T \Sigma^T (U^T)^T\) As \(U\) and \(V\) are orthogonal matrices, \(U^{T}U=I\) and \(V^TV=I\). Also, note that \((\Sigma^{\dagger})^T=\Sigma^{\dagger}\) and \(\Sigma^T=\Sigma\): \(\left(AA^{\dagger}\right)^{T} = U \Sigma V^T V \Sigma^{\dagger} U^T = AA^{\dagger}\)

Prove (d)

To prove (d), we take the transpose of \(A^{\dagger}A\): \(\left(A^{\dagger}A\right)^{T} = \left(V\Sigma^{\dagger}U^T U\Sigma V^T\right)^T\) Transpose reverses the order of multiplication: \(\left(A^{\dagger}A\right)^{T} = (V^T)^T \Sigma^T (U^T)^T (U^T)^T (\Sigma^{\dagger})^T V^T\) Using the properties of orthogonal matrices and noting that \((\Sigma^{\dagger})^T=\Sigma^{\dagger}\) and \(\Sigma^T=\Sigma\): \(\left(A^{\dagger}A\right)^{T} = V\Sigma^{\dagger} U^T U\Sigma V^T = A^{\dagger}A\)

Key concepts

Understanding Singular Value Decomposition (SVD)

If you've ever worked with matrices, understanding the Singular Value Decomposition (SVD) is crucial. It's a method of decomposing a matrix into three other matrices that provides a lot of insight into the structure of the original matrix.

SVD states that a matrix \( A \) can be broken down as follows:
\[ A = U\Sigma V^T \]
The matrices \( U \) and \( V \) are orthogonal, meaning their columns and rows are perpendicular to each other in multi-dimensional space, and they have a special property: multiplying them by their transposes yields an identity matrix, i.e., \( U^TU = V^TV = I \) where \( I \) is the identity matrix. The middle matrix, \( \Sigma \) (Sigma), is a diagonal matrix with singular values on its diagonal; these values are the square roots of the eigenvalues of \( A^TA \) and they provide important information regarding the 'influence' of different dimensions when the matrix \( A \) operates on a vector.

What makes SVD particularly powerful is that it's always possible for any matrix, even if \( A \) is not square. This decomposition is widely used in applications ranging from signal processing to statistics. For instance, SVD is often used to solve linear least squares problems, which are about finding the best-fit solution in an over-determined system of linear equations.

Linear Least Squares Problems and the Pseudo-Inverse

Linear least squares problems are one of the central themes in data fitting and numerical analysis. When dealing with an overdetermined system - where you have more equations than unknowns – it’s unlikely you’ll find a solution that perfectly satisfies all equations. Instead, you aim to find the 'best possible' solution that minimizes the errors or the squares of the residuals.

To solve these problems efficiently, we often turn to the pseudo-inverse of a matrix \( A \) denoted by \( A^\dagger \). The pseudo-inverse helps you find a solution that minimizes the sum of the squared differences between the observed values and those predicted by the model.

According to SVD, \( A^\dagger \) is defined as \( A^\dagger = V\Sigma^\dagger U^T \), where \( \Sigma^\dagger \) is the diagonal matrix obtained by taking the reciprocals of the non-zero singular values of \( A \) and leaving the other entries as zeros. This approach ensures that when there's no exact solution due to inconsistencies or noise in the data, you can still obtain a solution that optimizes the fit.

Properties of Orthogonal Matrices in Matrix Calculations

Orthogonal matrices, which pop out naturally in SVD, hold properties that are quite beneficial in matrix calculations. An orthogonal matrix is a square matrix \( Q \) such that its transpose is also its inverse: \( Q^TQ = QQ^T = I \) where \( I \) is the identity matrix. This property indicates the matrix preserves the length of vectors upon multiplication and implies that the matrix columns (and rows) form an orthonormal basis.

Here's why these properties are useful:

• Invariance under Transpose: The transpose of an orthogonal matrix is also orthogonal.
• Preservation of Vector Norms: Orthogonal matrices maintain the length (norm) and angle (dot product) of vectors when they are multiplied, making them particularly suitable for geometric transformations in computer graphics and signal processing.
• Stability in Numerical Calculations: Because of their special properties, orthogonal matrices do not amplify errors in numerical computations, which is crucial in sensitive calculations like those in physics simulations or optimization problems.

In relation to SVD, the orthogonality of \( U \) and \( V \) facilitates simplification in matrix operations, as we can easily 'cancel out' these matrices when they are multiplied with their respective transposes. This is precisely what happens in the SVD-based solutions to the linear least squares problems.