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Magazine Chapter 9: Problem 3
Find a reduced singular value decomposition of \(A\). [Note: Fach matrix appears in Exercise Set \(9.5,\) where you were asked to find its (unreduced) singular value decomposition.] \(A=\left[\begin{array}{rl}1 & 0 \\ 1 & 1 \\ -1 & 1\end{array}\right]\)
Short Answer
Expert verified
The reduced SVD of \( A \) is: \( U\Sigma V^T \) with \( U = \begin{bmatrix} 1/\sqrt{3} & 0 \\ 1/\sqrt{3} & 1/\sqrt{2} \\ -1/\sqrt{3} & 1/\sqrt{2} \end{bmatrix} \), \( \Sigma = \begin{bmatrix} \sqrt{3} & 0 \\ 0 & \sqrt{2} \end{bmatrix} \) and \( V^T = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \).
Step by step solution
01
Compute A^T A
We start by computing the matrix \( A^T A \) where \( A^T \) is the transpose of \( A \). For matrix \( A \), the transpose \( A^T \) becomes \( \begin{bmatrix} 1 & 1 & -1 \ 0 & 1 & 1 \end{bmatrix} \). Now, compute \( A^T A \): \[A^T A = \begin{bmatrix} 1 & 1 & -1 \ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \ 1 & 1 \ -1 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 0 \ 0 & 2 \end{bmatrix}\]
02
Find Eigenvalues of A^T A
Next, we find the eigenvalues of \( A^T A \), which are the singular values squared. The matrix \( A^T A \) is:\[\begin{bmatrix} 3 & 0 \ 0 & 2 \end{bmatrix}\]The eigenvalues are the diagonal values: \( \lambda_1 = 3 \) and \( \lambda_2 = 2 \). Therefore, the singular values are \( \sigma_1 = \sqrt{3} \) and \( \sigma_2 = \sqrt{2} \).
03
Compute Right Singular Vectors (V)
Solve the eigenvector equation \( (A^T A - \lambda I) \mathbf{v} = 0 \) for each eigenvalue.- For \( \lambda_1 = 3 \): \[ (A^T A - 3I) = \begin{bmatrix} 0 & 0 \ 0 & -1 \end{bmatrix} \] Eigenvector \( \mathbf{v}_1 = \begin{bmatrix} 1 \ 0 \end{bmatrix} \).- For \( \lambda_2 = 2 \): \[ (A^T A - 2I) = \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix} \] Eigenvector \( \mathbf{v}_2 = \begin{bmatrix} 0 \ 1 \end{bmatrix} \).Thus, \( V = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \).
04
Compute Left Singular Vectors (U)
To find \( U \), compute \( A \mathbf{v}_i / \sigma_i \) for each right singular vector \( \mathbf{v}_i \).- For \( \mathbf{v}_1 = \begin{bmatrix} 1 \ 0 \end{bmatrix} \), \( A \mathbf{v}_1 = \begin{bmatrix} 1 \ 1 \ -1 \end{bmatrix} \) and divide by \( \sigma_1 = \sqrt{3} \): \[ \mathbf{u}_1 = \frac{1}{\sqrt{3}} \begin{bmatrix} 1 \ 1 \ -1 \end{bmatrix} = \begin{bmatrix} 1/\sqrt{3} \ 1/\sqrt{3} \ -1/\sqrt{3} \end{bmatrix} \]- For \( \mathbf{v}_2 = \begin{bmatrix} 0 \ 1 \end{bmatrix} \), \( A \mathbf{v}_2 = \begin{bmatrix} 0 \ 1 \ 1 \end{bmatrix} \) and divide by \( \sigma_2 = \sqrt{2} \): \[ \mathbf{u}_2 = \frac{1}{\sqrt{2}} \begin{bmatrix} 0 \ 1 \ 1 \end{bmatrix} = \begin{bmatrix} 0 \ 1/\sqrt{2} \ 1/\sqrt{2} \end{bmatrix} \]Thus, \( U = \begin{bmatrix} 1/\sqrt{3} & 0 \ 1/\sqrt{3} & 1/\sqrt{2} \ -1/\sqrt{3} & 1/\sqrt{2} \end{bmatrix} \).
05
Form the Diagonal Matrix 危
Matrix \( \Sigma \) contains the singular values on the diagonal:\[\Sigma = \begin{bmatrix} \sqrt{3} & 0 \ 0 & \sqrt{2} \end{bmatrix}\]
06
Assemble the Reduced SVD
The reduced singular value decomposition of \( A \) is given by \( A = U\Sigma V^T \). We have:\[U = \begin{bmatrix} 1/\sqrt{3} & 0 \ 1/\sqrt{3} & 1/\sqrt{2} \ -1/\sqrt{3} & 1/\sqrt{2} \end{bmatrix}, \quad \Sigma = \begin{bmatrix} \sqrt{3} & 0 \ 0 & \sqrt{2} \end{bmatrix}, \quad V^T = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\]Thus, the reduced singular value decomposition is \( A = U\Sigma V^T \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Matrix Transposition
Matrix transposition is a straightforward yet crucial operation in linear algebra. When we transpose a matrix, we essentially "flip" it over its diagonal. This means that each element at position
For example, if we take a matrix \[A = \begin{bmatrix}1 & 0 \1 & 1 \-1 & 1\end{bmatrix}\]
and transpose it, we get
\[A^T = \begin{bmatrix}1 & 1 & -1 \0 & 1 & 1\end{bmatrix}\]
This method is fundamental, as it's often used in matrix operations like calculating the criterion matrix \(A^T A\).
In the context of Singular Value Decomposition (SVD), transposing a matrix is a critical first step.
- \((i, j)\)
- \((j, i)\)
For example, if we take a matrix \[A = \begin{bmatrix}1 & 0 \1 & 1 \-1 & 1\end{bmatrix}\]
and transpose it, we get
\[A^T = \begin{bmatrix}1 & 1 & -1 \0 & 1 & 1\end{bmatrix}\]
This method is fundamental, as it's often used in matrix operations like calculating the criterion matrix \(A^T A\).
In the context of Singular Value Decomposition (SVD), transposing a matrix is a critical first step.
Eigenvalues and Eigenvectors
Eigenvalues and eigenvectors are key concepts in linear algebra that give insight into the properties of matrices. Given a matrix
In mathematical terms: if
This concept is vital for SVD because SVD involves decomposing a matrix into its eigenvectors and eigenvalues.
In the given exercise, the key step is finding the eigenvalues of
- \(A\)
In mathematical terms: if
- \(A\vec{v} = \lambda \vec{v}\)
- \(\vec{v}\)
- \(\lambda\)
This concept is vital for SVD because SVD involves decomposing a matrix into its eigenvectors and eigenvalues.
In the given exercise, the key step is finding the eigenvalues of
- \(A^T A\), which helps in determining the singular values \(\sigma\).
Matrix Diagonalization
Matrix diagonalization involves converting a matrix into a diagonal form using its eigenvalues and eigenvectors. This process simplifies many matrix operations, especially when raising matrices to higher powers.
A matrix
In the Singular Value Decomposition process, diagonalization simplifies the process of finding the diagonal
A matrix
- \(A\)
- \(D\)
- \(P\)
In the Singular Value Decomposition process, diagonalization simplifies the process of finding the diagonal
- matrix \(\Sigma\)
Diagonal Matrix
A diagonal matrix is a special type of matrix where all non-diagonal elements are zero. It takes the form \[D = \begin{bmatrix}\sigma_1 & 0 & \cdots & 0 \0 & \sigma_2 & \cdots & 0 \\vdots & \vdots & \ddots & \vdots \0 & 0 & \cdots & \sigma_n\end{bmatrix}\]Here, the elements
In the context of SVD, the diagonal matrix
- \(\sigma_1, \sigma_2, \ldots, \sigma_n\)
In the context of SVD, the diagonal matrix
- \(\Sigma\)
- It plays a crucial role as it simplifies matrix multiplications
- and is essential when reconstructing the original matrix.
Reduced SVD
Reduced Singular Value Decomposition (SVD) focuses on simplifying matrices by only considering the non-zero singular values. It efficiently captures the essential properties of the matrix without incorporating all of its dimensions.
For a matrix
This form of decomposition is not only more computationally efficient
but is also helpful in applications like data reduction and noise filtering.
For a matrix
- \(A\)
- \(m \times n\), the reduced SVD simplifies it into \[A = U \Sigma V^T\]
- \(U\) and \(V\)
- \(\Sigma\)
This form of decomposition is not only more computationally efficient
but is also helpful in applications like data reduction and noise filtering.
