Chapter 9: Problem 2
Find the distinct singular values of \(A\) $$A=\left[\begin{array}{ll} 3 & 0 \\ 0 & 4 \end{array}\right]$$
Short Answer
Expert verified
The distinct singular values of \(A\) are 3 and 4.
Step by step solution
01
Find the Eigenvalues of A^TA
First, calculate the matrix product of the transpose of matrix \(A\) and \(A\) itself. Since \(A\) is a diagonal matrix, \(A^TA = A^2\). Thus, find \(A^2 = \left[\begin{array}{ll} 3^2 & 0 \ 0 & 4^2 \end{array}\right] = \left[\begin{array}{ll} 9 & 0 \ 0 & 16 \end{array}\right]\). The eigenvalues of this matrix are 9 and 16.
02
Calculate the Singular Values
Singular values of \(A\) are the square roots of the eigenvalues of \(A^TA\). Therefore, calculate the singular values as follows: \[ \sigma_1 = \sqrt{9} = 3, \quad \sigma_2 = \sqrt{16} = 4 \].
03
Determine the Distinct Singular Values
Since singular values cannot be negative, the distinct singular values obtained from the calculations are positive values only. Thus, list the distinct singular values: 3 and 4.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Eigenvalues
Eigenvalues are a fundamental concept in linear algebra. They can be thought of as special numbers associated with a square matrix. In a simple sense, if you have a matrix \(A\), an eigenvalue \(\lambda\) is found such that when \(A\) multiplies a vector \(\mathbf{v}\), the product is a scaled version of \(\mathbf{v}\) itself. This can be expressed through the equation:
\[ A\mathbf{v} = \lambda\mathbf{v} \]
Here, \(\mathbf{v}\) is not the zero vector, and \(\lambda\) represents the eigenvalue. Calculating eigenvalues usually involves solving the characteristic equation:
\[ A\mathbf{v} = \lambda\mathbf{v} \]
Here, \(\mathbf{v}\) is not the zero vector, and \(\lambda\) represents the eigenvalue. Calculating eigenvalues usually involves solving the characteristic equation:
- Formulated as \( \det(A - \lambda I) = 0 \)
- "\(\det\)" represents the determinant of the matrix.
- \(I\) is the identity matrix, which has ones on the diagonal and zeros elsewhere.
Diagonal Matrix
A diagonal matrix is one where all entries outside the main diagonal are zero. Consider matrix \(A\) as given in the problem:
\[A = \begin{bmatrix}3 & 0 \0 & 4\end{bmatrix}\]
This matrix is diagonal because non-diagonal entries are zero. Such matrices have straightforward properties and are easy to work with. Key properties include:
\[A = \begin{bmatrix}3 & 0 \0 & 4\end{bmatrix}\]
This matrix is diagonal because non-diagonal entries are zero. Such matrices have straightforward properties and are easy to work with. Key properties include:
- The eigenvalues of a diagonal matrix are simply the diagonal elements themselves.
- They simplify operations, such as matrix multiplication and finding determinants.
Matrix Transpose
The transpose of a matrix is found by "flipping" the matrix over its diagonal. Essentially, this means turning the rows of the original matrix into columns and vice versa. For matrix \(A\):
- The original matrix \(A = \begin{bmatrix} 3 & 0 \ 0 & 4 \end{bmatrix} \)
- The transpose, noted as \(A^T\), is simply \(A = A^T\) because \(A\) is diagonal.
- The eigenvalues of \(A\) and \(A^T\) are the same.
- The transpose of a diagonal matrix is also a diagonal matrix.
Eigenvalue Decomposition
Eigenvalue decomposition is a powerful tool in linear algebra where a matrix is decomposed into a set of eigenvectors and eigenvalues. When \(A\) is a square matrix, it can sometimes be expressed in the form:
\[ A = PDP^{-1} \]
In this equation:
\[ A = PDP^{-1} \]
In this equation:
- \(P\) is a matrix whose columns are the eigenvectors of \(A\).
- \(D\) is a diagonal matrix containing the eigenvalues of \(A\).
- \(P^{-1}\) is the inverse of \(P\).