91Ó°ÊÓ

In the context of Singular Value Decomposition (SVD), eigenvalues play a crucial role in determining the singular values of a matrix. When we decompose a matrix, we begin by finding the eigenvalues of the matrix product \( A^T A \). For our example matrix \( A = \begin{bmatrix} -2 & 0 \ 0 & 0 \end{bmatrix} \), this product translates to \( A^T A = \begin{bmatrix} 4 & 0 \ 0 & 0 \end{bmatrix} \). This step leads us to the eigenvalues \( \lambda = 4 \) and \( \lambda = 0 \).

• Finding eigenvalues involves solving the characteristic equation \( \text{det}(A^T A - \lambda I) = 0 \).
• The solutions to this equation provide the eigenvalues, which are later used to find singular values.

These eigenvalues are essential, as they lay the groundwork for constructing other matrices in the SVD, such as \( \Sigma \). It is important to note that the eigenvalues also determine the strength and nature of stretching or compressing applied by the matrix to vectors in its space.

Matrix \( V \) in Singular Value Decomposition is assembled from the eigenvectors of the matrix \( A^T A \). Each eigenvector corresponds to an eigenvalue, providing directionality to the transformations imposed by \( A \). For the given exercise, \( V \) is constructed as follows:

• For \( \lambda = 4 \) with eigenvector \( \mathbf{v}_1 = \begin{bmatrix} 1 \ 0 \end{bmatrix} \).
• For \( \lambda = 0 \) with eigenvector \( \mathbf{v}_2 = \begin{bmatrix} 0 \ 1 \end{bmatrix} \).

Thus, the matrix \( V \) becomes \( \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \), essentially representing a standard basis in the vector space.

Matrix \( V \) is orthogonal, meaning \( V^T V = I \), where \( I \) is the identity matrix. This orthogonality property is crucial because it simplifies computations involving \( V \), particularly in the reconstruction of the original matrix using its SVD components.

Matrix \( \Sigma \) in Singular Value Decomposition is a diagonal matrix comprising the singular values of the original matrix \( A \). These singular values are the square roots of the non-negative eigenvalues obtained from \( A^T A \). In the current exercise, since the eigenvalues are \( \lambda = 4 \) and \( \lambda = 0 \), we compute the singular values as:

• \( \sigma_1 = \sqrt{4} = 2 \)
• \( \sigma_2 = \sqrt{0} = 0 \)

Matrix \( \Sigma \) is then represented as \( \begin{bmatrix} 2 & 0 \ 0 & 0 \end{bmatrix} \).

This configuration places the singular values along its diagonal, revealing the scaling factors that \( A \) applies in the respective dimensions. The presence of zero singular values indicates a loss of dimension, hinting at a rank-deficient matrix in specific directions. Understanding \( \Sigma \) helps in illustrating the geometric interpretation of \( A \) as it stretches or compresses space.

Matrix \( U \) in the SVD captures the orthonormal basis for the output space of \( A \). It is determined using the equation \( A \mathbf{v}_i = \sigma_i \mathbf{u}_i \), where \( \mathbf{v}_i \) are the eigenvectors from matrix \( V \) and \( \sigma_i \) are the singular values.

For the exercise, computing \( U \) proceeds as:

• With \( \sigma_1 = 2 \), use \( A \mathbf{v}_1 = 2 \mathbf{u}_1 \), resulting in \( \mathbf{u}_1 = \begin{bmatrix} -1 \ 0 \end{bmatrix} \).
• With \( \sigma_2 = 0 \), choose \( \mathbf{u}_2 \) orthogonal to \( \mathbf{u}_1 \), giving \( \mathbf{u}_2 = \begin{bmatrix} 0 \ 1 \end{bmatrix} \).

Thus, \( U \) becomes \( \begin{bmatrix} -1 & 0 \ 0 & 1 \end{bmatrix} \). Matrix \( U \) is also orthogonal, ensuring that its columns form an orthonormal set and maintaining the orthogonality condition \( U^T U = I \). This property aids in simplifying manipulations and interpretations since transformations carried by \( U \) preserve vector norms.