Problem 1: a) A matrix A has the following spectrum d1 = 0, 12 = 51 13 = 2, da = 7. Is the matrix positive definite? Is it singular? Explain why or why not in each case. b) If the eigenvalues and eigenvectors of A are A; and u, respectively, what are the eigenvalues and eigenvectors of A", where n is a positive integer? c) Do the eigenvectors of any real symmetric n x n matrix A form a basis for n-space? Why or why not? d) Write the quadratic 207 + 13 - 213 + 2.1142 + 203 - 20103 = 4, in the form x?Ax = constant. e) Describe how to obtain the singular-value decomposition of an m x n matrix A. Be sure to indicate the sizes and types of any matrices, and clearly identify the singular values of A?Part (a): Is the matrix .A positive definite? Is it singular? We are given the eigenvalues of matrix A: Al = 0, 12 = 7, 13 = 2, 24= . Positive definite check: A matrix is positive definite if all of its eigenvalues are positive (i.e., greater than 0). Here, one of the eigenvalues is A1 = 0, which means the matrix is not positive definite. For a matrix to be positive definite, none of the eigenvalues should be zero or negative. Singular check: A matrix is singular if at least one of its eigenvalues is zero. Since 1 = 0, the matrix is singular. This means that the matrix is not invertible. Answer: . The matrix is not positive definite because one of its eigenvalues is zero. . The matrix is singular because one of its eigenvalues is zero. Part (b): Eigenvalues and eigenvectors of " We are asked to find the eigenvalues and eigenvectors of A", where n is a positive integer. Given: . Eigenvalues of A: A1 = 0, 12 = ;, 13 = 2, M = . Eigenvectors of A are u1, u2, u3, u4 . Eigenvalues of A": For any matrix A, if >; are the eigenvalues of A, the eigenvalues of A" are X, . Thus, the eigenvalues of 4" are: AT = 0" = 0, A7 = . Eigenvectors of A": The eigenvectors of A" are the same as the eigenvectors of . Therefore, the eigenvectors of A" are U1, u2, u3, u4. Answer. . The eigenvalues of A" are: . The eigenvectors of A" are the same as the eigenvectors of A.Part (c): Do the eigenvectors of any real symmetric n x n matrix form a basis for n-space? Yes, the eigenvectors of any real symmetric n x n matrix form a basis for n-space. Here's why: Symmetric matrices have real eigenvalues and their eigenvectors are orthogonal. . If there are n linearly independent eigenvectors for an n x n matrix, they span the entire n- dimensional space. . Therefore, these eigenvectors form a basis for the space. Answer: Yes, the eigenvectors of any real symmetric n x n matrix form a basis for n-space. This is because the eigenvectors are orthogonal and span the space. Part (d): Write the quadratic equation in the form XAX = constant We are given the quadratic equation: 2x + x - 2x3 + 2x122 + 2923 - 20123 = 4 We need to express it in the form x AX, where X = 1. Write down the terms: . 20102 -20103 2. Form the symmetric matrix A: The matrix . is constructed based on the coefficients of the quadratic terms: 3. Therefore, the quadratic form is: x Ax = (x1 22 23) (1 1 ) ()- Answer: The quadratic equation can be written as: x AX = 4 where A =Part (e): Singular value decomposition (SVD) of matrix .A To perform the singular value decomposition (SVD) of an m x n matrix A, we decompose . as: A = UEVI where: . U is an m x m orthogonal matrix (left singular vectors), . E is an m x n diagonal matrix containing the singular values of A, . V is an n x n orthogonal matrix (right singular vectors). . Sizes of matrices: If A is an m x n matrix, then: . U ism x m, . Lism x n, . V isn x n. . Singular values: The singular values of A are the square roots of the eigenvalues of ATA. Answer. To obtain the singular value decomposition of , follow the SVD process: Where the sizes of U. , and V/ depend on the dimensions of , and the singular values are the square roots of the eigenvalues of ATA