Set C = ones(6) + 7 * eye(6) and [X, D] = eig(C) (a) Even though = 7 is an eigenvalue of multiplicity 5,

Set
C = ones(6) + 7 * eye(6) and [X, D] = eig(C)
(a) Even though λ = 7 is an eigenvalue of multiplicity 5, the matrix C cannot be defective. Why? Explain. Check that C is not defective by computing the rank of X. Compute also XTX. What type of matrix is X? Explain. Compute also the rank of C - 7I. What can you conclude about the dimension of the eigenspace corresponding to λ = 7? Explain.
(b) The matrix C should be symmetric positive definite. Why? Explain. Thus C should have a Cholesky factorization LLT. The MATLAB command R = chol(C) will generate an upper triangular matrix R that is equal to LT. Compute R in this manner and set L = R′. Use MATLAB to verify that
C = LLT = RTR
(c) Alternatively one can determine the Cholesky factors from the LU factorization of C. Set
[L U] = lu(C)
and
D = diag(sqrt(diag(U))) and W = (L * D)′
How do R and W compare? This method of computing the Cholesky factorization is less efficient than the method MATLAB uses for its Chol function.