Linear Algebra with Applications (3rd Edition)

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. In other words, A has non-trivial nullspace, which means Ax = 0 0 0 = λ 3 x
where x is a non-zero vector and λ 3 must be zero. Therefore, the eigenvalues are λ 1 = 6, λ 2 = 3, λ 3 = 0. b) [ 10 marks] Find the corresponding eigenspaces. Sol. The first two results show that 1 2 1 and 1 - 1 1 are eigenvectors of A . We can find the third eigenvector that satisfies Ax = 0 via A 1 - 1 1 - A 2 - 1 0 = A - 1 0 1 = 0 that is - 1 0 1 is the third eigenvector. c) [ 5 marks] In each of the following questions, you must give a correct reason (based on the theory of eigenvalues and eigenvectors) to get full credit Is A a diagonalizable matrix? Is A an invertible matrix? Is A a projection matrix? Sol. A is diagonalizable as A has distinct eigenvalues (or because the eigenvectors are linearly independent). A is not invertible, as one of the eigenvalues is zero. A is not a projection matrix, as the eigenvalues of a projection matrix are either 1 or 0.

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