Suppose that all eigenvalues of an 8 times 8 matrix are real, and that among tho

Question

Suppose that all eigenvalues of an 8 times 8 matrix are real, and that among those are lambda = lambda_1 with algebraic multiplicity 1 and lambda = lambda_2 with algebraic multiplicity 3. Further suppose that the eigenvector associated with lambda_1 is v_1 and the eigenvectors associated with lambda_2 are w_1, w_2, and that there are no other linearly independent eigenvectors associated with either lambda_1 or lambda_2. Finally, suppose all other eigenvalues of the matrix have algebraic multiplicity 1. In this case, fill in the following values: The geometric multiplicity of lambda_1 = dim(ker(A - lambda_2)) = Is A diagonalizable?

Explanation / Answer

1.We know that the geometric multiplicity of an eigenvalue is the number of linearly independent eigenvectors associated with it.Apparently, then, the geometric multiplicity of the eigenvalue 1 is 1.

2. dim (Ker (A- 2I8)) = 2 as Ker (A- 2I8) is the set of solutions to the equation (A- 2I8)X = 0, which in turn is the eigespace of A ( set of eigenvectors) corresponding to the eigenvalue 2.

3. Since 2 has an algebraic multiplicity 3 and there are only 2 eigenvectors w1 and w2 associated with it, it means that the 3rd eigenvector associated with it is either w1 or w2. Thus Adoes not have 8 linearly independent distinct vectors. Hence A is not diagonalizable

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