Suppose that all eigenvalues of an 9 times 9 matrix are real, and that among tho

Question

Suppose that all eigenvalues of an 9 times 9 matrix are real, and that among those 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 vector v_1 and the eigenvectors associated with lambda_2 are vector omega_1, vector omega_2, vector omega_3, and that there are no other linearly independent eigenvectors associated with either lambda_1 or lambda_2. Finally, suppose all other eigenvectors 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

We know that the eigenvectors of a matrix A, associated with an eigenvalue , are the solutions to the equation Av = v or, (A – I)v = 0. Also, of these solutions, the set of linearly independent vectors, forms the basis for Ker (A-I).

The geometric multiplicity of 1 is the dimension of Ker (A-1I). Since there is only one linearly independent eigenvector associated with 1, there is only 1 non-zero vector in the basis for Ker (A-1I). Hence the geometric multiplicity of 1 is 1.

Since there are 3 linearly independent eigenvectors associated with 2, there 3 non-zero vectors in the basis for Ker (A-2I). Hence dim( Ker (A- 2I)) = 3.

Since the eigenvalue 2 has an algebraic multiplicity 3, the matrix A does not have 9 distinct eigenvalues. Hence A is not diagonalizable.

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