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A is a 9x9 matrix with 3 distinct eigenvalues. Two of the eigenvalues have geometric multiplicity 3; and one eigenvalues has geometric multiplicity 2. Is A diagonalizable? Justify your answer?
I know the A has to be diagonalizable because of the geometric multiplicity. I don't know how to solve this one?
In your next to last sentence, you say you know how to solve it and in your last sentence say you don't? There are two eigenvalues with geometric multiplicity 3, one with geometric multiplicity 2, and one with geometric multiplicity 1. That adds up to a total "geometric multiplicity" of 9. The "algebraic multplicity" of all eigenvectors is also 9. That is enough to say that the matrix is diagonalizable.
In more general terms, you have two eigenvalues each with three independent eigenvectors, one with two independent eigenvectors, and a fifth with one independent eigenvector. That is a total of 9 independent vectors. Since a 9 by 9 matrix acts on a 9 dimensional vector space so those nine eigenvectors form a basis. Writing the matrix in terms of those basis vectors gives a diagonal matrix.