Consider the matrix A below. First, show that A is diagonalizable by computing the geometric multiplicities of the eigenvalues and quoting the relevant theorem

Using a calculator, we find that A has three distinct eigenvalues, λ = 3, 2, −1, with λ = 2 having algebraic multiplicity two   , α_A (2) = 2.  The eigenvalues λ =3,-1 have algebraic multiplicity one, and so by Theorem ME we can conclude that their geometric multiplicities are one as well. Together with the computation of the geometric multiplicity of λ = 2 from Exercise EE.C20, we know

γ_A (3) = α_A (3) = 1                         γ_A (2) = α_A (2) = 2                         γ_A (−1) = α_A (−1) = 1

This satisfies the hypotheses of Theorem DMFE, and so we can conclude that A is diagonalizable.

A calculator will give us four eigenvectors of A, the two for λ = 2 being linearly independent presum- ably. Or, by hand, we could find basis vectors for the three eigenspaces. For λ = 3, 1 the eigenspaces have dimension one, and so any eigenvector for these eigenvalues will be multiples of the ones we use below. For λ = 2 there are many different bases for the eigenspace, so your answer could vary. Our eigenvectors are the basis vectors we would have obtained if we had actually constructed a basis in Exercise EE.C20 rather than just computing the dimension.

By the construction in the proof of Theorem DC, the required matrix S has columns that are four lin- early independent eigenvectors of A and the diagonal matrix has the eigenvalues on the diagonal (in the same order as the eigenvectors in S). Here are the pieces, “doing” the diagonalization,