136 CHAPTER 6. DIRECT SUMS AND BLOCK DIAGONAL MATRICES

Theorem 6.4.3 An n×n matrix is diagonalizable if and only if Fn has a basis of eigenvec-tors of A. Furthermore, you can take the matrix S described above, to be given as

S =(

s1 s2 · · · sn

)where here the sk are the eigenvectors in the basis for Fn. If A is diagonalizable, theeigenvalues of A are the diagonal entries of the diagonal matrix.

Proof: To say that A is diagonalizable, is to say that for some S,

S−1AS =

λ 1

. . .

λ n

the λ i being elements of F. This is to say that for S =

(s1 · · · sn

), sk being the kth

column,

A(

s1 · · · sn

)=(

s1 · · · sn

)λ 1

. . .

λ n

which is equivalent, from the way we multiply matrices and the above observation, that(

As1 · · · Asn

)=(

λ 1s1 · · · λ nsn

)which is equivalent to saying that the columns of S are eigenvectors and the diagonal matrixhas the eigenvectors down the main diagonal. Since S−1 is invertible, these eigenvectorsare a basis. Similarly, if there is a basis of eigenvectors, one can take them as the columnsof S and reverse the above steps, finally concluding that A is diagonalizable. ■

Corollary 6.4.4 Let A be an n×n matrix with minimum polynomial

p(λ ) =p

∏i=1

(λ −µ i)ki , the µ i being distinct.

Then A is diagonalizable if and only if each ki = 1.

Proof: Suppose first that it A is diagonalizable with a basis of eigenvectors {v1, · · · ,vn}with Avi = µ ivi. Since n≥ p, there may be some repeats here, a µ i going with more than

one vi. Say ki > 1. Now consider p̂(λ ) ≡ ∏pj=1, j ̸=i

(λ −µ j

)k j(λ −µ i) . Thus this is a

monic polynomial which has smaller degree than p(λ ) . If you have v ∈ Fn, since this is abasis, there are scalars ci such that v = ∑ j c jv j. Then p̂(A)v = 0. Since v is arbitrary, thisshows that p̂(A) = 0 contrary to the definition of the minimum polynomial being p(λ ).Thus each ki must be 1.

Conversely, if each ki = 1, then

Fn = ker(A−µ1I)⊕·· ·⊕ker(A−µ pI

)and you simply let β i be a basis for ker(A−µ iI) which consists entirely of eigenvectorsby definition of what you mean by ker(A−µ iI) . Then a basis of eigenvectors consists of{

β 1,β 2, · · · ,β p}

and so the matrix A is diagonalizable. ■