12.1. EIGENVALUES AND EIGENVECTORS OF A MATRIX 259

And now subtracting this from 12.11 yields

0 =k

∑i=1

ci (λ v−λ wi)wi

and by independence of the {w1, · · · ,wk} , this requires ci (λ v−λ wi) = 0 for each i. Sincethe eigenvalues are distinct, λ v−λ wi ̸= 0 and so each ci = 0. But from 12.10, this requiresv = 0 which is impossible because v is an eigenvector and

Eigenvectors are never equal to zero! ■

Definition 12.1.17 An n× n matrix A is called nondefective if and only if there exists abasis of eigenvectors for Fn.

In fact the geometric multiplicity is never larger than the algebraic multiplicity. Let Adenote an n×n matrix in what follows and we assume there is an eigenvalue λ and F willdenote the field of scalars.

Theorem 12.1.18 Let λ be an eigenvalue for an n× n matrix. Then its geometric multi-plicity is never larger than its algebraic multiplicity.

Proof: Let {v1, · · · ,vr} be a basis for the eigenspace corresponding to some λ . Thenby Theorem 8.5.21, there is a longer list {v1, · · · ,vr,u1, · · · ,un−r} which is a basis for Fn.

Thus the matrix S≡(

v1 · · · vr u1 · · · un−r

)is invertible. Say its inverse is

S−1 =(

a1 · · · ar b1 · · · bn−r

)T

where aTi v j = δ i j. Then S−1AS must be(a1 · · · ar b1 · · · bn−r

)T (λv1 · · · λvr Au1 · · · Aun−r

)and so S−1AS is of the form (

D M0 N

)(*)

where D is an r× r diagonal matrix having λ down the main diagonal, M an r× (n− r) ,and N a (n− r)× (n− r). Now

det(S−1AS−µI

)= det

(S−1 (A−µI)S

)= det

(S−1)det(S)det(A−µI) = det(A−µI)

and so this matrix in ∗ has the same eigenvalues with the same multiplicities as the matrixA. So consider the characteristic polynomial with variable µ of the matrix in ∗. Expandingrepeatedly along the first column, one obtains the characteristic polynomial is of the form

q(µ) = (µ−λ )r det(µI−N) = 0

and so the multiplicity of λ is at least r. ■This yields easily the following corollary which ties this theorem to Theorem 12.1.16.