Page 101 - Linear Algebra Done Right

Chapter 5. Eigenvalues and Eigenvectors
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this follows immediately from the definition of the matrix of an opera-
tor with respect to a basis. Thus an operator T ∈L(V) has a diagonal
matrix with respect to some basis of V if and only if V has a basis
consisting of eigenvectors of T.
If an operator has a diagonal matrix with respect to some basis,
then the entries along the diagonal are precisely the eigenvalues of the
operator; this follows from 5.18 (or you may want to find an easier
proof that works only for diagonal matrices).
Unfortunately not every operator has a diagonal matrix with respect
to some basis. This sad state of affairs can arise even on complex vector
2
spaces. For example, consider T ∈L(C ) defined by
5.19 T(w, z) = (z, 0).

As you should verify, 0 is the only eigenvalue of this operator and
the corresponding set of eigenvectors is the one-dimensional subspace
2
{(w, 0) ∈ C : w ∈ C}. Thus there are not enough linearly independent
2
eigenvectors of T to form a basis of the two-dimensional space C .
Hence T does not have a diagonal matrix with respect to any basis
2
of C .
The next proposition shows that if an operator has as many distinct
eigenvalues as the dimension of its domain, then the operator has a di-
agonal matrix with respect to some operator. However, some operators
with fewer eigenvalues also have diagonal matrices (in other words, the
converse of the next proposition is not true). For example, the operator
3
T defined on the three-dimensional space F by
T(z 1 ,z 2 ,z 3 ) = (4z 1 , 4z 2 , 5z 3 )

has only two eigenvalues (4 and 5), but this operator has a diagonal
matrix with respect to the standard basis.

Later we will find other 5.20 Proposition: If T ∈L(V) has dim V distinct eigenvalues, then
conditions that imply T has a diagonal matrix with respect to some basis of V.
that certain operators
have a diagonal matrix Proof: Suppose that T ∈L(V) has dim V distinct eigenvalues
with respect to some λ 1 ,...,λ dim V . For each j, let v j ∈ V be a nonzero eigenvector cor-
basis (see 7.9 and 7.13). responding to the eigenvalue λ j . Because nonzero eigenvectors cor-
responding to distinct eigenvalues are linearly independent (see 5.6),
(v 1 ,...,v dim V ) is linearly independent. A linearly independent list of