Independence of eigenvectors with distinct eigenvalues
Eigenvectors belonging to distinct eigenvalues are linearly independent, and so are unions of bases of distinct eigenspaces.
Theorem 4.13
Suppose that an matrix has distinct eigenvalues . Then the corresponding eigenvectors are linearly independent.
Proof
Suppose were linearly dependent, so for scalars not all zero. Take the minimal for which there are with . Applying ,
a linear combination of eigenvectors with non-zero coefficients, contradicting the minimality of .
Bases of distinct eigenspaces
If is a basis of the eigenspace associated to and are distinct eigenvalues, then is linearly independent: eigenvectors taken from within a single eigenspace behave like one eigenvector under the minimality argument, since each contributes vectors with the same eigenvalue.
Related
Stated in
- Theorem 4.13§4.3.1 Linearly Independent Eigenvectors