Caution: Not every linear transformation has an eigenvalues! ... Thus every non-zero vector in L⊥ is an eigenvector with eigenvalue 0. If we let v be a basis for L and w be a basis for L⊥, then ( v, w) is an eigenbasis for p. Caution: Not every linear transformation has an eigenbasis!

Does every linear operator have an eigenvalue?

Every linear operator on an n-dimensional vector space has n-distinct eigenvalues. If a real matrix has one eigenvector, then it has an infinite number of eigenvectors.

Do eigenvalues always have eigenvectors?

Since a nonzero subspace is infinite, every eigenvalue has infinitely many eigenvectors. (For example, multiplying an eigenvector by a nonzero scalar gives another eigenvector.) On the other hand, there can be at most n linearly independent eigenvectors of an n × n matrix, since R n has dimension n .