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 all linear operators have an eigenvalue?

On a finite-dimensional vector space V over the complex numbers, it should be obvious that any linear operator must have eigenvalues, although some or all of those eigenvalues might be zero.

Do all linear functions have eigenvectors?

Every real matrix has an eigenvalue, but it may be complex. In fact, a field K is algebraically closed iff every matrix with entries in K has an eigenvalue. You can use the companion matrix to prove one direction. ... Thus a matrix has eigenvectors if and only if the characteristic polynomial has at least one root.