Let \(A\) be an \(n \times n\) matrix.
Define what is meant by the eigenvalues and eigenvectors of \(A\).
Show that if \(v\) and \(w\) are eigenvectors of \(A\) corresponding to distinct eigenvalues \(\lambda\) and \(\mu\), then \(v\) and \(w\) are linearly independent.
Solution
A scalar \(\lambda\) is an eigenvalue of \(A\) if there exists a nonzero vector \(v\) such that \[ Av=\lambda v. \] Such a nonzero vector is an eigenvector corresponding to \(\lambda\).
Suppose \(Av=\lambda v\) and \(Aw=\mu w\), where \(v,w\ne0\) and \(\lambda\ne\mu\). If \[ av+bw=0, \] then applying \(A\) gives \[ a\lambda v+b\mu w=0. \] Subtracting \(\lambda\) times the original relation yields \[ b(\mu-\lambda)w=0. \] Since \(w\ne0\) and \(\mu-\lambda\ne0\), we get \(b=0\). Then \(av=0\), so \(a=0\). Thus \(v\) and \(w\) are linearly independent.