Let A be an n times n matrix and lambda be an eigenvalue of

Let A be an n times n matrix and lambda be an eigenvalue of A. Show that E_lambda, eigenspace of A associated with lambda, is a subspace of R^n.

Solution

Let A be a nxn matrix and let be an eigenvalue of A. Then E = {v Rⁿ : Av = v} where v is a n-vector (an eigenvector of A associated with the eigenvalue ). Let u and v be two arbitrary vectors in E and let p be an arbitrary scalar. Then, Au = u and Av = v. Hence A(u+v) = Au + Av = u+ v = (u+v). Thus E is closed under vector addition. Further A(pu) = p(Au)=p(u) = (pu). Thus E is closed under scalar multiplication. Hence E is a vector space, and, therefore, a subspace of Rⁿ.