Prove that an eigenspace associated to an eigenvector lambda

Prove that an eigenspace associated to an eigenvector lambda of a matrix A is a subspace

Solution

Suppose is an eigenvalue of A. Then E = {v R n such that Av = v} is called the eigenspace of A corresponding to the eigenvalue . We know that 0 E as A(0) = (0).

Suppose that x , y E _. Then x and y are two eigenvectors of A corresponding to the eigenvalue . Then A ( x + y) = Ax + Ay = x + y = )x + y). Then, either x + y = 0 or x + y is an eigenvector of A. In either case (x + y ) E _.Thus E _.is closed under addition. Suppose that c C and that x E _, i.e. x is an eigenvector of A. Then, A (cx) = c (Ax) = cx = (cx). Thus, either cx = 0 or, cx is an eigenvector of A. In either case, cx E and thus E is closed under scalar multiplication. Thus E is a subspace.