I posted many pictures for them same question to let you und

I posted many pictures for them same question to let you understand the question clearly.

Problem 2 Prove: . If A is Hermitian and full-rank, the basis of eigenvectors may be chosen to be mutually orthogonal. The eigenvalues are real. . The eigenvectors of A are the same as the cigenvectors of A. Suppose that A and B are n x n matrices that commute (that is, AB = BA) and suppose that distinct eigenvalues. (a) Show that if Bv Av then BAv = Av (b) Show that every eigenvector for B is also an eigenvector for A.

Solution

Solution : 3 )

a )   

Proof. This follows from the fact that AB = BA. Indeed,

BAv = ABv = A(v) = Av

since scalar multiplication commutes with matrix multiplication.

b )

Proof. Suppose v is an eigenvector of B with eigenvalue . By part (a), we have BAv = Av. So either Av = 0 or Av is also an eigenvector of B with eigenvalue . Since B has n distinct eigenvalues, they all have multiplicity 1 which means that all of the eigenspaces of B are one-dimensional. Since v and Av both lie in the one- dimensional eigenspace of B corresponding to the eigenvalue , v and Av must be linearly dependent. Since v 0, this means that Av = µv for some scalar µ. Therefore, v is an eigenvector of A corresponding to the eigenvalue µ.