Hey n22.
Are you familiar with the diagonalization theorem?
Diagonalizable matrix - Wikipedia, the free encyclopedia
Hi ,
I would really appreciate if someone could help me out here.thanks.
Prove that if an nxn matrix A has n distinct eigenvalues ,then the set of n eigenvectors formed by
choosing one non zero vector from each eigenspace of A is a basis of R^n
Hey n22.
Are you familiar with the diagonalization theorem?
Diagonalizable matrix - Wikipedia, the free encyclopedia
Not "any two vectors", one vector from each eigenspace. Do you know what an eigenspace is? You should have already proved that if v satisfies then so does any multiple of it: . From that it follows that the set of all eigenvectors, corresponding to a given eigenvalue (together with the 0 vector) is a subspace.
It is also easy to see that if two eigenvectors correspond to different eigenvalues, they are independent (one is not a multiple of the other). So if A has n different eigenvalues, I has n different eigenspaces.