If A is nxn diagonalizable matrix, then each vector in can be written as a linear comination of eigenvectors of A.
This is true but how do I prove this?
Since A is diagonalizable, there exist a diagonal matrix D and an invertible matrix P such that ,
Suppose is an eigenvalue of A. That means that there exist a non-zero vector v such that . Let . Since P is invertible, u is non-zero also and . But now, .
That is, is also an eigenvalue of the diagonal matrix D and, since a diagonal matix has its eigenvalues on its diagonal, is one of the numbers on the diagonal of D. Further the eigenvectors of a diagonal matrix are simply the standard basis vectors <1, 0, 0...>, etc.
That means that set of all eigenvectors u, corresponding to the eigenvalues of D are independent and their number is equal to the dimension of the space.. Now, v= Pu and since P is invertible, it maps independent sets of vectors into independent sets of vectors. Since their number is equal to the dimension of the space, it follow that they are a basis for the space.