If A is nxn diagonalizable matrix, then each vector in $\displaystyle \mathbb{R}^n$ 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 $\displaystyle A= PDP^{-1}$, $\displaystyle D= P^{-1}AP$
Suppose $\displaystyle \lambda$ is an eigenvalue of A. That means that there exist a non-zero vector v such that $\displaystyle Av= \lambda v$. Let $\displaystyle u= P^{-1}v$. Since P is invertible, u is non-zero also and $\displaystyle v= Pu$. But now, $\displaystyle Du= P^{-1}APu= P^{-1}Av= \lambda P^{-1}v= \lambda u$.
That is, $\displaystyle \lambda$ 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.