Let $A$ be an $n \times n$ matrix, and let $v_1,...,v_n$ be a basis of $R^n$ so that each $v_i$ is an eigenvector of $A$. Prove that $A$ diagonalizable.
Does the diagonalization of $A = QDQ^{-1}$ anything to do with this?
Any help would be appreciated, thanks.
The diagonalization theorem states that for an $n \times n$ matrix to be diagonalizable, the matrix must have $n$ linearly independent eigenvectors to form the matrix $Q$. In your case, there are $n$ distinct eigenvectors ($v_1,...,v_n$). Furthermore, the fact that they form a basis indicates that they are linearly independent (by the definition of a basis). Therefore, we have shown that the conditions given meet the requirement to be a diagonalizable matrix.