Let $\displaystyle T$ be a linear operator on the n-dimensional vector space $\displaystyle V$, and suppose that $\displaystyle T$ has $\displaystyle n$ distinct characteristic values. Prove that $\displaystyle T$ is diagonalizable
Let $\displaystyle T$ be a linear operator on the n-dimensional vector space $\displaystyle V$, and suppose that $\displaystyle T$ has $\displaystyle n$ distinct characteristic values. Prove that $\displaystyle T$ is diagonalizable
T is diagonalisable if V has a basis consisting of eigenvectors of T. But eigenvectors associated with distinct eigenvalues are linearly independent. If all n eigenvalues are distinct then there must be n linearly independent eigenvectors, and they form a basis.