Proposition (Enough Diagonal Elements Imply Diagonalizability)

Let $V$ be a finite dimensional vector space with $dim(V)=n$, and $T\in\mathscr{L}(V)$. If $T$ has $n$ distinct eigenvalues, then there is a basis such that $\mathcal{M}(T)$ is Diagonal.

[!prp] Equivalent conditions for diagonalizability Let $V$ be a Finite Dimensional Vector Space, where $dim(V)=n$, and $T\in\mathscr{L}(V)$. If $\lambda_1,\cdots,\lambda_m$ are all distinct eigenvalues of $T$, then the following statements are equivalent.

1. $T$ is
2. There is a basis for $V$ consisting of eigenvectors of $T$
3. There exists 1-dimensional subspaces $W_1,\cdots,W_n$ of $V$, each of which are invariant under $T$, such that $V=W_1\oplus\cdots\oplus W_n$.
4. $V$ is a direct sum of eigenspaces. That is, $$V=\mbox{Ker}(T-\lambda_{1}I_{V})\oplus\cdots\oplus\mbox{Ker}(T-\lambda_{m}I_{V})$$
5. $$\mbox{dim}(V)=\mbox{dim}(\mbox{Ker}(T-\lambda_{1}I_{V}))\oplus\cdots\oplus\mbox{dim}(\mbox{Ker}(T-\lambda_{m}I_{V}))$$