Nice Case of Diagonalization

Theorem

• Suppose $V$ is a n-dim $\mathbb{R}$ vector space
• If $T:V\to V$ has $n$ distinct $\mathbb{R}$ eigenvalues, then $T$ is Diagonalizable

Proof

1. Suppose $T:V\to V$ has $n$ distinct real eigenvalues
2. For each eigenvalue, ${\lambda }_i$, we can find a non-zero eigenvector $v_i$ s.t $T(v_i)={\lambda }_i{v}_i$ by definition of eigenvalue
3. Because eigenvalues are distinct by Eigenvectors of Distinct Eigenvalues are Independent thoerem, $\{v_1,\dots ,v_n\}$ is Linearly Independent
4. It follows that $\alpha =\{v_1,\dots ,v_n\}$ is a basis of $V$ because $V$ is $n$ dimensional and $|a|=n$.
5. Then, this basis

$$[T{]}_{\alpha }^{\alpha }=\left[\begin{array}{c}{\lambda }_1\dots 0\\ 0\dots {\lambda }_n\end{array}\right]$$

Therefore, $T$ is diagonizable