I have a problem with this result
a) Prove $T:V\rightarrow V$ is diagonalizable iff exists a basis $B$ of $V$ consisting unically of eigenvectors.
My work
a) $\implies$ Let $B=\{v_1,...,v_n\}$ a basis of $V$. As $T$ is diagonalizable then the matrix associated to $T$, $T_{BB}=A$ is a diagonal matrix whose each diagonal entry $A_{ii}=\lambda_i$ with $1 \leq i\leq n$
Moreover, as $T_{BB}=A$ then $T(v_i)=\lambda v_i$
In consequence $v_i$ are eigenvectors of $T$
I don't know how to prove the other implication. Can someone help me?
If $\;\{v_1,...,v_n\}\;$ is a basis of eigenvectors corresponding to eigenvalues $\;\lambda_1,...,\lambda_n\;$ (some care here: the $\;\lambda\,'s$ could not be all different!), then the matrix of $\;T\;$ corresponding to this basis is, as you can check at once
$$\begin{pmatrix}\lambda_1&0&\ldots&0\\0&\lambda_2&0\ldots&0\\\ldots&\ldots&\ldots&\ldots\\0&0&\ldots&\lambda_n\end{pmatrix}$$