Let $\mathbb{K}$ be a field, $1\leq n\in \mathbb{N}$ and let $V$ be a $\mathbb{K}$-vector space with $\dim_{\mathbb{R}}V=n$.
Let $\phi :V\rightarrow V$ be a linear map.
I want to show that $\phi$ is diagonalizable if and only if there is a basis $(b_1, \ldots , b_n)$ of $V$ and $\lambda_1, \ldots, \lambda_n\in \mathbb{K}$ such that $\phi (b_i)=\lambda_i b_i$.
Could you give me a hint how we could show this equivalence?
We say that a linear operator $T:V\to V$ is diagonalizable if and only if there exits an ordered basis $\mathcal{B} = \{b_{1},b_{2},\ldots,b_{n}\}$ of $V$ such that $[T]_{\mathcal{B}}$ is a diagonal matrix.
Based on such definition, let us prove the implication $(\Rightarrow)$ first.
If $T$ is diagonalizable, then there is a basis $\mathcal{B} = \{b_{1},b_{2},\ldots,b_{n}\}$ such that \begin{align*} [T]_{\mathcal{B}} = \begin{bmatrix} \lambda_{1} & 0 & \cdots & 0\\ 0 & \lambda_{2} & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & \lambda_{n} \end{bmatrix} \end{align*}
But we also know that $[T]_{\mathcal{B}} = [[T(b_{1})]_{\mathcal{B}},[T(b_{2})]_{\mathcal{B}},\ldots,[T(b_{n})]_{\mathcal{B}}]$.
Consequently, $T(b_{j}) = \lambda_{j}b_{j}$ and we are done with the first part.
Let us prove the converse implication $(\Leftarrow)$ now.
Let us suppose there exists a basis $\mathcal{B} = \{b_{1},b_{2},\ldots,b_{n}\}$ and scalars $\lambda_{j}\in\textbf{F}$ s.t. $T(b_{j}) = \lambda_{j}b_{j}$.
Thus it results that $$[T]_{\mathcal{B}} = [[T(b_{1})]_{\mathcal{B}},[T(b_{2})]_{\mathcal{B}},\ldots,[T(b_{n})]_{\mathcal{B}}] = [[\lambda_{1} b_{1}]_{\mathcal{B}},[\lambda_{2}b_{2}]_{\mathcal{B}},\ldots,[\lambda_{n}b_{n}]_{\mathcal{B}}] = \text{diag}(\lambda_{1},\lambda_{2},\ldots,\lambda_{n})$$
whence we conclude that $T$ is diagonalizable, and we are done.
Hopefully this helps.