If a basis for matrix $A$ is $B = \{b_1, b_2, b_3, \ldots, b_n\}$ and $A$ has $n$ distinct eigenvalues $(λ_1, λ_2, λ_3, \ldots, λ_n)$, then $A$ is diagonalizable with $A = PDP^{-1}$, where $P$ has column vectors $[b_1 b_2 b_3 \ldots b_n]$ and $D$ is a diagonal matrix with the eigenvalues $(λ_1, λ_2, λ_3, \ldots, λ_n)$.
TRUE, SOMETIMES, or FALSE?
If you change the wording to "$B=\{b_1,\dots,b_n\}$ is a basis of $\mathbb{R}^n$ (or $\mathbb{C}^n$) consisting of eigenvectors of $A$, and $\lambda_i$ is the eigenvalue corresponding to $b_i$ for $i=1,\dots,n$", then the statement is always true. You don't even need that the eigenvalues are distinct.