= Solution
Let $v_1,\ldots,v_n$ be the <linearly independent> <eigenvectors>, and form $V=(v_1\ \cdots\ v_n)$. This <matrix> is invertible and $AV=VD$, where $D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$. Hence
$$
\boxed{V^{-1}AV=D,}
$$
which proves <diagonalizability>.
If $B=P^{-1}AP$, then $P^{-1}v_j$ are linearly independent <eigenvectors> of $B$ with the same <eigenvalues>. Explicitly, with $W=P^{-1}V$,
$$
\boxed{W^{-1}BW=V^{-1}P(P^{-1}AP)P^{-1}V=D.}
$$
Thus <matrix similarity> preserves diagonalizability.
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