Two matrices $A$ and $B$ are similar if there exists a nonsingular (invertible) matrix $S$ such that
\[S^{-1}BS=A.\]

A matrix $A$ is diagonalizable if $A$ is similar to a diagonal matrix. Namely, $A$ is diagonalizable if there exist a nonsingular matrix $S$ and a diagonal matrix $D$ such that
\[S^{-1}AS=D.\]

Some useful facts are

If $S$ and $T$ are invertible matrices, then we have
\[(TS)^{-1}=S^{-1}T^{-1}.\]
(Note the order of the product.)

A matrix is nonsingular if and only if its determinant is nonzero.

Proof.

Since the matrix $A$ is diagonalizable, there exist a nonsingular matrix $S$ and a diagonal matrix $D$ such that
\[S^{-1}AS=D. \tag{*}\]
Also, since $B$ is similar to $A$, there exist a nonsingular matrix $T$ such that
\[T^{-1}BT=A. \tag{**}\]

Inserting the expression of $A$ from (**) into the equality (*), we obtain
\begin{align*}
D&=S^{-1}(T^{-1}BT)S\\
&=(S^{-1}T^{-1})B(TS)\\
&=(TS)^{-1}B(TS) \tag{***}.
\end{align*}

Now let us put $U:=TS$. Then the matrix $U$ is nonsingular.
(This is because we have
\[\det(U)=\det(TS)=\det(T)\det(S)\neq 0\]
since $T$ and $S$ are nonsingular matrices, hence their determinants are not zero.)

Therefore from (***) we have
\[D=U^{-1}BU,\]
where $D$ is a diagonal matrix and $U$ is a nonsingular matrix.
Thus $B$ is a diagonalizable matrix.

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2 & 4 & 3 \\
-4 &-6 &-3 \\
3 & 3 & 1
\end{bmatrix}.\]
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[…] For a proof, see the post “A matrix similar to a diagonalizable matrix is also diagonalizable“. […]