[proofplan] The proof is a direct algebraic manipulation of the eigenvalue equation $Aw = \lambdaw$. We left-multiply by $S$, then insert the identity $S^{-1}S$ between $A$ and $w$ to recognise $(SAS^{-1})(Sw) = \lambda(Sw)$. Each step is an equivalence (since $S$ is nonsingular), establishing the biconditional. [/proofplan] [step:Rewrite the eigenvalue equation $Aw = \lambdaw$ in terms of $SAS^{-1}$ and $Sw$] Suppose $Aw = \lambdaw$. Since $S$ is nonsingular, $S^{-1}$ exists and $S^{-1}S = I$. Left-multiplying both sides of $Aw = \lambdaw$ by $S$: \begin{align*} SAw = \lambda Sw. \end{align*} Inserting the identity $I = S^{-1}S$ between $A$ and $w$ on the left-hand side: \begin{align*} S A (S^{-1} S) w = (SAS^{-1})(Sw) = \lambda(Sw). \end{align*} Setting $v := Sw$, this reads $(SAS^{-1})v = \lambdav$, so $v = Sw$ is an eigenvector of $SAS^{-1}$ with eigenvalue $\lambda$. Note that $v \neq 0$ because $S$ is nonsingular and $w \neq 0$. [guided] We want to show that the similarity transform $A \mapsto SAS^{-1}$ maps eigenvectors of $A$ to eigenvectors of $SAS^{-1}$, with the eigenvalue unchanged. The key observation is that every step in the chain of equalities is reversible because $S$ is invertible. Starting from $Aw = \lambdaw$, we left-multiply both sides by $S$: \begin{align*} SAw = \lambda Sw. \end{align*} Now we want to see $SAS^{-1}$ acting on $Sw$. We insert $I = S^{-1}S$ between $A$ and $w$: \begin{align*} SA(S^{-1}S)w = (SAS^{-1})(Sw) = \lambda(Sw). \end{align*} Setting $v := Sw$, we have $(SAS^{-1})v = \lambdav$, so $v$ is an eigenvector of $SAS^{-1}$ with eigenvalue $\lambda$. Since $S$ is nonsingular and $w \neq 0$ (being an eigenvector), we have $v = Sw \neq 0$, confirming that $v$ is a legitimate eigenvector. For the converse, suppose $(SAS^{-1})v = \lambdav$ with $v \neq 0$. Set $w := S^{-1}v$, so that $v = Sw$. Substituting: \begin{align*} (SAS^{-1})(Sw) = \lambda(Sw) \implies SAw = \lambda Sw. \end{align*} Left-multiplying by $S^{-1}$: $Aw = \lambdaw$. Since $S^{-1}$ is nonsingular and $v \neq 0$, we have $w = S^{-1}v \neq 0$, so $w$ is an eigenvector of $A$ with eigenvalue $\lambda$. This establishes the biconditional. [/guided] [/step]