To prove this we start with the eigenvalue problem and a similarity transformed matrix \( \mathbf{B} \). $$ \mathbf{A}\mathbf{x}=\lambda\mathbf{x} \hspace{1cm} \mathrm{and}\hspace{1cm} \mathbf{B}= \mathbf{S}^T \mathbf{A}\mathbf{S}. $$ We multiply the first equation on the left by \( \mathbf{S}^T \) and insert \( \mathbf{S}^{T}\mathbf{S} = \mathbf{I} \) between \( \mathbf{A} \) and \( \mathbf{x} \). Then we get $$ \begin{equation} (\mathbf{S}^T\mathbf{A}\mathbf{S})(\mathbf{S}^T\mathbf{x})=\lambda\mathbf{S}^T\mathbf{x} , \tag{3} \end{equation} $$ which is the same as $$ \mathbf{B} \left ( \mathbf{S}^T\mathbf{x} \right ) = \lambda \left (\mathbf{S}^T\mathbf{x}\right ). $$ The variable \( \lambda \) is an eigenvalue of \( \mathbf{B} \) as well, but with eigenvector \( \mathbf{S}^T\mathbf{x} \).