We specialize to a symmetric \( 3\times 3 \) matrix \( \mathbf{A} \). We start the process as follows (assuming that \( a_{23}=a_{32} \) is the largest non-diagonal) with \( c=\cos{\theta} \) and \( s=\sin{\theta} \) $$ \mathbf{B} = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & c & -s \\ 0 & s & c \end{array} \right)\left( \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} \right) \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & c & s \\ 0 & -s & c \end{array} \right). $$ We will choose the angle \( \theta \) in order to have \( a_{23}=a_{32}=0 \). We get (symmetric matrix) $$ \mathbf{B} =\left( \begin{array}{ccc} a_{11} & a_{12}c -a_{13}s& a_{12}s+a_{13}c \\ a_{12}c -a_{13}s & a_{22}c^2+a_{33}s^2 -2a_{23}sc& (a_{22}-a_{33})sc +a_{23}(c^2-s^2) \\ a_{12}s+a_{13}c & (a_{22}-a_{33})sc +a_{23}(c^2-s^2) & a_{22}s^2+a_{33}c^2 +2a_{23}sc \end{array} \right). $$ Note that \( a_{11} \) is unchanged! As it should.