The Frobenius norm of an orthogonal transformation is always preserved. The Frobenius norm is defined as $$ \mathrm{norm}(\mathbf{A})_F = \sqrt{\sum_{i=1}^n\sum_{j=1}^n |a_{ij}|^2}. $$ This means that for our \( 2\times 2 \) case we have $$ 2a_{kl}^2+a_{kk}^2+a_{ll}^2 = b_{kk}^2+b_{ll}^2, $$ which leads to $$ \mathrm{off}(\mathbf{B})^2 = \mathrm{norm}(\mathbf{B})_F^2-\sum_{i=1}^nb_{ii}^2=\mathrm{off}(\mathbf{A})^2-2a_{kl}^2, $$ since $$ \mathrm{norm}(\mathbf{B})_F^2-\sum_{i=1}^nb_{ii}^2=\mathrm{norm}(\mathbf{A})_F^2-\sum_{i=1}^na_{ii}^2+(a_{kk}^2+a_{ll}^2 -b_{kk}^2-b_{ll}^2). $$ This results means that the matrix \( \mathbf{A} \) moves closer to diagonal form for each transformation.