following c × ( r + 1 − c ) matrix: Dividing the matrix into two parts, we find that Now consider the following matrix: Similarly, we have From the above two equations, we have N ( r + 1 , c ) = N ( r + 1 , r + 1 − c ). By induction, we obtain the lemma. N ( r + 1 , r + 1 − c ) = N ( c + 1 , 1) N ( r, r − c ) . 1 0 · · · 0 0 0 1 · · · 0 0 . . . . . . . 0 0 · · · 1 0 □ · · · □ □ · · · □ . . . . . □ · · · □ 0 0 · · · 0 1 ︸ ︷︷ ︸ r $^{+1}$− c □ · · · □ ︸ ︷︷ ︸ c N ( r + 1 , c ) = N ( c + 1 , c ) N ( r, c ) . 1 0 · · · 0 □ 0 1 · · · 0 □ . . . . . . . 0 0 · · · 1 □ ︸ ︷︷ ︸ c +1 □ · · · □ □ · · · □ . . . . . □ · · · □ ︸ ︷︷ ︸ $^{r}$−c On the other hand, it can be observed that the selection operator W in the previous subsection can be decomposed as a series of unitary row operators of type 2, type 3, and type 4, for the matrix A determines a series of type 4 unitary row operators, the matrix B determines a series of type 3 unitary row operators, and the symmetric matrix M , satisfying 25