By W. Waliszewski

Warsaw 1965 Rozprawy Matematyczne XLV. Sm.4to., 40pp., unique revealed wraps. Uncut. VG, gentle soiling.

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**Example text**

2) First let Co contain two distinct lines L 1, L 2 which are nonorthogonal. Then every pair of distinct lines K 1, K 2 in Co must be nonorthogonal. Otherwise we would have distinct K], K 2 in Co with q(K 1, K 2 ) = 0. Pick a line J of V that's not in the coset Co' If q(L p J) = 0, then it follows from Witt's Theorem that there is a ~ in PSPn such that ~K] = L" ~K2 = J, and this disrupts the partition. L 2 = J, and this is again absurd. So, indeed, every pair of distinct lines in Co is nonorthogonal.

D. 14. Let char F = 2 and let A be any symmetric matrix over F with A =1= O. Then there is an invertible matrix T over F such that with A' invertible and diagonal if A is not alternating, and A' of the form 1 + 1 0 + 1 1 0 if A is alternating. PROOF. If we add a multiple of one column of A to another and then add the same multiple of the first corresponding row to the second corresponding row, or if we interchange two columns of A and then interchange the corresponding rows, or if we multiply a column of A by a nonzero scalar and then multiply the corresponding row by the same scalar, then, in each of these cases, it is easily seen that the matrix obtained is a nonzero symmetric matrix of the form ITAT for some invertible matrix T over F.

The permutation group PSPn( V) acting on the set of lines primitive when n ;;.. 4. r of V is PROOF. (1) We must consider a partition gJ of e which contains at least two cosets such that at least one coset, say Co, has at least two lines. And we must find an element of PSPn (V) that will disrupt this partition. Suppose, if possible. there is no such element. (2) First let Co contain two distinct lines L 1, L 2 which are nonorthogonal. Then every pair of distinct lines K 1, K 2 in Co must be nonorthogonal.