Linear Algebra

280 Chapter Three. Maps Between SpacesWe want to show that the rank of the matrix of coefficients, the number ofrows in a maximal linearly independent set, is 2n + 1. The first n rows of thematrix of coefficients add to the same vector as the second n rows, the vector ofall ones. So a maximal linearly independent must omit at least one row. We willshow that the set of all rows but the first {⃗ρ 2 . . . ⃗ρ 2n+2 } is linearly independent.So consider this linear relationship.c 2 ⃗ρ 2 + · · · + c 2n ⃗ρ 2n + c 2n+1 ⃗ρ 2n+1 + c 2n+2 ⃗ρ 2n+2 = ⃗0 (*)Now it gets messy. In the final two rows, in the first n columns, is a subrowthat is all zeros except that it starts with a one in column 1 and a subrowthat is all zeros except that it ends with a one in column n. With ⃗ρ 1 not in(∗), each of those columns contains only two ones and so we can conclude thatc 2n+1 = −c n+1 as well as that c 2n+2 = −c 2n .Next consider the columns between those two — in the n = 3 illustrationabove this includes only the second column while in the n = 4 matrix it includesboth the second and third columns. Each such column has a single one. That is,for each column index j ∈ {2 . . . n − 2} the column consists of only zeros exceptfor a one in row n + j, and hence c n+j = 0.On to the next block of columns, from n + 1 through 2n. Column n + 1 hasonly two ones (because n 3 the ones in the last two rows do not fall in the firstcolumn of this block). Thus c 2 = −c n+1 and therefore c 2 = c 2n+1 . Likewise,from column 2n we conclude that c 2 = −c 2n and so c 2 = c 2n+2 .Because n 3 there is at least one column between column n + 1 andcolumn 2n − 1. In at least one of those columns a one appears in ⃗ρ 2n+1 . If aone also appears in that column in ⃗ρ 2n+2 then we have c 2 = −(c 2n+1 + c 2n+2 )(recall that c n+j = 0 for j ∈ {2 . . . n − 2}). If a one does not appear in thatcolumn in ⃗ρ 2n+2 then we have c 2 = −c 2n+1 . In either case c 2 = 0, and thusc 2n+1 = c 2n+2 = 0 and c n+1 = c 2n = 0.If the next block of n-many columns is not the last then similarly concludefrom its first column that c 3 = c n+1 = 0.Keep this up until we reach the last block of columns, those numbered(n − 1)n + 1 through n 2 . Because c n+1 = · · · = c 2n = 0 column n 2 gives thatc n = −c 2n+1 = 0.Therefore the rank of the matrix is 2n + 1, as required.The classic source on normal magic squares is [Ball & Coxeter]. More onthe Lo Shu square is at [Wikipedia Lo Shu Square]. The proof given here beganwith [Ward].Exercises1 Let M be a 3×3 magic square with magic number s.(a) Prove that the sum of M’s entries is 3s.(b) Prove that s = 3 · m 2,2 .(c) Prove that m 2,2 is the average of the entries in its row, its column, and ineach diagonal.