Prove that the determinant of a skew symmetric matrix of even order is a perfect square

Grade:12

1 Answers

SHAIK AASIF AHAMED
askIITians Faculty 74 Points
9 years ago
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see that for an [nXn] skew symmetric marix A, with n odd, we have [det A=0] , which is againa perfect square. if n is even, let [A=[aij] we have [aii=0] and [aij=-aji.] the proof is by induction on even numbers, but for a more general case, i.e. matrices with entriesin [{Q}.] for n = 2, it's clear. for [n4] : if [a1j=0] , for all j, then the determinant of A is 0 and sowe're done.
otherwise [a1j$\neq$0] for some [j > 1.] for simplicity suppose j = 2. then using columnoperations we can change the first row to [(0,a12, 0, 0, ... ,0).] similarly using row operations wecan change the first column to [(0, -a12, 0, 0, ... ,0).]
note that the first row will remain the same,i.e. [(0,a12, 0, 0, ... ,0).] call this new matrix [A'] . clearly [det A=det A'.] now in [A'] , if we ignore
the first two rows and columns, we'll get an [(n-2)X(n-2)] matrix, call it [A''] , which is againskew symmetric.on the other hand, it's easy to
see that [detA=det A' = a212det A''.] but by induction [det A'' = r^2] , for some [r in Q}.] thus[det A = a12^2 r^2.] since the entries of A are integers, we have [det A in {Z}.] thus [a12^2 r^2 = m^2] , forsome integer [m.]

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