Let V be an n-dimensional vector space over the field F and let T be a linear transformation from V into V?

Question

such that the range and null space of T are identical. Prove that n is even.

Answer 1

Since range (T) = ker (T), we have that rank (T) = dim (range (T)) = dim (ker (T)) = nullity (T). From this and the rank-nullity theorem, we have that n = dim (V) = rank (T) + nullity (T) = rank (T) + rank (T) = 2*rank (T). Thus, n is a multiple of 2, so n is even. Q.E.D.

Answer 2

Ditto for Awms. you're employing a sledge hammer to force a thumb tack. From AB = I, taking transposes we get B^T A^T = I^T = I. as a result, det(B^T)*det(A^T) = det(I) = a million, so the two det(A^T) and det(B^T) are non-0, so A^T and B^T are invertible. apart from, B^T A^T = I says that B^T is the inverse of A^T; it extremely is, B^T = (A^T)^(-a million). What did we depart out?