Non-square Determinant Theorems

Determinant Theorems Associated with non-square matrices PhL 3.22.16 Suppose A is non-square. We know that both AAT and ATA are square, though of different sizes. Suppose A is m x n (m rows, n columns, tall, m > n). Then we know AAT = m x m matrix = large matrix ATA = n x n matrix = small matrix Theorem 1: If C = AB, then CT = BTAT even if A and B are non-square, as long as they conform. Proof: Cik = AijBjk = Bjk Aij = (BT)kj(AT)ji = (BTAT)ki = [(BTAT)T]ik Therefore C = (BTAT)T and CT = BTAT . QED Theorem 2: (rank theorems) A and B are conforming matrices where the common dimension is n (a) rank(AB) ≤ min{rank(A),rank(B) } (4.10) "theorems on rank.pdf" (b) rank(A†A) = rank(AA†) = rank(A) for any complex matrix A (4.12) "same" (c) rank(ATA) = rank(AAT) = rank(A) for any real matrix A (special case of above) (d) rank(A) + rank(B) ≤ rank(AB) + n (Sylvester's Inequality, see web) (e) rank(AB) + rank(BC) ≤ rank(ABC) + rank(B) (Frobenius Inequality, see web) Proofs: The first three have simple proofs in a pdf I saved. The last two have proofs are on the web. As far as I know, the first three theorems do not have names. Theorem 3: Assume that A is an m x n matrix where m ≠ n. Then: (a) If m > n, then det(AAT) = 0 and thus AAT, although square, is not invertible. (b) If m < n, then det(ATA) = 0 and thus ATA, although square, is not invertible. (c) In general, det(AAT) ≠ det(ATA) in the above two cases. (d) If m = n, then det(AAT) = det(ATA) = det2(A) and the two agree. Proof: (a) By Theorem 2 (c) we know that rank(AAT) = rank(A) ≤ n since n is the smaller dimension of A. But for m > n the matrix AAT is an m x m matrix shown in the graphic above. If it has rank ≤ n, then it does not have full rank, and thus it must have det(AAT) = 0. (b) replace A→AT and (b) becomes (a). (c) counterexample given below In case (a) if instead we have m < n, then rank(AAT) = rank(A) ≤ m but now AAT is mxm so all we know in this case is that rank(AAT) ≤ m so it could be full rank, or it could be less than full rank. In general then you will get det(AAT) ≠ 0 in this case. So in our theorem, the claim that det(..) = 0 only applies to the case of m and n shown, and the opposite case will not in general have det(..) = 0. Comment: I could not prove this theorem just writing out the determinant expansion for det(AAT). It did not seem obvious that det(AAT) = 0 for m > n. For example in the special case that m = 3 and n = 2 we have AAT being a 3x3 matrix, so det(AAT) = εabc(AAT)1a(AAT)2b(AAT)3c = εabc A1jAajA2kAbk A3rAcr = A1jA2kA3r εabcAajAbkAcr = etc = ??? Example: Here is a general matrix with m = 3 rows and n = 2 columns so m > n : We compute AAT and det(AAT): which agrees with Theorem 3 (a). Now compute ATA and det(ATA) and get some general non-zero result for the determinant. In this case Implication: Suppose the R matrix has m rows and n columns and is tall with m > n. Then det(RRT) = 0 and thus RRT, although square, is not invertible Paradox?: In wedge doc I say that RRT = 1 for any R, where this is all say down-tilt. If we ignore the italic T, this seems to contradict the fact that det(RRT) = 0 for m > n. Please explain. Resolution: RT simply is not the same as RT for downtilt or uptilt, so the above discussion simply does not apply. RRT = 1 is the orthog relation which remains true from the chain rule. Theorem 4: det(CAAT) = det(AATC) where C is a conforming square matrix Proof: Obvious since one can write det(CAAT) = det(C[AAT]) = det(C)det(AAT) =det([AAT]C) = det(AATC) QED Theorem 5: det(CAAT) = det(AATCT) where C is a conforming square matrix Proof : Let D = CAAT . Then det(CAAT) = det(D) = det(DT) = det(AATCT) QED Theorem 6: det(CAAT) = det(ATCA) ?? cyclic forward one position // not true Proof: If m > n and C is m x m, then CA has the same shape as A, so at least the matrices conform on both sides of the above. But the theorem is not valid. Here is a simple counterexample: if C = 1, we know in general that det(AAT) ≠ det(ATA). When m ≠ n in fact one of these determinants is 0 and the other generally is not, as shown in Theorem 3. So expunge the notion that you are allowed to cyclic permute things in a determinant with non-square matrices and maintain the determinant value.