Matrix Reference Manual
Proofs Section 3: Matrix Properties

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 3.1 det([A[m#m], B[m#n]; C[n#m], D[n#n]]) = det(A)*det(D-CA-1B) = det(D)*det(A-BD-1C) [A B; C D] = [A   0; C   I] [I   A-1B; 0   D-CA-1B] = [I   B; 0   D] [A-BD-1C   0; D-1C   I] The result follows since the determinant of a block triangular matrix is the product of the determinant of the diagonal blocks. 3.2 det(I[n#n] + A[m#n]TB[m#n])  = det(I[m#m] + ABT) = det(I[n#n] + BTA) = det(I[m#m] + BAT) det(I + ATB) = det((I + ATB)T) = det(I + BTA)  [since det XT = det X] det([I[m#m], -B; AT, I[n#n]]) = det(I)*det(I + ATIB) = det(I)*det(I + BIAT)   [3.1] Hence det(I + ATB) = det(I + BAT) = det((I + BAT)T) = det(I + ABT) 3.3 det(I+xyT) =  1+yTx det(I+xyT) = det(I[1#1]+yTx)   [3.2] = 1+yTx 3.4 det(A+xyT) = det(A) × (1+yTA-1x) det(A+xyT) = det(A) * det(I+A-1xyT) = det(A) * (1+yTA-1x)   [3.3] 3.5 If  M = [A[m#m], B; C, D[n#n]]  then M-1 = [Q-1, -Q-1BD-1; -D-1CQ-1, D-1(I+CQ-1BD-1)] where Q =A-BD-1C and also M-1 = [A-1(I+BP-1CA-1), -A-1BP-1; -P-1CA-1, P-1] where P[n#n]=D-CA-1B [A, B; C, D] [Q-1, -Q-1BD-1; -D-1CQ-1, D-1(I+CQ-1BD-1)] =[AQ-1-BD-1CQ-1, -AQ-1BD-1+BD-1(I+CQ-1BD-1); CQ-1-DD-1CQ-1, -CQ-1BD-1+DD-1(I+CQ-1BD-1)] =[(A-BD-1C)Q-1, -AQ-1BD-1+BD-1+BD-1CQ-1BD-1; CQ-1-CQ-1, -CQ-1BD-1+(I+CQ-1BD-1)] =[I, -(A-BD-1C)Q-1BD-1+BD-1; 0, I] = [I, -IBD-1+BD-1; 0, I] = [I, 0; 0, I] = I Define W = [0, I[n#n]; I[m#m], 0] and hence W-1 = [0, I[m#m]; I[n#n], 0] and WMW-1 = [D, C; B, A] From above WM-1W-1 = (WMW-1)-1 = [P-1, -P-1CA-1; -A-1BP-1, A-1(I+BP-1CA-1)] Hence M-1 = W-1[P-1, -P-1CA-1; -A-1BP-1, A-1(I+BP-1CA-1)]W = [A-1(I+BP-1CA-1), -A-1BP-1; -P-1CA-1, P-1] 3.6 If S is +ve semidefinite Hermitian, then |aHSb|2 <= aHSa×bHSb for any a, b. Also |si,j| <= sqrt(si,isj,j) [a b]HS[a b] is +ve semidefinite Hermitian and so its determinant is non-negative det([a b]HS[a b]) =  det(aHSa aHSb; bHSa bHSb) = aHSa×bHSb - |aHSb|2 >= 0 If we take a = ei and b = ej, then we get |si,j|2 <= si,isj,j 3.7 If B is +ve definite and A is +ve semidefinite then B-1A is diagonalizable and has non-negative eigenvalues. If S is the +ve definite hermitian square root of B-1 (i.e. S2B=I) then B-1A = S (SAS) S-1 so B-1A and SAS and so have the same eignevalues. SAS = SHAS and so is +ve semidefinite and so has non-negative eigenvalues and, since it is hermitian, is unitarily diagonalizable.

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