2 x 2 Example 2: Matrix Cube(Y=X3 2 X 2 Example 3: Matrix Inverse(Y=X) 2 X 2 Example 4: Linear Transformation (Y= AX+B) 2 X 2 Example 5: The LU Decomposition(X=LU) 2 X 2 Example 6: A Symmetric Decomposition (S= DMD) 2 X 2 Example 7: Traceless Symmetric = Polar Decomposition(S=QAQ, tr S=O) 2 X 2 Example 8: The Symmetric Eigenvalue Problem(S=QAQ) 2 X 2 Example 9: Symmetric Congruence(Y= ATSA) Discussion: Example 1: Matrix Square(Y=X2) With A≈/pq and Y=X2 the jacobian matrix of interest is 01aY1 +s aYr rq0 0Y12 On this first example we label the columns and rows so that the elements correspond to the definition J=(OX. Later we will omit the labels. We invite readers to compare with Equation (3). We see that the Jacobian matrix and the differential contain the same information. We can compute then det=4(p+s),(sp-gr)=4(tr X)2 det(X) Notice that breakdown occurs if x is singular or has trace zero Example 2: Matrix Cube(Y=x) With X p q andY=X p2+2 erp+sr p2+2qr+(p+s) rp+2sr p(p+s)+2 qr+8 pq+2qs pg +2 gs (p+s)+sr 2 qr+3s so that det J=9(sp-gr)(gr +p2+82+sp)2-9(det X)(tr X2+(tr X)2)2 Breakdown occurs if X is singular or if the eigenvalue ratio is a complex cube root of unit Example 3: Matrix Inverse (Y=X) and Y=X-I  " # " # 2 × 2 Example 2: Matrix Cube (Y = X3) 2 × 2 Example 3: Matrix Inverse (Y = X−1) 2 × 2 Example 4: Linear Transformation (Y = AX + B) 2 × 2 Example 5: The LU Decomposition (X = LU) 2 × 2 Example 6: A Symmetric Decomposition (S = DMD) 2 × 2 Example 7: Traceless Symmetric = Polar Decomposition (S = QΛQT , tr S = 0) 2 × 2 Example 8: The Symmetric Eigenvalue Problem (S = QΛQT ) 2 × 2 Example 9: Symmetric Congruence (Y = ATSA) Discussion: Example 1: Matrix Square (Y = X2) " # With X = p q and Y = X2 the Jacobian matrix of interest is r s ∂p ∂r ∂q ∂s   2p q r 0 ∂Y11   r p + s 0 r   ∂Y21 J =  q 0 p + s q  ∂Y12 0 q r 2s ∂Y22 On this first example we label the columns and rows so that the elements correspond to the definition J = ∂Yij . Later we will omit the labels. We invite readers to compare with Equation (3). We see that ∂Xkl the Jacobian matrix and the differential contain the same information. We can compute then det J = 4(p + s) 2(sp − qr) = 4(tr X) 2 det(X). Notice that breakdown occurs if X is singular or has trace zero. Example 2: Matrix Cube (Y = X3) p q With X = and Y = X3 r s  3 p2 + 2 qr pq + q (p + s) 2 rp + sr qr     2 rp + sr p2 + 2 qr + (p + s) s r2 rp + 2 sr    J =  ,  2 pq + qs q2 p (p + s) + 2 qr + s2 pq + 2 qs    qr pq + 2 qs r (p + s) + sr 2 qr + 3 s2 so that 2 2 det J = 9(sp − qr) 2(qr + p + s + sp) 2 = 9 (det X) 2(tr X2 + (tr X) 2) 2 . 4 Breakdown occurs if X is singular or if the eigenvalue ratio is a complex cube root of unity. Example 3: Matrix Inverse (Y = X−1) p q With X = and Y = X−1 r s  2  −s qs sr −qr    sr −ps 2 rp  1  −r  J =  , det X2 ×  qs −q2 −ps pq    − 2 qr pq rp −p