Linear-Algebra.ppt
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Linear-Algebra.ppt
- Εφαρμοσμένα Μαθηματικά Μ. Γιαμαρέλου & Γ. Μπίσκος Γραμμική Άλγεβρα Linear Algebra
- Linear Systems of Algebraic Equations Solve a system of n simultaneous linear algebraic equations for n unknowns, x1, x2, … , xn: a1,1x1 + a1,2 x2 + + a1,n xn = b1 a2,1x1 + a2,1x2 + + a2,nxn = b2 an,1x1 + an,1x2 + + an,n xn = bm • Problems in science, engineering, economics, etc. • Systems of non-linear equations → linear equations • Numerical methods needed for systems larger than 4×4
- Linear Systems of Algebraic Equations Solve a system of n simultaneous linear algebraic equations for n unknowns, x1, x2, … , xn: a11x1 + a12x2 + + a1n xn = b1 a21x1 + a21x2 + + a2n xn = b2 an1x1 + an1x2 + + ann xn = bm The system has infinitely many solutions The system has a single unique solution The system has no solution
- Linear Systems in a Matrix/Vector form Ax = B a11 a12 a1 j a1n a21 a22 a2 j a2n ai1 ai2 aij ain an1 an2 anj ann é ë ê ê ê ê ê ê ê ù û ú ú ú ú ú ú ú x1 x2 xi xn é ë ê ê ê ê ê ê ê ù û ú ú ú ú ú ú ú = b1 b2 bi bn é ë ê ê ê ê ê ê ê ù û ú ú ú ú ú ú ú
- Solving the Example in MATLAB In MATLAB we can solve the example with the single command x = Ab. To do that we need to type: >> A = [1 1 1; 2 1 3; 3 1 6]; >> b = [4; 7; 2]; >> x = Ab x = 19.0000 -7.0000 -8.0000
- Elimination methods for solving Linear Systems • Gaussian elimination The basic idea is to use a sequence of operations that converts the original system into a simpler, but equivalent system that may be solved easily. a11x1 + a12x2 + + a1n xn = b1 a21x1 + a21x2 + + a2n xn = b2 an1x1 + an1x2 + + ann xn = bm
- Straightforward Systems a1,1x1 = b1 a2,1x2 = b2 an,n xn = bn A = a1,1 0 0 0 0 a2,2 0 0 0 0 ai,i 0 0 0 0 an,n é ë ê ê ê ê ê ê ê ê ù û ú ú ú ú ú ú ú ú
- Straightforward Systems a1,1x1 + a1,2 x2 + + a1,n xn = b1 + a2,1x2 + + a2,n xn = b2 an,n xn = bn a1,1x1 = b1 a2,1x1 + a2,2 x2 = b2 an,1x1 + an,1x2 + + an,nxn = bn A = a1,1 a1,2 a1, j a1,n 0 a2,2 a2 j a2,n 0 0 ai,i ai,n 0 0 0 an,n é ë ê ê ê ê ê ê ê ê ù û ú ú ú ú ú ú ú ú A = a1,1 0 0 0 a2,1 a2,2 0 0 ai,1 ai,2 ai,i 0 an,1 an,2 an, j an,n é ë ê ê ê ê ê ê ê ê ù û ú ú ú ú ú ú ú ú
- Back/Forward-substitution xn = bn an,n , xi = bi - ai, j xj j=i+1 j=n å ai,i Upper triangular: x1 = b1 a1,1 , xi = bi - ai, j xj j=1 j=i-1 å ai,i Lower triangular: Diagonal: xi = bi ai,i ,
- Elimination methods for solving Linear Systems • Gaussian elimination The basic idea is to use a sequence of operations that converts the original system into a simpler, but equivalent system that may be solved easily. a1,1x1 + a1,2 x2 + + a1,n xn = b1 a2,1x1 + a2,1x2 + + a2,nxn = b2 an,1x1 + an,1x2 + + an,n xn = bm
- Elementary Row Calculations • Adding two rows (i.e., two equations) • Multiplying a scalar to a row • Combining the two k c x i + k aj1x1 + aj2x2 + + ajn xn = bj + ak1x1 + ak2x2 + + akn xn = bk (ai1 + ak1)x1 + (ai2 + ak2)x2 + + (ain + akn )xn = bi + bk cai1x1 + cai2x2 + + cain xn = cbi (cai1 + ak1)x1 + (cai2 + ak2)x2 + + (cain + akn )xn = cbi + bk
- Elementary Row Calculations ¢ A = a1,1 a1,2 a1,n a2,1 a2,2 a2,n ai,1 ai,2 ai,n (cai,1 + ak,1) (cai,2 + ak,2 ) (cai,n + ak,i ) an,1 an,2 an,n é ë ê ê ê ê ê ê ê ê ê ê ê ê ù û ú ú ú ú ú ú ú ú ú ú ú ú c x i + k → k
- Augmented Matrix [A, b] = a11 a12 a1 j a1n a21 a22 a2 j a2n ai1 ai2 aij ain an1 an2 anj ann b1 b2 bi bn é ë ê ê ê ê ê ê ê ù û ú ú ú ú ú ú ú
- Gaussian Elimination [A, b] = a11 a12 a1 j a1n 0 a22 (2,1) a2 j (2,1) a2n (2,1) 0 0 aij (i,i-1) ain (i,i-1) 0 0 0 ann (n,n-1) b1 b2 (2,1) bi (i.i-1) bn (n,n-1) é ë ê ê ê ê ê ê ê ù û ú ú ú ú ú ú ú
- Algorithm for Gaussian Elimination Given that matrix A is n×n we have: for j = 1:N-1 (iterate over columns from left to right) for i = j+1:N (iterate over rows below diagonal) if ai,j = 0 STOP (to avoid division by zero) λ = ai,j/ai,i for k = j:N (iterate in row i from column j to right) ai,k = ai,k – λaj,k end bi = bi – λbj end end Number of FLOPS = 2/3N3
- Gauss-Jordan Elimination [A, b]= d1,1 0 0 0 0 d2,2 0 0 0 0 di, j 0 0 0 0 an,n b1 b2 bi bn é ë ê ê ê ê ê ê ê ê ê ù û ú ú ú ú ú ú ú ú ú
- Partial Pivoting [A, b]= a1,1 a1,2 a1, j a1,n a2,1 - l2,1a1,1 a2,2 - l2,1a1,2 a2, j - l21a1j a2,n - l2,1a1,n ai,1 ai,2 ai, j ai,n an,1 an,2 an, j an,n b1 b2 - l2,1b1 bi bn é ë ê ê ê ê ê ê ê ê ù û ú ú ú ú ú ú ú ú l21 = a21 a11 l21 = ±¥ However if , then a1,1 = 0
- Partial Pivoting [A, b]= ai,1 ai,2 ai, j ai,n a2,1 a2,2 a2, j a2,n a1,1 a1,2 a1, j a1,n an,1 an,2 an, j an,n bi b2 b1 bn é ë ê ê ê ê ê ê ê ê ù û ú ú ú ú ú ú ú ú
- Algorithm for GE including partial pivoting for j = 1:N-1 (iterate over columns from left to right) select row j such that |aj,i| = max {|ai,i|, |ai+1,i|, . . . , |aN,1|} if aj,i = 0, no unique solution exists, STOP if j ≠ i, interchange rows i and j of augmented matrix for i = j+1:N (iterate over rows below diagonal) λ = ai,j/ai,I for k = j:N (iterate in row i from column j to right) ai,k = ai,k – λaj,k end bi = bi – λbj end end
- Matrix Factorization: LU decomposition • Advantageous when we have systems that have the same left-hand-side (i.e., only vector b) • With Gaussian elimination we would have to solve all the systems using the same number of calculations.
- • Let us assume that we can decompose A into a product of a lower L and upper U triangular matrix: we can obtain the solution by solving two triangular problems Matrix Factorization: LU decomposition Ax = LUx = b Lc = b Ux = c
- LU Decomposition using Gauss Elimination a1,1 a1,2 a1, j a1,n a2,1 a2,2 a2 j a2,n ai,1 ai,2 ai, j ai,n an,1 an,2 an, j an,n é ë ê ê ê ê ê ê ê ê ù û ú ú ú ú ú ú ú ú = 1 0 0 0 l2,1 1 0 0 li,1 li,2 1 0 ln,1 ln,2 ln, j 1 é ë ê ê ê ê ê ê ê ê ù û ú ú ú ú ú ú ú ú a1,1 a1,2 a1, j a1,n 0 ¢ a2,2 ¢ a2 j ¢ a2,n 0 0 a(i) i, j a(i) i,n 0 0 0 a(n) n,n é ë ê ê ê ê ê ê ê ê ê ù û ú ú ú ú ú ú ú ú ú A = LU
- LU Decomposition with Pivoting • Pivoting may be required for the successful execution of the Gaussian elimination procedure in finding L and U • To store all the information about the pivoting we use a permutation matrix P so that PA = LU
- Matrix Inversion In = 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 é ë ê ê ê ê ê ê ù û ú ú ú ú ú ú
- Calculating the Inverse of a Matrix a1,1 a1,2 a1, j a1,n a2,1 a2,2 a2 j a2,n ai,1 ai,2 ai, j ai,n an,1 an,2 an, j an,n é ë ê ê ê ê ê ê ê ê ù û ú ú ú ú ú ú ú ú x1,1 x1,2 x1, j x1,n x2,1 x2,2 x2 j x2,n xi,1 xi,2 xi, j xi,n xn,1 xn,2 xn, j xn,n é ë ê ê ê ê ê ê ê ê ù û ú ú ú ú ú ú ú ú = 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 é ë ê ê ê ê ê ê ê ù û ú ú ú ú ú ú ú a1,1 a1,2 a1, j a1,n a2,1 a2,2 a2 j a2,n ai,1 ai,2 ai, j ai,n an,1 an,2 an, j an,n é ë ê ê ê ê ê ê ê ê ù û ú ú ú ú ú ú ú ú x1,1 x1,2 xi,1 xn,1 é ë ê ê ê ê ê ê ê ê ê ê ù û ú ú ú ú ú ú ú ú ú ú = 1 0 0 0 é ë ê ê ê ê ê ê ê ù û ú ú ú ú ú ú ú 1st system:
- Solving Systems of Linear Eqs. in Matlab
- Solving the Example in MATLAB In MATLAB we can solve the example with the single command x = Ab. To do that we need to type: >> A = [1 1 1; 2 1 3; 3 1 6]; >> b = [4; 7; 2]; >> x = Ab x = 19.0000 -7.0000 -8.0000
- Solving the Example in MATLAB In MATLAB we can solve the example with the single command x = Ab. To do that we need to type: >> A = [1 1 1; 2 1 3; 3 1 6]; >> b = [4; 7; 2]; >> Ainv = inv(A); >> x = Ainv*b x = 19.0000 -7.0000 -8.0000
- Solving the Example in MATLAB In MATLAB we can solve a linear system using LU decomposition with the command lu as follows: >> A = [1 1 1; 2 1 3; 3 1 6]; >> b = [4; 7; 2]; >> [L, U, P] = lu(A) L = … U = … P = … >> c = L(P*b) c = … >> x = Uc x = …
- Solving the Example in MATLAB In MATLAB we can perform LU factorization with the command lu: >> A = [1 1 1; 2 1 3; 3 1 6]; >> [L, U, P] = lu(A) L = …, U = …, P = … >> b = [4; 7; 2]; >> x = U(L(P*b)) x = 19.0000 -7.0000 -8.0000
- The Determinant A = a11 a12 a21 a22 = a11a22 - a21a12 A = a11 a12 a13 a21 a22 a23 a31 a32 a33 = a11 a22 a23 a32 a33 - a12 a21 a23 a31 a33 + a13 a21 a22 a31 a32 If det(A) ≠ 0, (i.e., A is nonsingular) then the system has a unique solution: Cramer’s method x1 = det(Ax1) det(A) , x2 = det(Ax2) det(A) , , xn = det(Axn ) det(A) ,