Getting started with Matlab
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Learn the basics of Matlab and its numerical methods for linear algebra, including solving equations, calculating determinants, LU decomposition, QR decomposition, and singular value decomposition.
Getting started with Matlab
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Getting started with Matlab Numerical Methods Appendix B http://www.mathworks.com/access/helpdesk/help/techdoc/learn_matlab/learn_matlab.html
Linear Algebra with Matlab Introduction of Basic Matlab functions
Solve Ax=b Matlab x= A\b
trace(A) • >> A=[1 3 6;2 1 9; 3 6 1] • A = • 1 3 6 • 2 1 9 • 3 6 1 • >> trace(A) • ans = • 3
rank() • >> B=[1 2 3; 3 4 7; 4 -3 1;-2 5 3; 1 -7 6] • B = • 1 2 3 • 3 4 7 • 4 -3 1 • -2 5 3 • 1 -7 6 • >> rank (B) • ans = • 3 Number of independent rows
Reduced row echelon form;rref(B) • >> B=[1 2 3; 3 4 7; 4 -3 1;-2 5 3; 1 -7 6]; • >> rref(B) • ans = • 1 0 0 • 0 1 0 • 0 0 1 • 0 0 0 • 0 0 0
inv(A) • >> A=rand(3,3) • A = • 0.1389 0.6038 0.0153 • 0.2028 0.2722 0.7468 • 0.1987 0.1988 0.4451 • >> inv(A) • ans = • -0.8783 -8.5418 14.3617 • 1.8694 1.8898 -3.2348 • -0.4429 2.9695 -2.7204 • >> A*inv(A) • ans = ???
det(A) • >> A=rand(3,3) • A = • 0.9318 0.8462 0.6721 • 0.4660 0.5252 0.8381 • 0.4186 0.2026 0.0196 • >> det(A) • ans = • 0.0562
>> A=rand(3,3) A = 0.9318 0.8462 0.6721 0.4660 0.5252 0.8381 0.4186 0.2026 0.0196 >> b=rand(3,1) b = 0.1509 0.6979 0.3784 >> x = A\b x = 3.4539 -5.4966 2.3564 >> A*x ans = 0.1509 0.6979 0.3784 Ax=b; x=A\b
>> tic >> toc elapsed_time = 2.1630 >> tic >> x=toc x = 2.5240 tic, toc, elapsed_time
time.m • A=rand(1000,1000); • tic • inv(A); • time_to_inverse_A=toc
output functions • disp(‘strings to be shown on screen’); • fprintf(‘As C language %8.2f\n’, ver1); • >> ver1=1.3333 • >> fprintf('As C language %8.2f\n', ver1); • As C language 1.33
v=[3, 4] norm(v,1) ans = 7 norm(v) ans = 5 norm(v,3) ans = 4.4979 norm(v,inf) ans = 4 ||V||1=(|v1|+|V2|+…|Vn|) ||V||2=(|v1|2+|V2|2+…|Vn|2 ) -2 ||V||3=(|v1|3+|V2|3+…|Vn|3 ) -3 ||V||inf=(|v1|∞+|V2| ∞ +…|Vn|∞) - ∞ norm(V,n)
If Ax=b has solution • Then • Ax =0 only when x=0 • det(A) ≠0 • reff(A) = I • rank(A)=n
>> A=rand(3) A = 0.9991 0.8848 0.4642 0.3593 0.4178 0.2477 0.3566 0.0836 0.1263 >> [L1,U]=lu(A) L1 = 1.0000 0 0 0.3596 -0.4291 1.0000 0.3569 1.0000 0 U = 0.9991 0.8848 0.4642 0 -0.2322-0.0394 0 0 0.0638 >> [L,U,P]=lu(A) L = 1.0000 0 0 0.3569 1.0000 0 0.3596 -0.4291 1.0000 U = 0.9991 0.8848 0.4642 0 -0.2322 -0.0394 0 0 0.0638 P = 1 0 0 0 0 1 0 1 0 LU decomposition
>> A=[2 3 4; 3 6 7; 4 7 10]; >> P=chol(A) P = 1.4142 2.1213 2.8284 0 1.2247 0.8165 0 0 1.1547 >> P'*P-A ans = 1.0e-015 * 0.4441 0 0 0 0 0 0 0 0 A is positive definite symmetric A=PTP 11.1.2 Cholesky decomposition
A = 0.8138 0.7576 0.2240 0.1635 0.0536 0.8469 0.0567 0.5092 0.0466 >> [Q,R]=qr(A) Q = -0.9781 -0.0246 -0.2065 -0.1965 -0.2161 0.9564 -0.0682 0.9761 0.2065 R = -0.8319 -0.7862 -0.3887 0 0.4667 -0.1430 0 0 0.7734 Q Orthogonal matrix R Upper triangle matrix QR decomposition
>> A = [1 2 3 4 5; 5 9 2 3 4; 2 2 3 4 2] >> [U,S,V]=svd(A) U = -0.4581 0.6831 -0.5688 -0.7993 -0.5966 -0.0727 -0.3890 0.4213 0.8193 S = 14.0087 0 0 0 0 0 5.1976 0 0 0 0 0 1.9346 0 0 V = -0.3735 -0.2803 0.3650 -0.4827 -0.6447 -0.6344 -0.6080 -0.0793 0.2604 0.3920 -0.2955 0.4079 0.3133 -0.5529 0.5852 -0.4130 0.5056 0.4052 0.6063 -0.2051 -0.4473 0.3601 -0.7734 -0.1609 -0.2149 >> U*S*V' ans = 1.0000 2.0000 3.0000 4.0000 5.0000 5.0000 9.0000 2.0000 3.0000 4.0000 2.0000 2.0000 3.0000 4.0000 2.0000 A=USVT A .. m x n U .. m x m S .. m x n (singular value) V .. n x n Singular Value Decomposition
Reduce 3 columns • V = • -0.3735 -0.2803 0 0 0 • -0.6344 -0.6080 0 0 0 • -0.2955 0.4079 0 0 0 • -0.4130 0.5056 0 0 0 • -0.4473 0.3601 0 0 0 • >> U*S*V' • ans = • 1.4016 1.9127 3.3447 4.4458 4.1490 • 5.0514 8.9888 2.0441 3.0570 3.8912 • 1.4215 2.1258 2.5035 3.3579 3.2258
A = 0 0 0 0 0 0 1 0 0 1 >> Aplus=pinv(A) Aplus = 0 0 0 1 0 0 0 0 0 1 >> Aplus*A ans = 1 0 0 1 >> A*Aplus ans = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 Since A is not n by n, there is no inverse A-1. Ax=b can be solved by pseudo-inverse A+. x = A+ * b Pseudo-inverse
>> A=eye(3).*20 A = 20 0 0 0 20 0 0 0 20 >> [cond(A) rcond(A)] ans = 1 1 cond(A) =1 A is perfectly conditioned stable cond(A) > large number A is ill-conditioned too sensitive rcond is a rapid version of cond rcond~1/cond cond(A), rcond(A)
Condition of a Hilbert matrix • >> hilb(4) • ans = • 1.0000 0.5000 0.3333 0.2500 • 0.5000 0.3333 0.2500 0.2000 • 0.3333 0.2500 0.2000 0.1667 • 0.2500 0.2000 0.1667 0.1429 • >> cond(hilb(4)) • ans = • 1.5514e+004
if, else, and elseif • if A > B • 'greater' • elseif A < B • 'less' • elseif A == B • 'equal' • else • error('Unexpected situation') • end
switch and case switch (rem(n,4)==0) + (rem(n,2)==0) case 0 M = odd_magic(n) case 1 M = single_even_magic(n) case 2 M = double_even_magic(n) otherwise error('This is impossible') end
for for n = 3:32 r(n) = rank(magic(n)); end r r Display the result
While a = 0; fa = -Inf; b = 3; fb = Inf; while b-a > eps*b x = (a+b)/2; fx = x^3-2*x-5; if sign(fx) == sign(fa) a = x; fa = fx; else b = x; fb = fx; end end x (bisection method) The result is a root of the polynomial x^3 - 2x - 5 x = 2.09455148154233
continue fid = fopen('magic.m','r'); count = 0; while ~feof(fid) line = fgetl(fid); if isempty(line) | strncmp(line,'%',1) continue end count = count + 1; end disp(sprintf('%d lines',count));
break a = 0; fa = -Inf; b = 3; fb = Inf; while b-a > eps*b x = (a+b)/2; fx = x^3-2*x-5; if fx == 0 break elseif sign(fx) == sign(fa) a = x; fa = fx; else b = x; fb = fx; end end x
try - catch try statement ... statement catch statement ... statement end
return function d = det(A) %DET det(A) is the determinant of A. if isempty(A) d = 1; return else ... end return terminates the current sequence of commands and returns control to the invoking function
