lgebra Linear e Aplicaes RECTANGULAR SYSTEMS AND ECHELON FORMS - PowerPoint PPT Presentation

Álgebra Linear e Aplicações

RECTANGULAR SYSTEMS AND ECHELON FORMS

Rectangular systems • General linear system a 11 x 1 + a 12 x 2 + · · · a 1 n x n = b 1 a 21 x 1 + a 22 x 2 + · · · a 2 n x n = b 2 . . . a m 1 x 1 + a m 2 x 2 + · · · a mn x n = b m • What happens to elimination when m ≠ n ?

More on matrix notation • Matrix of size m × n m = n : square matrix m � n : rectangular matrix   a 11 a 12 a 1 n · · · a 21 a 22 a 2 n · · ·   A =  = [ a ij ] m × n  . . .  ... . . .   . . .  a m 1 a m 2 a mn · · · � a i 1 � A i ∗ = row vector a i 2 a in · · ·   a 1 j a 2 j   A ∗ j = column vector   .   . .   a mj

Gaussian Elimination example • Consider the system     1 1 2 2 1 1 3 3 3 3 0 0 0 0 − 2 − 2 − 2 − 2 − 2 − 2     x 1 + 2 x 2 + x 3 + 3 x 4 + 3 x 5 = 5     0 0 0 0 2 2 2 2 2 2     2 x 1 + 4 x 2 + 4 x 4 + 4 x 5 = 6 0 0 0 0 − 2 − 2 − 2 − 2 1 1 x 1 + 2 x 2 + 3 x 3 + 5 x 4 + 5 x 5 = 9 2 x 1 + 4 x 2 + 4 x 4 + 7 x 5 = 9   1 2 1 3 3 0 0 − 2 − 2 − 2   • Gaussian Elimination   0 0 0 0 0   0 0 0 0 3     1 2 1 3 3 1 2 1 3 3 2 4 0 4 4 0 0 − 2 − 2 − 2         1 2 3 5 5 0 0 0 0 3     2 4 0 4 7 0 0 0 0 0

Modified Gaussian Elimination • Look for a column that • Stop if all rows below has a viable pivot last pivot are zero       ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ 0 0 0 0 0 s s s s s s s ∗ ∗ ∗             0 0 0 0 0 0 0 s s s s s s s ∗       0 0 0 0 0 0 0 0 s s s s s s s • If needed, exchange rows to move pivot up     ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ 0 0 0 0 0 s s ∗ s s         0 0 0 0 0 ∗ s s s s     0 0 0 0 s s s s s s

Row Echelon Form   ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ 0 0 ∗ ∗ ∗ ∗ ∗ ∗     0 0 0 ∗ ∗ ∗ ∗ ∗     0 0 0 0 0 0 ∗ ∗     0 0 0 0 0 0 0 0   0 0 0 0 0 0 0 0 • Two conditions • All rows below a zero-row are also zero • All elements below or left of pivots are zero • Elements not unique, but form is unique!

Rank of a Matrix • Let A have row echelon form E • rank ( A ) = number of pivots • rank ( A ) = number of non-zero rows in E • rank ( A ) = number of basic columns in A • Basic column is a column at a pivotal position   1 2   A ∗ 1 =     1 2 1 3 3 1 2 1 3 3   1   2 4 0 4 4 0 0 − 2 − 2 − 2     2 A =  E =     1 2 3 5 5 0 0 0 0 3        1 3 2 4 0 4 7 0 0 0 0 0 0 4     A ∗ 3 = A ∗ 5 =     3 5     0 7

Modified Gauss-Jordan example • Back to our system     1 2 1 3 3 1 2 0 2 2 0 0 − 2 − 2 − 2 0 0 1 1 1         0 0 0 0 3 0 0 0 0 1     0 0 0 0 0 0 0 0 0 0   1 2 1 3 3   1 2 0 2 0 0 0 1 1 1 0 0 1 1 0       0 0 0 0 3   0 0 0 0 1     0 0 0 0 0 0 0 0 0 0   1 2 0 2 2 0 0 1 1 1     0 0 0 0 3   0 0 0 0 0

Reduced Row Echelon Form • Echelon form is result of Gaussian Elimination • Gauss-Jordan leads to reduced row echelon form   1 0 0 0 ∗ ∗ ∗ ∗ 0 0 1 0 0 ∗ ∗ ∗     0 0 0 1 0 ∗ ∗ ∗     0 0 0 0 0 0 1 ∗     0 0 0 0 0 0 0 0   0 0 0 0 0 0 0 0 • Notation: E A is reduced row echelon form of A • Reduced form is unique (proof later on) • Useful for theoretical results

Why basic columns are basic • Non-basic columns are combinations of basic columns, as made obvious by reduced form   1 2 0 2 0 E ∗ 2 = 2 E ∗ 1 0 0 1 1 0   E A =   0 0 0 0 1 E ∗ 4 = 2 E ∗ 1 + E ∗ 3   0 0 0 0 0 • Row operations preserve column relationships   1 2 1 3 3 2 4 0 4 4 A ∗ 2 = 2 A ∗ 1   A =   1 2 3 5 5   A ∗ 4 = 2 A ∗ 1 + A ∗ 3 2 4 0 4 7

Column relationships in A and E A         1 0 0 µ 1 0 1 0 µ 2                 . . . . . . . .         . . . .         E ∗ k = = µ 1 + µ 2 + · · · + µ j         0 0 1 µ j                 . . . . . . . .         . . . .         0 0 0 0 = µ 1 E ∗ b 1 + µ 2 E ∗ b 2 + · · · + µ j E ∗ b j are the basic columns to the left of E * k • E ∗ b i • Same relationships hold for and A * k A ∗ b i A ∗ k = µ 1 A ∗ b 1 + µ 2 A ∗ b 2 + · · · + µ j A ∗ b j • Why?

SOLVING RECTANGULAR SYSTEMS

Consistency • Rectangular system is consistent if it has at least one solution • It is inconsistent if it has no solution Consistent system Inconsistent system

Using Gaussian elimination instead • Reduce [ A | b ] to [ E | c ] in row echelon form   ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ 0 0 ∗ ∗ ∗ ∗ ∗ ∗ ∗     0 0 0 ∗ ∗ ∗ ∗ ∗ ∗     0 0 0 0 0 0 ∗ ∗ ∗     0 0 0 0 0 0 0 0 α    . . . . . . . . .  . . . . . . . . . . . . . . . . . . • If , we have a problem α 6 = 0 0 x 1 + 0 x 2 + · · · + 0 x n = α • Otherwise, back-substitution works!

Equivalent consistency criteria • When row reducing [ A | b ], a row of the following form never appears � 0 α � α 6 = 0 0 0 · · · • b is a combination of the basic columns in A • rank [ A | b ] = rank ( A )

Homogeneous systems • An homogeneous system is of the form a 11 x 1 + a 12 x 2 + · · · a 1 n x n = 0 a 21 x 1 + a 22 x 2 + · · · a 2 n x n = 0 . . . a m 1 x 1 + a m 2 x 2 + · · · a mn x n = 0 • Can it be inconsistent ? • Values of ? α 6 = 0 • Trivial solution! x 1 = x 2 = · · · = x m = 0

Homogeneous system example • Consider the system • Equivalent to x 1 + 2 x 2 + 2 x 3 + 3 x 4 = 0 x 1 + 2 x 2 + 2 x 3 + 3 x 4 = 0 2 x 1 + 4 x 2 + x 3 + 3 x 4 = 0 − 3 x 3 − 3 x 4 = 0 3 x 1 + 6 x 2 + x 3 + 4 x 4 = 0 • 4 unknowns and only • In row echelon form 2 equations! • Pick basic variables     1 2 2 3 1 2 2 3 2 4 1 3 0 0 − 3 − 3 • pivotal positions     3 6 1 4 0 0 − 5 − 5 • as functions of free variables   1 2 2 3 0 0 − 3 − 3 x 3 = − x 4   0 0 0 0 x 1 = − 2 x 2 − 2 x 3 − 3 x 4 = − 2 x 2 − x 4

Representating the general solution • Pick any values for x 2 and x 4 , and you create a new solution to the system x 1 = − 2 x 2 − x 4         − 2 x 2 − x 4 − 2 − 1 x 1 x 2 = x 2 1 0 x 2 x 2          =  = x 2  + x 4         0 − 1 x 3 − x 4 x 3 = − x 4      0 1 x 4 x 4 x 4 = x 4 • Every solution is a combination of two particular solutions     − 2 − 1 1 0     h 1 = h 2 =     0 − 1     0 1

General homogeneous system • Consider an m × n system [ A | 0 ] • How many basic variables? • r = rank ( A ) • How many free variables? • n – r • General form of solution is x = x f 1 h 1 + x f 2 h 2 + · · · + x f n − r h n − r

Unicity of homogeneous solution • Consistency has already been shown • Trivial solution x 1 = x 2 = · · · = x m = 0 • Look at general solution x = x f 1 h 1 + x f 2 h 2 + · · · + x f n − r h n − r • When is it unique? • When there are no free variables • When n - r = 0 • i.e., when rank ( A ) = n

Nonhomogeneous systems • A system is non-homogeneous a 11 x 1 + a 12 x 2 + · · · a 1 n x n = b 1 a 21 x 1 + a 22 x 2 + · · · a 2 n x n = b 2 . . . a m 1 x 1 + a m 2 x 2 + · · · a mn x n = b m • whenever for at least one i b i 6 = 0 • Solution follows the same procedure as the homogeneous case

Solving nonhomogeneous systems • Start with [ A | b ] and reduce to [ E | c ] • Identify the basic and free variables • Apply back-substitution to solve for basic variables as function of free variables • Write the result in the form x = p + x f 1 h 1 + x f 2 h 2 + · · · + x f n − r h n − r • The only difference is the term p that was not needed in the homogeneous case