Linear Programming

Lecture 1: Linear Algebra Review Linear Programming 1 / 24 1 Linear Algebra Review

2 Linear Algebra Review

3 Block Structured Matrices

4 Gaussian Elimination Matrices

5 Gauss-Jordan Elimination (Pivoting)

Lecture 1: Linear Algebra Review Linear Programming 2 / 24 columns rows   a1•  a2•  = a a ... a =   •1 •2 •n  .   .  am•

   T  a11 a21 ... am1 a•1 T  a12 a22 ... am2   a•2  AT =   =   = aT aT ... aT  . . .. .   .  1• 2• m•  . . . .   .  T a1n a2n ... amn a•n

Matrices in Rm×n

A ∈ Rm×n

  a11 a12 ... a1n  a21 a22 ... a2n  A =    . . .. .   . . . .  am1 am2 ... amn

Lecture 1: Linear Algebra Review Linear Programming 3 / 24 rows   a1•  a2•  =    .   .  am•

Matrices in Rm×n

A ∈ Rm×n

columns   a11 a12 ... a1n  a21 a22 ... a2n  A =   = a a ... a  . . .. .  •1 •2 •n  . . . .  am1 am2 ... amn

Lecture 1: Linear Algebra Review Linear Programming 3 / 24    T  a11 a21 ... am1 a•1 T  a12 a22 ... am2   a•2  AT =   =   = aT aT ... aT  . . .. .   .  1• 2• m•  . . . .   .  T a1n a2n ... amn a•n

Matrices in Rm×n

A ∈ Rm×n

columns rows     a11 a12 ... a1n a1•  a21 a22 ... a2n   a2•  A =   = a a ... a =    . . .. .  •1 •2 •n  .   . . . .   .  am1 am2 ... amn am•

Lecture 1: Linear Algebra Review Linear Programming 3 / 24  T  a•1 T  a•2  =   = aT aT ... aT  .  1• 2• m•  .  T a•n

Matrices in Rm×n

A ∈ Rm×n

  a11 a21 ... am1  a12 a22 ... am2  AT =    . . .. .   . . . .  a1n a2n ... amn

Lecture 1: Linear Algebra Review Linear Programming 3 / 24 T T T = a1• a2• ... am•

Matrices in Rm×n

A ∈ Rm×n

   T  a11 a21 ... am1 a•1 T  a12 a22 ... am2   a•2  AT =   =    . . .. .   .   . . . .   .  T a1n a2n ... amn a•n

Lecture 1: Linear Algebra Review Linear Programming 3 / 24 Matrices in Rm×n

A ∈ Rm×n

Lecture 1: Linear Algebra Review Linear Programming 3 / 24  a11   a12   a1n   a21   a22   a2n  = x  + x  + ··· + x   1  .  2  .  n  .   .   .   .  am1 am2 amn

= x1 a•1 + x2 a•2 + ··· + xn a•n

A linear combination of the columns.

Matrix Vector Multiplication

A column space view of matrix vector multiplication.

 a11 a12 ... a1n   x1   a21 a22 ... a2n   x2       . . .. .   .   . . . .   .  am1 am2 ... amn xn

Lecture 1: Linear Algebra Review Linear Programming 4 / 24  a12   a1n   a22   a2n  + x  + ··· + x   2  .  n  .   .   .  am2 amn

= x1 a•1 + x2 a•2 + ··· + xn a•n

A linear combination of the columns.

Matrix Vector Multiplication

A column space view of matrix vector multiplication.

 a11 a12 ... a1n   x1   a11   a21 a22 ... a2n   x2   a21      = x    . . .. .   .  1  .   . . . .   .   .  am1 am2 ... amn xn am1

Lecture 1: Linear Algebra Review Linear Programming 4 / 24  a1n   a2n  + ··· + x   n  .   .  amn

= x1 a•1 + x2 a•2 + ··· + xn a•n

A linear combination of the columns.

Matrix Vector Multiplication

A column space view of matrix vector multiplication.

 a11 a12 ... a1n   x1   a11   a12   a21 a22 ... a2n   x2   a21   a22      = x  + x    . . .. .   .  1  .  2  .   . . . .   .   .   .  am1 am2 ... amn xn am1 am2

Lecture 1: Linear Algebra Review Linear Programming 4 / 24 = x1 a•1 + x2 a•2 + ··· + xn a•n

A linear combination of the columns.

Matrix Vector Multiplication

A column space view of matrix vector multiplication.

 a11 a12 ... a1n   x1   a11   a12   a1n   a21 a22 ... a2n   x2   a21   a22   a2n      = x  + x  + ··· + x    . . .. .   .  1  .  2  .  n  .   . . . .   .   .   .   .  am1 am2 ... amn xn am1 am2 amn

Lecture 1: Linear Algebra Review Linear Programming 4 / 24 A linear combination of the columns.

Matrix Vector Multiplication

A column space view of matrix vector multiplication.

= x1 a•1 + x2 a•2 + ··· + xn a•n

Lecture 1: Linear Algebra Review Linear Programming 4 / 24 Matrix Vector Multiplication

A column space view of matrix vector multiplication.

= x1 a•1 + x2 a•2 + ··· + xn a•n

A linear combination of the columns.

Lecture 1: Linear Algebra Review Linear Programming 4 / 24 Range of A

m n Ran (A) = {y ∈ R | ∃ x ∈ R such that y = Ax }

Ran (A) = the linear span of the columns of A

The Range of a Matrix

Let A ∈ Rm×n (an m × n matrix having real entries).

Lecture 1: Linear Algebra Review Linear Programming 5 / 24 Ran (A) = the linear span of the columns of A

The Range of a Matrix

Let A ∈ Rm×n (an m × n matrix having real entries).

Range of A

m n Ran (A) = {y ∈ R | ∃ x ∈ R such that y = Ax }

Lecture 1: Linear Algebra Review Linear Programming 5 / 24 The Range of a Matrix

Let A ∈ Rm×n (an m × n matrix having real entries).

Range of A

m n Ran (A) = {y ∈ R | ∃ x ∈ R such that y = Ax }

Ran (A) = the linear span of the columns of A

Lecture 1: Linear Algebra Review Linear Programming 5 / 24 The linear span of v1,..., vk :

Span [v1,..., vk ] = {y | y = ξ1v1 + ξ2v2 + ··· + ξk vk , ξ1, . . . , ξk ∈ R}

The subspace orthogonal to v1,..., vk :

⊥ n {v1,..., vk } = {z ∈ R | z • vi = 0, i = 1,..., k }

⊥ ⊥ Facts: {v1,..., vk } = Span [v1,..., vk ] ⊥ h ⊥i Span [v1,..., vk ] = Span [v1,..., vk ]

Two Special Subspaces

n Let v1,..., vk ∈ R .

Lecture 1: Linear Algebra Review Linear Programming 6 / 24 The subspace orthogonal to v1,..., vk :

⊥ n {v1,..., vk } = {z ∈ R | z • vi = 0, i = 1,..., k }

⊥ ⊥ Facts: {v1,..., vk } = Span [v1,..., vk ] ⊥ h ⊥i Span [v1,..., vk ] = Span [v1,..., vk ]

Two Special Subspaces

n Let v1,..., vk ∈ R .

The linear span of v1,..., vk :

Span [v1,..., vk ] = {y | y = ξ1v1 + ξ2v2 + ··· + ξk vk , ξ1, . . . , ξk ∈ R}

Lecture 1: Linear Algebra Review Linear Programming 6 / 24 ⊥ ⊥ Facts: {v1,..., vk } = Span [v1,..., vk ] ⊥ h ⊥i Span [v1,..., vk ] = Span [v1,..., vk ]

Two Special Subspaces

n Let v1,..., vk ∈ R .

The linear span of v1,..., vk :

Span [v1,..., vk ] = {y | y = ξ1v1 + ξ2v2 + ··· + ξk vk , ξ1, . . . , ξk ∈ R}

The subspace orthogonal to v1,..., vk :

⊥ n {v1,..., vk } = {z ∈ R | z • vi = 0, i = 1,..., k }

Lecture 1: Linear Algebra Review Linear Programming 6 / 24 ⊥ ⊥ {v1,..., vk } = Span [v1,..., vk ] ⊥ h ⊥i Span [v1,..., vk ] = Span [v1,..., vk ]

Two Special Subspaces

n Let v1,..., vk ∈ R .

The linear span of v1,..., vk :

Span [v1,..., vk ] = {y | y = ξ1v1 + ξ2v2 + ··· + ξk vk , ξ1, . . . , ξk ∈ R}

The subspace orthogonal to v1,..., vk :

⊥ n {v1,..., vk } = {z ∈ R | z • vi = 0, i = 1,..., k }

Facts:

Lecture 1: Linear Algebra Review Linear Programming 6 / 24 ⊥ h ⊥i Span [v1,..., vk ] = Span [v1,..., vk ]

Two Special Subspaces

n Let v1,..., vk ∈ R .

The linear span of v1,..., vk :

Span [v1,..., vk ] = {y | y = ξ1v1 + ξ2v2 + ··· + ξk vk , ξ1, . . . , ξk ∈ R}

The subspace orthogonal to v1,..., vk :

⊥ n {v1,..., vk } = {z ∈ R | z • vi = 0, i = 1,..., k }

⊥ ⊥ Facts: {v1,..., vk } = Span [v1,..., vk ]

Lecture 1: Linear Algebra Review Linear Programming 6 / 24 Two Special Subspaces

n Let v1,..., vk ∈ R .

The linear span of v1,..., vk :

Span [v1,..., vk ] = {y | y = ξ1v1 + ξ2v2 + ··· + ξk vk , ξ1, . . . , ξk ∈ R}

The subspace orthogonal to v1,..., vk :

⊥ n {v1,..., vk } = {z ∈ R | z • vi = 0, i = 1,..., k }

⊥ ⊥ Facts: {v1,..., vk } = Span [v1,..., vk ] ⊥ h ⊥i Span [v1,..., vk ] = Span [v1,..., vk ]

Lecture 1: Linear Algebra Review Linear Programming 6 / 24 Matrix Vector Multiplication

A row space view of matrix vector multiplication.

       Pn  a11 a12 ... a1n x1 a1• • x i=1 a1i xi Pn  a21 a22 ... a2n   x2   a2• • x   i=1 a2i xi      =   =    . . .. .   .   .   .   . . . .   .   .   .  Pn am1 am2 ... amn xn am• • x i=1 ami xi

The dot product of x with the rows of A.

Lecture 1: Linear Algebra Review Linear Programming 7 / 24 Null Space of A n Nul (A) = {x ∈ R | Ax = 0}

Nul (A) = subspace orthogonal to the rows of A ⊥ = Span [a1•, a2•,..., am•] ⊥ = Ran AT

Fundamental Theorem of the Alternative:

⊥ ⊥ Nul (A) = Ran AT Ran (A) = Nul AT

The Null Space of a Matrix

Let A ∈ Rm×n (an m × n matrix having real entries).

Lecture 1: Linear Algebra Review Linear Programming 8 / 24 Nul (A) = subspace orthogonal to the rows of A ⊥ = Span [a1•, a2•,..., am•] ⊥ = Ran AT

Fundamental Theorem of the Alternative:

⊥ ⊥ Nul (A) = Ran AT Ran (A) = Nul AT

The Null Space of a Matrix

Let A ∈ Rm×n (an m × n matrix having real entries). Null Space of A n Nul (A) = {x ∈ R | Ax = 0}

Lecture 1: Linear Algebra Review Linear Programming 8 / 24 ⊥ = Span [a1•, a2•,..., am•] ⊥ = Ran AT

Fundamental Theorem of the Alternative:

⊥ ⊥ Nul (A) = Ran AT Ran (A) = Nul AT

The Null Space of a Matrix

Let A ∈ Rm×n (an m × n matrix having real entries). Null Space of A n Nul (A) = {x ∈ R | Ax = 0}

Nul (A) = subspace orthogonal to the rows of A

Lecture 1: Linear Algebra Review Linear Programming 8 / 24 ⊥ = Ran AT

Fundamental Theorem of the Alternative:

⊥ ⊥ Nul (A) = Ran AT Ran (A) = Nul AT

The Null Space of a Matrix

Let A ∈ Rm×n (an m × n matrix having real entries). Null Space of A n Nul (A) = {x ∈ R | Ax = 0}

Nul (A) = subspace orthogonal to the rows of A ⊥ = Span [a1•, a2•,..., am•]

Lecture 1: Linear Algebra Review Linear Programming 8 / 24 Fundamental Theorem of the Alternative:

⊥ ⊥ Nul (A) = Ran AT Ran (A) = Nul AT

The Null Space of a Matrix

Let A ∈ Rm×n (an m × n matrix having real entries). Null Space of A n Nul (A) = {x ∈ R | Ax = 0}

Nul (A) = subspace orthogonal to the rows of A ⊥ = Span [a1•, a2•,..., am•] ⊥ = Ran AT

Lecture 1: Linear Algebra Review Linear Programming 8 / 24 The Null Space of a Matrix

Let A ∈ Rm×n (an m × n matrix having real entries). Null Space of A n Nul (A) = {x ∈ R | Ax = 0}

Nul (A) = subspace orthogonal to the rows of A ⊥ = Span [a1•, a2•,..., am•] ⊥ = Ran AT

Fundamental Theorem of the Alternative:

⊥ ⊥ Nul (A) = Ran AT Ran (A) = Nul AT

Lecture 1: Linear Algebra Review Linear Programming 8 / 24  3 −4 1 1 0 0   2 2 0 0 1 0    B I3×3 = −1 0 0 0 0 1 =   02×3 C  0 0 0 2 1 4  0 0 0 1 0 3

where  3 −4 1  2 1 4 B = 2 2 0 , C =   1 0 3 −1 0 0

Block Structured Matrices

 3 −4 1 1 0 0   2 2 0 0 1 0    A= −1 0 0 0 0 1     0 0 0 2 1 4  0 0 0 1 0 3

Lecture 1: Linear Algebra Review Linear Programming 9 / 24 B I = 3×3 02×3 C

where  3 −4 1  2 1 4 B = 2 2 0 , C =   1 0 3 −1 0 0

Block Structured Matrices

 3 −4 1 1 0 0   3 −4 1 1 0 0   2 2 0 0 1 0   2 2 0 0 1 0      A= −1 0 0 0 0 1 = −1 0 0 0 0 1       0 0 0 2 1 4   0 0 0 2 1 4  0 0 0 1 0 3 0 0 0 1 0 3

Lecture 1: Linear Algebra Review Linear Programming 9 / 24 where  3 −4 1  2 1 4 B = 2 2 0 , C =   1 0 3 −1 0 0

Block Structured Matrices

 3 −4 1 1 0 0   3 −4 1 1 0 0   2 2 0 0 1 0   2 2 0 0 1 0      B I3×3 A= −1 0 0 0 0 1 = −1 0 0 0 0 1 =     02×3 C  0 0 0 2 1 4   0 0 0 2 1 4  0 0 0 1 0 3 0 0 0 1 0 3

Lecture 1: Linear Algebra Review Linear Programming 9 / 24 Block Structured Matrices

where  3 −4 1  2 1 4 B = 2 2 0 , C =   1 0 3 −1 0 0

Lecture 1: Linear Algebra Review Linear Programming 9 / 24 Can we exploit the structure of A?

Multiplication of Block Structured Matrices

Consider the matrix product AM, where

 1 2   3 −4 1 1 0 0  0 4 2 2 0 0 1 0      −1 −1  A =  −1 0 0 0 0 1  and M =      2 −1   0 0 0 2 1 4       4 3  0 0 0 1 0 3   −2 0

Lecture 1: Linear Algebra Review Linear Programming 10 / 24 Multiplication of Block Structured Matrices

Consider the matrix product AM, where

 1 2   3 −4 1 1 0 0  0 4 2 2 0 0 1 0      −1 −1  A =  −1 0 0 0 0 1  and M =      2 −1   0 0 0 2 1 4       4 3  0 0 0 1 0 3   −2 0 Can we exploit the structure of A?

Lecture 1: Linear Algebra Review Linear Programming 10 / 24 BI X A = 3×3 so take M = , 02×3 C Y

 1 2   2 −1  where X =  0 4  , and Y =  4 3  . −1 −1 −2 0

Multiplication of Block Structured Matrices

Consider the matrix product AM, where

Lecture 1: Linear Algebra Review Linear Programming 11 / 24 X so take M = , Y

 1 2   2 −1  where X =  0 4  , and Y =  4 3  . −1 −1 −2 0

Multiplication of Block Structured Matrices

Consider the matrix product AM, where

BI A = 3×3 02×3 C

Lecture 1: Linear Algebra Review Linear Programming 11 / 24  1 2   2 −1  where X =  0 4  , and Y =  4 3  . −1 −1 −2 0

Multiplication of Block Structured Matrices

Consider the matrix product AM, where

BI X A = 3×3 so take M = , 02×3 C Y

Lecture 1: Linear Algebra Review Linear Programming 11 / 24 Multiplication of Block Structured Matrices

Consider the matrix product AM, where

BI X A = 3×3 so take M = , 02×3 C Y

 1 2   2 −1  where X =  0 4  , and Y =  4 3  . −1 −1 −2 0

Lecture 1: Linear Algebra Review Linear Programming 11 / 24 BX + Y = CY

  2 −11   2 −1     2 12  +  4 3      −1 −2 −2 0  =        4 1  −4 −1

 4 −12   6 15    =  1 −2  .    4 1  −4 −1

Multiplication of Block Structured Matrices

BI X AM = 3×3 02×3 C Y

Lecture 1: Linear Algebra Review Linear Programming 12 / 24   2 −11   2 −1     2 12  +  4 3      −1 −2 −2 0  =        4 1  −4 −1

 4 −12   6 15    =  1 −2  .    4 1  −4 −1

Multiplication of Block Structured Matrices

BI X BX + Y AM = 3×3 = 02×3 C Y CY

Lecture 1: Linear Algebra Review Linear Programming 12 / 24  4 −12   6 15    =  1 −2  .    4 1  −4 −1

Multiplication of Block Structured Matrices

BI X BX + Y AM = 3×3 = 02×3 C Y CY

  2 −11   2 −1     2 12  +  4 3      −1 −2 −2 0  =        4 1  −4 −1

Lecture 1: Linear Algebra Review Linear Programming 12 / 24 Multiplication of Block Structured Matrices

BI X BX + Y AM = 3×3 = 02×3 C Y CY

  2 −11   2 −1     2 12  +  4 3      −1 −2 −2 0  =        4 1  −4 −1

 4 −12   6 15    =  1 −2  .    4 1  −4 −1

Lecture 1: Linear Algebra Review Linear Programming 12 / 24 The solution set is either empty, a single point, or an inﬁnite set

n If a solution x0 ∈ R exists, then the set of solutions is given by

x0 + Nul (A) .

Solving Systems of Linear equations

Let A ∈ Rm×n and b ∈ Rm. Find all solutions x ∈ Rn to the system Ax = b.

Lecture 1: Linear Algebra Review Linear Programming 13 / 24 empty, a single point, or an inﬁnite set

n If a solution x0 ∈ R exists, then the set of solutions is given by

x0 + Nul (A) .

Solving Systems of Linear equations

Let A ∈ Rm×n and b ∈ Rm. Find all solutions x ∈ Rn to the system Ax = b.

The solution set is either .

Lecture 1: Linear Algebra Review Linear Programming 13 / 24 a single point, or an inﬁnite set

n If a solution x0 ∈ R exists, then the set of solutions is given by

x0 + Nul (A) .

Solving Systems of Linear equations

Let A ∈ Rm×n and b ∈ Rm. Find all solutions x ∈ Rn to the system Ax = b.

The solution set is either empty, .

Lecture 1: Linear Algebra Review Linear Programming 13 / 24 or an inﬁnite set

n If a solution x0 ∈ R exists, then the set of solutions is given by

x0 + Nul (A) .

Solving Systems of Linear equations

Let A ∈ Rm×n and b ∈ Rm. Find all solutions x ∈ Rn to the system Ax = b.

The solution set is either empty, a single point, .

Lecture 1: Linear Algebra Review Linear Programming 13 / 24 n If a solution x0 ∈ R exists, then the set of solutions is given by

x0 + Nul (A) .

Solving Systems of Linear equations

Let A ∈ Rm×n and b ∈ Rm. Find all solutions x ∈ Rn to the system Ax = b.

The solution set is either empty, a single point, or an inﬁnite set.

Lecture 1: Linear Algebra Review Linear Programming 13 / 24 Solving Systems of Linear equations

Let A ∈ Rm×n and b ∈ Rm. Find all solutions x ∈ Rn to the system Ax = b.

The solution set is either empty, a single point, or an inﬁnite set.

n If a solution x0 ∈ R exists, then the set of solutions is given by

x0 + Nul (A) .

Lecture 1: Linear Algebra Review Linear Programming 13 / 24 This process is called Gaussian elimination.

The three elementary row operations. 1 Interchange any two rows. 2 Multiply any row by a non-zero constant. 3 Replace any row by itself plus a multiple of any other row.

These elementary row operations can be interpreted as multiplying the augmented matrix on the left by a special nonsingular matrix.

Gaussian Elimination and the 3 Elementary Row Operations

We solve the system Ax = b by transforming the augmented matrix

[ A | b]

into upper echelon form using the three elementary row operations.

Lecture 1: Linear Algebra Review Linear Programming 14 / 24 The three elementary row operations. 1 Interchange any two rows. 2 Multiply any row by a non-zero constant. 3 Replace any row by itself plus a multiple of any other row.

These elementary row operations can be interpreted as multiplying the augmented matrix on the left by a special nonsingular matrix.

Gaussian Elimination and the 3 Elementary Row Operations

We solve the system Ax = b by transforming the augmented matrix

[ A | b]

into upper echelon form using the three elementary row operations.

This process is called Gaussian elimination.

Lecture 1: Linear Algebra Review Linear Programming 14 / 24 1 Interchange any two rows. 2 Multiply any row by a non-zero constant. 3 Replace any row by itself plus a multiple of any other row.

These elementary row operations can be interpreted as multiplying the augmented matrix on the left by a special nonsingular matrix.

Gaussian Elimination and the 3 Elementary Row Operations

We solve the system Ax = b by transforming the augmented matrix

[ A | b]

into upper echelon form using the three elementary row operations.

This process is called Gaussian elimination.

The three elementary row operations.

Lecture 1: Linear Algebra Review Linear Programming 14 / 24 2 Multiply any row by a non-zero constant. 3 Replace any row by itself plus a multiple of any other row.

These elementary row operations can be interpreted as multiplying the augmented matrix on the left by a special nonsingular matrix.

Gaussian Elimination and the 3 Elementary Row Operations

We solve the system Ax = b by transforming the augmented matrix

[ A | b]

into upper echelon form using the three elementary row operations.

This process is called Gaussian elimination.

The three elementary row operations. 1 Interchange any two rows.

Lecture 1: Linear Algebra Review Linear Programming 14 / 24 3 Replace any row by itself plus a multiple of any other row.

These elementary row operations can be interpreted as multiplying the augmented matrix on the left by a special nonsingular matrix.

Gaussian Elimination and the 3 Elementary Row Operations

We solve the system Ax = b by transforming the augmented matrix

[ A | b]

into upper echelon form using the three elementary row operations.

This process is called Gaussian elimination.

The three elementary row operations. 1 Interchange any two rows. 2 Multiply any row by a non-zero constant.

Lecture 1: Linear Algebra Review Linear Programming 14 / 24 These elementary row operations can be interpreted as multiplying the augmented matrix on the left by a special nonsingular matrix.

Gaussian Elimination and the 3 Elementary Row Operations

We solve the system Ax = b by transforming the augmented matrix

[ A | b]

into upper echelon form using the three elementary row operations.

This process is called Gaussian elimination.

The three elementary row operations. 1 Interchange any two rows. 2 Multiply any row by a non-zero constant. 3 Replace any row by itself plus a multiple of any other row.

Lecture 1: Linear Algebra Review Linear Programming 14 / 24 Gaussian Elimination and the 3 Elementary Row Operations

We solve the system Ax = b by transforming the augmented matrix

[ A | b]

into upper echelon form using the three elementary row operations.

This process is called Gaussian elimination.

The three elementary row operations. 1 Interchange any two rows. 2 Multiply any row by a non-zero constant. 3 Replace any row by itself plus a multiple of any other row.

These elementary row operations can be interpreted as multiplying the augmented matrix on the left by a special nonsingular matrix.

Lecture 1: Linear Algebra Review Linear Programming 14 / 24 Multiplying any 4 × n matrix on the left by the exchange matrix

 1 0 0 0   0 0 0 1     0 0 1 0  0 1 0 0

will exchange the second and fourth rows of the matrix.

(multiplication of a m × 4 matrix on the right by this exchanges the second and fourth columns.) A permutation matrix is obtained by permuting the columns of the identity matrix.

Exchange and Permutation Matrices

An exchange matrix is given by permuting any two columns of the identity.

Lecture 1: Linear Algebra Review Linear Programming 15 / 24 (multiplication of a m × 4 matrix on the right by this exchanges the second and fourth columns.) A permutation matrix is obtained by permuting the columns of the identity matrix.

Exchange and Permutation Matrices

An exchange matrix is given by permuting any two columns of the identity.

Multiplying any 4 × n matrix on the left by the exchange matrix

 1 0 0 0   0 0 0 1     0 0 1 0  0 1 0 0

will exchange the second and fourth rows of the matrix.

Lecture 1: Linear Algebra Review Linear Programming 15 / 24 A permutation matrix is obtained by permuting the columns of the identity matrix.

Exchange and Permutation Matrices

An exchange matrix is given by permuting any two columns of the identity.

Multiplying any 4 × n matrix on the left by the exchange matrix

 1 0 0 0   0 0 0 1     0 0 1 0  0 1 0 0

will exchange the second and fourth rows of the matrix.

(multiplication of a m × 4 matrix on the right by this exchanges the second and fourth columns.)

Lecture 1: Linear Algebra Review Linear Programming 15 / 24 Exchange and Permutation Matrices

An exchange matrix is given by permuting any two columns of the identity.

Multiplying any 4 × n matrix on the left by the exchange matrix

 1 0 0 0   0 0 0 1     0 0 1 0  0 1 0 0

will exchange the second and fourth rows of the matrix.

(multiplication of a m × 4 matrix on the right by this exchanges the second and fourth columns.) A permutation matrix is obtained by permuting the columns of the identity matrix.

Lecture 1: Linear Algebra Review Linear Programming 15 / 24 Left Multiplication of A: When multiplying A on the left by an m × m matrix M, it is often useful to think of this as an action on the rows of A.

For example, left multiplication by a permutation matrix permutes the rows of the matrix.

However, mechanically, left multiplication corresponds to matrix vector multiplication on the columns.

MA = [Ma•1 Ma•2 ··· Ma•n]

Notes on Matrix Multiplication

m×n Let A = [aij ]m×n ∈ R .

Lecture 1: Linear Algebra Review Linear Programming 16 / 24 When multiplying A on the left by an m × m matrix M, it is often useful to think of this as an action on the rows of A.

For example, left multiplication by a permutation matrix permutes the rows of the matrix.

However, mechanically, left multiplication corresponds to matrix vector multiplication on the columns.

MA = [Ma•1 Ma•2 ··· Ma•n]

Notes on Matrix Multiplication

m×n Let A = [aij ]m×n ∈ R .

Left Multiplication of A:

Lecture 1: Linear Algebra Review Linear Programming 16 / 24 For example, left multiplication by a permutation matrix permutes the rows of the matrix.

However, mechanically, left multiplication corresponds to matrix vector multiplication on the columns.

MA = [Ma•1 Ma•2 ··· Ma•n]

Notes on Matrix Multiplication

m×n Let A = [aij ]m×n ∈ R .

Left Multiplication of A: When multiplying A on the left by an m × m matrix M, it is often useful to think of this as an action on the rows of A.

Lecture 1: Linear Algebra Review Linear Programming 16 / 24 However, mechanically, left multiplication corresponds to matrix vector multiplication on the columns.

MA = [Ma•1 Ma•2 ··· Ma•n]

Notes on Matrix Multiplication

m×n Let A = [aij ]m×n ∈ R .

Left Multiplication of A: When multiplying A on the left by an m × m matrix M, it is often useful to think of this as an action on the rows of A.

For example, left multiplication by a permutation matrix permutes the rows of the matrix.

Lecture 1: Linear Algebra Review Linear Programming 16 / 24 Notes on Matrix Multiplication

m×n Let A = [aij ]m×n ∈ R .

Left Multiplication of A: When multiplying A on the left by an m × m matrix M, it is often useful to think of this as an action on the rows of A.

For example, left multiplication by a permutation matrix permutes the rows of the matrix.

However, mechanically, left multiplication corresponds to matrix vector multiplication on the columns.

MA = [Ma•1 Ma•2 ··· Ma•n]

Lecture 1: Linear Algebra Review Linear Programming 16 / 24 Right Multiplication of A: When multiplying A on the right by an n × n matrix N, it is often useful to think of this as an action on the columns of A.

For example, right multiplication by a permutation matrix permutes the columns of the matrix.

 a N  However, mechanically, right 1•  a2•N  multiplication corresponds to left AN =    .  matrix vector multiplication on  .  the rows. am•N

Notes on Matrix Multiplication

m×n Let A = [aij ]m×n ∈ R .

Lecture 1: Linear Algebra Review Linear Programming 17 / 24 When multiplying A on the right by an n × n matrix N, it is often useful to think of this as an action on the columns of A.

For example, right multiplication by a permutation matrix permutes the columns of the matrix.

 a N  However, mechanically, right 1•  a2•N  multiplication corresponds to left AN =    .  matrix vector multiplication on  .  the rows. am•N

Notes on Matrix Multiplication

m×n Let A = [aij ]m×n ∈ R .

Right Multiplication of A:

Lecture 1: Linear Algebra Review Linear Programming 17 / 24 For example, right multiplication by a permutation matrix permutes the columns of the matrix.

 a N  However, mechanically, right 1•  a2•N  multiplication corresponds to left AN =    .  matrix vector multiplication on  .  the rows. am•N

Notes on Matrix Multiplication

m×n Let A = [aij ]m×n ∈ R .

Right Multiplication of A: When multiplying A on the right by an n × n matrix N, it is often useful to think of this as an action on the columns of A.

Lecture 1: Linear Algebra Review Linear Programming 17 / 24  a N  However, mechanically, right 1•  a2•N  multiplication corresponds to left AN =    .  matrix vector multiplication on  .  the rows. am•N

Notes on Matrix Multiplication

m×n Let A = [aij ]m×n ∈ R .

Right Multiplication of A: When multiplying A on the right by an n × n matrix N, it is often useful to think of this as an action on the columns of A.

For example, right multiplication by a permutation matrix permutes the columns of the matrix.

Lecture 1: Linear Algebra Review Linear Programming 17 / 24   a1•N  a2•N  AN =    .   .  am•N

Notes on Matrix Multiplication

m×n Let A = [aij ]m×n ∈ R .

Right Multiplication of A: When multiplying A on the right by an n × n matrix N, it is often useful to think of this as an action on the columns of A.

For example, right multiplication by a permutation matrix permutes the columns of the matrix.

However, mechanically, right multiplication corresponds to left matrix vector multiplication on the rows.

Lecture 1: Linear Algebra Review Linear Programming 17 / 24 Notes on Matrix Multiplication

m×n Let A = [aij ]m×n ∈ R .

Right Multiplication of A: When multiplying A on the right by an n × n matrix N, it is often useful to think of this as an action on the columns of A.

For example, right multiplication by a permutation matrix permutes the columns of the matrix.

 a N  However, mechanically, right 1•  a2•N  multiplication corresponds to left AN =    .  matrix vector multiplication on  .  the rows. am•N

Lecture 1: Linear Algebra Review Linear Programming 17 / 24 This can be accomplished by left matrix multiplication as follows.

Gaussian Elimination Matrices

The key step in Gaussian elimination is to transform a vector of the form

 a   α  , b

where a ∈ Rk , 0 6= α ∈ R, and b ∈ Rn−k−1, into one of the form  a   α  . 0

Lecture 1: Linear Algebra Review Linear Programming 18 / 24 Gaussian Elimination Matrices

The key step in Gaussian elimination is to transform a vector of the form

 a   α  , b

where a ∈ Rk , 0 6= α ∈ R, and b ∈ Rn−k−1, into one of the form  a   α  . 0

This can be accomplished by left matrix multiplication as follows.

Lecture 1: Linear Algebra Review Linear Programming 18 / 24 a α 0

  =   .

Gaussian Elimination Matrices

a ∈ Rk , 0 6= α ∈ R, and b ∈ Rn−k−1

    Ik×k 0 0 a  0 1 0   α  −1 0 −α b I(n−k−1)×(n−k−1) b

Lecture 1: Linear Algebra Review Linear Programming 19 / 24 a α 0

Gaussian Elimination Matrices

a ∈ Rk , 0 6= α ∈ R, and b ∈ Rn−k−1

      Ik×k 0 0 a  0 1 0   α  =   . −1 0 −α b I(n−k−1)×(n−k−1) b

Lecture 1: Linear Algebra Review Linear Programming 19 / 24 α 0

Gaussian Elimination Matrices

a ∈ Rk , 0 6= α ∈ R, and b ∈ Rn−k−1

      Ik×k 0 0 a a  0 1 0   α  =   . −1 0 −α b I(n−k−1)×(n−k−1) b

Lecture 1: Linear Algebra Review Linear Programming 19 / 24 0

Gaussian Elimination Matrices

a ∈ Rk , 0 6= α ∈ R, and b ∈ Rn−k−1

      Ik×k 0 0 a a  0 1 0   α  =  α  . −1 0 −α b I(n−k−1)×(n−k−1) b

Lecture 1: Linear Algebra Review Linear Programming 19 / 24 Gaussian Elimination Matrices

a ∈ Rk , 0 6= α ∈ R, and b ∈ Rn−k−1

      Ik×k 0 0 a a  0 1 0   α  =  α  . −1 0 −α b I(n−k−1)×(n−k−1) b 0

Lecture 1: Linear Algebra Review Linear Programming 19 / 24 This matrix is invertible with inverse   Ik×k 0 0  0 1 0  . −1 0 α b I(n−k−1)×(n−k−1)

Note that a Gaussian elimination matrix and its inverse are both lower triangular matrices.

Gaussian Elimination Matrices

The matrix   Ik×k 0 0  0 1 0  −1 0 −α b I(n−k−1)×(n−k−1) is called a Gaussian elimination matrix.

Lecture 1: Linear Algebra Review Linear Programming 20 / 24 Note that a Gaussian elimination matrix and its inverse are both lower triangular matrices.

Gaussian Elimination Matrices

The matrix   Ik×k 0 0  0 1 0  −1 0 −α b I(n−k−1)×(n−k−1) is called a Gaussian elimination matrix.

This matrix is invertible with inverse   Ik×k 0 0  0 1 0  . −1 0 α b I(n−k−1)×(n−k−1)

Lecture 1: Linear Algebra Review Linear Programming 20 / 24 Gaussian Elimination Matrices

The matrix   Ik×k 0 0  0 1 0  −1 0 −α b I(n−k−1)×(n−k−1) is called a Gaussian elimination matrix.

This matrix is invertible with inverse   Ik×k 0 0  0 1 0  . −1 0 α b I(n−k−1)×(n−k−1)

Note that a Gaussian elimination matrix and its inverse are both lower triangular matrices.

Lecture 1: Linear Algebra Review Linear Programming 20 / 24 A subset S of Rn×n is said to be a sub-algebra of Rn×n if S is a subspace of Rn×n, S is closed wrt matrix multiplication, and if M ∈ S is invertible, then M−1 ∈ S.

Matrix Sub-Algebras

Lower (upper) triangular matrices in Rn×n are said to form a sub-algebra of Rn×n.

Lecture 1: Linear Algebra Review Linear Programming 21 / 24 S is a subspace of Rn×n, S is closed wrt matrix multiplication, and if M ∈ S is invertible, then M−1 ∈ S.

Matrix Sub-Algebras

Lower (upper) triangular matrices in Rn×n are said to form a sub-algebra of Rn×n.

A subset S of Rn×n is said to be a sub-algebra of Rn×n if

Lecture 1: Linear Algebra Review Linear Programming 21 / 24 S is closed wrt matrix multiplication, and if M ∈ S is invertible, then M−1 ∈ S.

Matrix Sub-Algebras

Lower (upper) triangular matrices in Rn×n are said to form a sub-algebra of Rn×n.

A subset S of Rn×n is said to be a sub-algebra of Rn×n if S is a subspace of Rn×n,

Lecture 1: Linear Algebra Review Linear Programming 21 / 24 if M ∈ S is invertible, then M−1 ∈ S.

Matrix Sub-Algebras

Lower (upper) triangular matrices in Rn×n are said to form a sub-algebra of Rn×n.

A subset S of Rn×n is said to be a sub-algebra of Rn×n if S is a subspace of Rn×n, S is closed wrt matrix multiplication, and

Lecture 1: Linear Algebra Review Linear Programming 21 / 24 Matrix Sub-Algebras

Lower (upper) triangular matrices in Rn×n are said to form a sub-algebra of Rn×n.

A subset S of Rn×n is said to be a sub-algebra of Rn×n if S is a subspace of Rn×n, S is closed wrt matrix multiplication, and if M ∈ S is invertible, then M−1 ∈ S.

Lecture 1: Linear Algebra Review Linear Programming 21 / 24 1 1 2 0 2 − 2 0 2 5

 1 1 2  A =  2 4 2  . −1 1 3

 1 0 0  Eliminate the ﬁrst column with a G1 =  −2 1 0  Gaussian elimination matrix. 1 0 1

 1 0 0   1 1 2    G1A =  −2 1 0   2 4 2  =   1 0 1 −1 1 3

Gaussian Elimination in Practice

Transformation to echelon (upper triangular) form.

Lecture 1: Linear Algebra Review Linear Programming 22 / 24 1 1 2 0 2 − 2 0 2 5

 1 0 0  Eliminate the ﬁrst column with a G1 =  −2 1 0  Gaussian elimination matrix. 1 0 1

 1 0 0   1 1 2    G1A =  −2 1 0   2 4 2  =   1 0 1 −1 1 3

Gaussian Elimination in Practice

 1 1 2  Transformation to echelon (upper A =  2 4 2  . triangular) form. −1 1 3

Lecture 1: Linear Algebra Review Linear Programming 22 / 24 1 1 2 0 2 − 2 0 2 5

 1 0 0  G1 =  −2 1 0  1 0 1

 1 0 0   1 1 2    G1A =  −2 1 0   2 4 2  =   1 0 1 −1 1 3

Gaussian Elimination in Practice

 1 1 2  Transformation to echelon (upper A =  2 4 2  . triangular) form. −1 1 3

Eliminate the ﬁrst column with a Gaussian elimination matrix.

Lecture 1: Linear Algebra Review Linear Programming 22 / 24 1 1 2 0 2 − 2 0 2 5

 1 0 0   1 1 2    G1A =  −2 1 0   2 4 2  =   1 0 1 −1 1 3

Gaussian Elimination in Practice

 1 1 2  Transformation to echelon (upper A =  2 4 2  . triangular) form. −1 1 3

 1 0 0  Eliminate the ﬁrst column with a G1 =  −2 1 0  Gaussian elimination matrix. 1 0 1

Lecture 1: Linear Algebra Review Linear Programming 22 / 24 1 1 2 0 2 − 2 0 2 5

  =  

Gaussian Elimination in Practice

 1 1 2  Transformation to echelon (upper A =  2 4 2  . triangular) form. −1 1 3

 1 0 0  Eliminate the ﬁrst column with a G1 =  −2 1 0  Gaussian elimination matrix. 1 0 1

 1 0 0   1 1 2  G1A =  −2 1 0   2 4 2  1 0 1 −1 1 3

Lecture 1: Linear Algebra Review Linear Programming 22 / 24 1 1 2 0 2 − 2 0 2 5

Gaussian Elimination in Practice

 1 1 2  Transformation to echelon (upper A =  2 4 2  . triangular) form. −1 1 3

 1 0 0  Eliminate the ﬁrst column with a G1 =  −2 1 0  Gaussian elimination matrix. 1 0 1

 1 0 0   1 1 2    G1A =  −2 1 0   2 4 2  =   1 0 1 −1 1 3

Lecture 1: Linear Algebra Review Linear Programming 22 / 24 1 2 0 2 − 2 0 2 5

Gaussian Elimination in Practice

 1 1 2  Transformation to echelon (upper A =  2 4 2  . triangular) form. −1 1 3

 1 0 0  Eliminate the ﬁrst column with a G1 =  −2 1 0  Gaussian elimination matrix. 1 0 1

 1 0 0   1 1 2   1  G1A =  −2 1 0   2 4 2  =   1 0 1 −1 1 3

Lecture 1: Linear Algebra Review Linear Programming 22 / 24 2 0 2 − 2 0 2 5

Gaussian Elimination in Practice

 1 1 2  Transformation to echelon (upper A =  2 4 2  . triangular) form. −1 1 3

 1 0 0  Eliminate the ﬁrst column with a G1 =  −2 1 0  Gaussian elimination matrix. 1 0 1

 1 0 0   1 1 2   1 1  G1A =  −2 1 0   2 4 2  =   1 0 1 −1 1 3

Lecture 1: Linear Algebra Review Linear Programming 22 / 24 0 2 − 2 0 2 5

Gaussian Elimination in Practice

 1 1 2  Transformation to echelon (upper A =  2 4 2  . triangular) form. −1 1 3

 1 0 0  Eliminate the ﬁrst column with a G1 =  −2 1 0  Gaussian elimination matrix. 1 0 1

 1 0 0   1 1 2   1 1 2  G1A =  −2 1 0   2 4 2  =   1 0 1 −1 1 3

Lecture 1: Linear Algebra Review Linear Programming 22 / 24 2 − 2 0 2 5

Gaussian Elimination in Practice

 1 1 2  Transformation to echelon (upper A =  2 4 2  . triangular) form. −1 1 3

 1 0 0  Eliminate the ﬁrst column with a G1 =  −2 1 0  Gaussian elimination matrix. 1 0 1

 1 0 0   1 1 2   1 1 2  G1A =  −2 1 0   2 4 2  =  0  1 0 1 −1 1 3

Lecture 1: Linear Algebra Review Linear Programming 22 / 24 − 2 0 2 5

Gaussian Elimination in Practice

 1 1 2  Transformation to echelon (upper A =  2 4 2  . triangular) form. −1 1 3

 1 0 0  Eliminate the ﬁrst column with a G1 =  −2 1 0  Gaussian elimination matrix. 1 0 1

 1 0 0   1 1 2   1 1 2  G1A =  −2 1 0   2 4 2  =  0 2  1 0 1 −1 1 3

Lecture 1: Linear Algebra Review Linear Programming 22 / 24 0 2 5

Gaussian Elimination in Practice

 1 1 2  Transformation to echelon (upper A =  2 4 2  . triangular) form. −1 1 3

 1 0 0  Eliminate the ﬁrst column with a G1 =  −2 1 0  Gaussian elimination matrix. 1 0 1

 1 0 0   1 1 2   1 1 2  G1A =  −2 1 0   2 4 2  =  0 2 − 2  1 0 1 −1 1 3

Lecture 1: Linear Algebra Review Linear Programming 22 / 24 Gaussian Elimination in Practice

 1 1 2  Transformation to echelon (upper A =  2 4 2  . triangular) form. −1 1 3

 1 0 0  Eliminate the ﬁrst column with a G1 =  −2 1 0  Gaussian elimination matrix. 1 0 1

 1 0 0   1 1 2   1 1 2  G1A =  −2 1 0   2 4 2  =  0 2 − 2  1 0 1 −1 1 3 0 2 5

Lecture 1: Linear Algebra Review Linear Programming 22 / 24 1 1 2 0 2 − 2 0 0 7

 1 0 0  G2 =  0 1 0  0 −1 1

 1 0 0   1 1 2     0 1 0   0 2 −2  =   0 −1 1 0 2 5

Gaussian Elimination in Practice

Now do Gaussian eliminiation on the second column.  1 1 2   0 2 −2  0 2 5

Lecture 1: Linear Algebra Review Linear Programming 23 / 24 1 1 2 0 2 − 2 0 0 7

 1 0 0   1 1 2     0 1 0   0 2 −2  =   0 −1 1 0 2 5

Gaussian Elimination in Practice

Now do Gaussian eliminiation on the second column.  1 1 2   1 0 0   0 2 −2  G2 =  0 1 0  0 2 5 0 −1 1

Lecture 1: Linear Algebra Review Linear Programming 23 / 24 1 1 2 0 2 − 2 0 0 7

Gaussian Elimination in Practice

Now do Gaussian eliminiation on the second column.  1 1 2   1 0 0   0 2 −2  G2 =  0 1 0  0 2 5 0 −1 1

 1 0 0   1 1 2     0 1 0   0 2 −2  =   0 −1 1 0 2 5

Lecture 1: Linear Algebra Review Linear Programming 23 / 24 1 2 0 2 − 2 0 0 7

Gaussian Elimination in Practice

Now do Gaussian eliminiation on the second column.  1 1 2   1 0 0   0 2 −2  G2 =  0 1 0  0 2 5 0 −1 1

 1 0 0   1 1 2   1   0 1 0   0 2 −2  =   0 −1 1 0 2 5

Lecture 1: Linear Algebra Review Linear Programming 23 / 24 1 2 2 − 2 0 7

Gaussian Elimination in Practice

Now do Gaussian eliminiation on the second column.  1 1 2   1 0 0   0 2 −2  G2 =  0 1 0  0 2 5 0 −1 1

 1 0 0   1 1 2   1   0 1 0   0 2 −2  =  0  0 −1 1 0 2 5 0

Lecture 1: Linear Algebra Review Linear Programming 23 / 24 2 2 − 2 0 7

Gaussian Elimination in Practice

Now do Gaussian eliminiation on the second column.  1 1 2   1 0 0   0 2 −2  G2 =  0 1 0  0 2 5 0 −1 1

 1 0 0   1 1 2   1 1   0 1 0   0 2 −2  =  0  0 −1 1 0 2 5 0

Lecture 1: Linear Algebra Review Linear Programming 23 / 24 2 − 2 0 7

Gaussian Elimination in Practice

Now do Gaussian eliminiation on the second column.  1 1 2   1 0 0   0 2 −2  G2 =  0 1 0  0 2 5 0 −1 1

 1 0 0   1 1 2   1 1 2   0 1 0   0 2 −2  =  0  0 −1 1 0 2 5 0

Lecture 1: Linear Algebra Review Linear Programming 23 / 24 − 2 0 7

Gaussian Elimination in Practice

Now do Gaussian eliminiation on the second column.  1 1 2   1 0 0   0 2 −2  G2 =  0 1 0  0 2 5 0 −1 1

 1 0 0   1 1 2   1 1 2   0 1 0   0 2 −2  =  0 2  0 −1 1 0 2 5 0

Lecture 1: Linear Algebra Review Linear Programming 23 / 24 0 7

Gaussian Elimination in Practice

Now do Gaussian eliminiation on the second column.  1 1 2   1 0 0   0 2 −2  G2 =  0 1 0  0 2 5 0 −1 1

 1 0 0   1 1 2   1 1 2   0 1 0   0 2 −2  =  0 2 − 2  0 −1 1 0 2 5 0

Lecture 1: Linear Algebra Review Linear Programming 23 / 24 Gaussian Elimination in Practice

Now do Gaussian eliminiation on the second column.  1 1 2   1 0 0   0 2 −2  G2 =  0 1 0  0 2 5 0 −1 1

 1 0 0   1 1 2   1 1 2   0 1 0   0 2 −2  =  0 2 − 2  0 −1 1 0 2 5 0 0 7

Lecture 1: Linear Algebra Review Linear Programming 23 / 24 0 1 0

  =   .

What is the inverse of this matrix?   Ik×k a 0  0 α 0  . 0 b I(n−k−1)×(n−k−1)

Gauss-Jordan Elimination, or Pivot Matrices

What happens in the following multiplication?

 −1    Ik×k −α a 0 a  0 α−1 0   α  −1 0 −α b I(n−k−1)×(n−k−1) b

Lecture 1: Linear Algebra Review Linear Programming 24 / 24 0 1 0

What is the inverse of this matrix?   Ik×k a 0  0 α 0  . 0 b I(n−k−1)×(n−k−1)

Gauss-Jordan Elimination, or Pivot Matrices

What happens in the following multiplication?

 −1      Ik×k −α a 0 a  0 α−1 0   α  =   . −1 0 −α b I(n−k−1)×(n−k−1) b

Lecture 1: Linear Algebra Review Linear Programming 24 / 24 1 0

What is the inverse of this matrix?   Ik×k a 0  0 α 0  . 0 b I(n−k−1)×(n−k−1)

Gauss-Jordan Elimination, or Pivot Matrices

What happens in the following multiplication?

 −1      Ik×k −α a 0 a 0  0 α−1 0   α  =   . −1 0 −α b I(n−k−1)×(n−k−1) b

Lecture 1: Linear Algebra Review Linear Programming 24 / 24 0

What is the inverse of this matrix?   Ik×k a 0  0 α 0  . 0 b I(n−k−1)×(n−k−1)

Gauss-Jordan Elimination, or Pivot Matrices

What happens in the following multiplication?

 −1      Ik×k −α a 0 a 0  0 α−1 0   α  =  1  . −1 0 −α b I(n−k−1)×(n−k−1) b

Lecture 1: Linear Algebra Review Linear Programming 24 / 24 What is the inverse of this matrix?   Ik×k a 0  0 α 0  . 0 b I(n−k−1)×(n−k−1)

Gauss-Jordan Elimination, or Pivot Matrices

What happens in the following multiplication?

 −1      Ik×k −α a 0 a 0  0 α−1 0   α  =  1  . −1 0 −α b I(n−k−1)×(n−k−1) b 0

Lecture 1: Linear Algebra Review Linear Programming 24 / 24   Ik×k a 0  0 α 0  . 0 b I(n−k−1)×(n−k−1)

Gauss-Jordan Elimination, or Pivot Matrices

What happens in the following multiplication?

 −1      Ik×k −α a 0 a 0  0 α−1 0   α  =  1  . −1 0 −α b I(n−k−1)×(n−k−1) b 0

What is the inverse of this matrix?

Lecture 1: Linear Algebra Review Linear Programming 24 / 24 Gauss-Jordan Elimination, or Pivot Matrices

What happens in the following multiplication?

 −1      Ik×k −α a 0 a 0  0 α−1 0   α  =  1  . −1 0 −α b I(n−k−1)×(n−k−1) b 0

What is the inverse of this matrix?   Ik×k a 0  0 α 0  . 0 b I(n−k−1)×(n−k−1)

Lecture 1: Linear Algebra Review Linear Programming 24 / 24