CS475/CS675 Lecture 1: May 3, 2016 Basic Theory of Linear Algebra - - PowerPoint PPT Presentation

cs475 cs675 lecture 1 may 3 2016
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CS475/CS675 Lecture 1: May 3, 2016 Basic Theory of Linear Algebra - - PowerPoint PPT Presentation

CS475/CS675 Lecture 1: May 3, 2016 Basic Theory of Linear Algebra Reading: [TB] chapt 1 (p. 110), chapt 20 (p. 147152) CS475/CS675 (c) 2016 P. Poupart & J. Wan 1 Range Definition: the range of A is defined as


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CS475/CS675 Lecture 1: May 3, 2016

Basic Theory of Linear Algebra Reading: [TB] chapt 1 (p. 1‐10), chapt 20 (p. 147‐152)

CS475/CS675 (c) 2016 P. Poupart & J. Wan 1

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Range

  • Definition: the range of A is defined as

– : for some

  • Theorem:

– space spanned by the columns of … ⋯ ∑

  • Thus

is also called the column space of

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Rank

  • Definition:

– Column rank = dimension of column space – Row rank = dimension of row space

  • Theorem: column rank = row rank
  • Thus we simply call it the rank of ,

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Full Rank

  • Definition: An

matrix is of full rank if

  • Thus, if

, , then a full rank matrix has n independent column vectors

  • Definition: A nonsingular (invertible) matrix is a

square matrix of full rank

  • Definition: The null space of ,

is defined as

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Matrix inverse

  • Matrix inverse
  • Interesting identity:

Proof:

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Updating Matrix Inverses

  • Sherman‐Morrison‐Woodbury formula

A UV

where

(

n ).

  • Thus a rank

correction to results in a rank correction of the inverse k

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Example

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How to compute ?

  • In numerical linear algebra, NEVER compute

and

then

.

  • We always consider as the solution of the eqn:
  • We compute by solving the equation by Gaussian

Elimination

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Gaussian Elimination (GE)

  • Big picture of GE:

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GE algorithm

For 1,2, … , 1 For 1, … , / ( for 1, … , end end End At the end, is solved by back substitution Update Update RHS

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LU Factorization

  • Theorem:

where

lower , unit diagonal upper

  • mult

1 1

. . .

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Solve

  • Let

, then we have

– Solve by forward solve – Solve by back solve

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Forward solve algorithm

For 1,2, … , For 1,2, … , 1 end end

  • CS475/CS675 (c) 2016 P. Poupart & J. Wan

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Complexity

  • 1 flop = + / ‐ / x /
  • Consider forward solve

– For each , the ‐loop performs 2 1 flops – Total flops ∑ 2 2

  • 2 ∑

∑ 2

  • 2
  • 2

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Complexity (continued)

  • (exercise)
  • For large , factorization is more expensive than

forward or backward solves

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