208 - Proof v Rn, 0 form a generating set for Rm. Proof (a)...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
1 Proof v R n , Proof “(a) (d)” By Theorem 1.6, rank A = m if and only if A v = b has at least one solution b R m . 0 form a generating set for R m . Example: with with the reduced row echelon form and rank 2 3 T is not onto. The system of linear equations A x = b may be written as T A ( x) = b . A x = b has a solution if and only if b is in the range of T A . T A is not onto if and only if b such that A x = b is inconsistent .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
one-to-one not one-to-one An equivalent condition for a function f to be one-to-one is that f ( u ) = f ( v ) implies u = v . 0 0 0 0 z Proof “(b) (c)” By the definitions of null space and L.I. . “(c) (d)” By Theorem 1.8. “(a) (b)” It is known that T ( 0 ) = 0 for all linear transforma- tion and T is one-to-one. “(b) (a)” If u , v R n are such that T ( u ) = T ( v ). T ( u v ) = T ( u ) T ( v ) = 0 u v = 0 . 11
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 10/16/2010 for the course EE 155 taught by Professor Fong during the Spring '09 term at National Taiwan University.

Page1 / 5

208 - Proof v Rn, 0 form a generating set for Rm. Proof (a)...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online