571_5.2 - 5.2 ORTHOGONAL SUBSPACES KIAM HEONG KWA pp....

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 5.2 ORTHOGONAL SUBSPACES KIAM HEONG KWA pp. 214-215 Two subspaces X,Y R n are said to be orthogonal , denoted X Orthogonal sub- spaces Y , if h x , y i = x T y = 0 for all x X and all y Y . If X Y , then X Y = { } . To see this, let z X Y , so that k z k 2 = h z , z i = 0 and thus z = . p. 215 Let Y be a subspace of R n . The orthogonal complement of Y is Orthogonal com- plement the set Y = { x R n |h x , y i = x T y = 0 for all y Y } . Clearly, Y , so that Y 6 = . Let x 1 , x 2 Y and let , R . Then for any y Y , h x 1 + x 2 , y i = h x 1 , y i + h x 2 , y i = 0 , so that x 1 + x 2 Y . This shows that Y is a subspace of R n . Exercise 1. Prove that if Y R n is a subspace, then Y ( Y ) . Example 1 (Exercise 5.2.8 in the text) . If S = Span( x 1 , x 2 , , x k ) R n , then for any y R n , y S if and only if y x i for all i { 1 , 2 , ,k } . Proof. If y S , then y x i in view of the definition of S and the fact that x i S for all i { 1 , 2 , ,k } . Conversely, suppose that y x i for all i { 1 , 2 , ,k } . Let x S but otherwise arbitrary. Since S = Span( x 1 , x 2 , , x k ) , there are scalars c i s such that x = k i =1 c i x i . Hence h y , x i = * y , k X i =1 c i x i + = k X i =1 c i h y , x i i = k X i =1 c i 0 = 0 , indicating that y S . pp. 215-216 Let A R m n . Recall that a vector w R m is in the column space of A if and only if A x = w for a vector x R n . Considering A as a map from R n into R m , it follows that such a vector b is in the Date : July 12, 2011. 1 2 KIAM HEONG KWA range R ( A ) of A . In other words, the column space of A is in fact the range of A as a map: R ( A ) = { w R m | A x = w for some x R n } = the column space of A. Likewise, the column space of A T is the range of A T as a map: R ( A T ) = { y R n | A T z = y for some z R m } = the column space of A T . Note that y R ( A T ) if and only if y T is in the row space of A . More important is the fact that Theorem 5.2.1: Fundamental Sub- spaces Theorem N ( A ) = R ( A T ) and N ( A T ) = R ( A ) ....
View Full Document

Page1 / 7

571_5.2 - 5.2 ORTHOGONAL SUBSPACES KIAM HEONG KWA pp....

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