Lect2-09-web

# Lect2-09-web - MATH 304 Fall 2011 Linear Algebra Homework...

This preview shows pages 1–8. Sign up to view the full content.

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

View Full Document

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

View Full Document

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

View Full Document

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH 304, Fall 2011 Linear Algebra Homework assignment #7 (due Thursday, October 27) All problems are from Leon’s book. Section 4.1: 4, 7c, 7d, 17c, 19a Section 4.2: 2c, 3c, 4a, 6, 15 MATH 304 Linear Algebra Lecture 16: Matrix transformations (continued). Matrix of a linear transformation. Linear transformation Definition. Given vector spaces V 1 and V 2 , a mapping L : V 1 → V 2 is linear if L ( x + y ) = L ( x ) + L ( y ), L ( r x ) = rL ( x ) for any x , y ∈ V 1 and r ∈ R . Basic properties of linear mappings: • L ( r 1 v 1 + ··· + r k v k ) = r 1 L ( v 1 ) + ··· + r k L ( v k ) for all k ≥ 1, v 1 , . . . , v k ∈ V 1 , and r 1 , . . . , r k ∈ R . • L ( 1 ) = 2 , where 1 and 2 are zero vectors in V 1 and V 2 , respectively. • L ( − v ) = − L ( v ) for any v ∈ V 1 . Matrix transformations Any m × n matrix A gives rise to a transformation L : R n → R m given by L ( x ) = A x , where x ∈ R n and L ( x ) ∈ R m are regarded as column vectors. This transformation is linear . Example. L x y z = 1 0 2 3 4 7 0 5 8 x y z . Let e 1 = (1 , , 0), e 2 = (0 , 1 , 0), e 3 = (0 , , 1) be the standard basis for R 3 . We have that L ( e 1 ) = (1 , 3 , 0), L ( e 2 ) = (0 , 4 , 5), L ( e 3 ) = (2 , 7 , 8). Thus L ( e 1 ) , L ( e 2 ) , L ( e 3 ) are columns of the matrix. Problem. Find a linear mapping L : R 3 → R 2 such that L ( e 1 ) = (1 , 1), L ( e 2 ) = (0 , − 2), L ( e 3 ) = (3 , 0), where e 1 , e 2 , e 3 is the standard basis for R 3 . L ( x , y , z ) = L ( x e 1 + y e 2 + z e 3 ) = xL ( e 1 ) + yL ( e 2 ) + zL ( e 3 ) = x (1 , 1) + y (0 , − 2) + z (3 , 0) = ( x + 3 z , x − 2 y ) L ( x , y , z ) = parenleftbigg x + 3 z x − 2 y parenrightbigg = parenleftbigg 1 0 3 1 − 2 0 parenrightbigg x y z Columns of the matrix are vectors L ( e 1 ) , L ( e 2 ) , L ( e 3 ). Theorem Suppose L : R n → R m is a linear map. Then there exists an m × n matrix A such that L ( x ) = A x for all x ∈ R n . Columns of A are vectors L ( e 1 ) , L ( e 2 ) , . . . , L ( e n ), where e 1 , e 2 , . . . , e n is the standard basis for R n . y = A x ⇐⇒ y 1 y 2 ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 34

Lect2-09-web - MATH 304 Fall 2011 Linear Algebra Homework...

This preview shows document pages 1 - 8. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online