235_course_notes_chapter2

# 235_course_notes_chapter2 - Chapter 2 Linear Mappings 2.1...

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Unformatted text preview: Chapter 2 Linear Mappings 2.1 General Linear Mappings Linear Mappings L : V → W We now look at linear mappings whose domain is a vector space V and whose codomain is a vector space W . REMARK It is important not to assume that any results that held for linear mappings L : R n → R m also hold for linear mappings L : V → W . DEFINITION Linear Mapping Let V and W be vector spaces. A mapping L : V → W is a linear mapping if L ( svectorx + tvector y ) = sL ( vectorx ) + tL ( vector y ) for all vectorx, vector y ∈ V and s, t ∈ R . EXAMPLE 1 Let L : P 3 → R 2 be defined by L ( a + bx + cx 2 + dx 3 ) = bracketleftbigg a + d b + c bracketrightbigg . Prove that L is linear. Solution: Let a 1 + b 1 x + c 1 x 2 + d 1 x 3 , a 2 + b 2 x + c 2 x 2 + d 2 x 3 ∈ P 3 and s, t ∈ R , then L ( s ( a 1 + b 1 x + c 1 x 2 + d 1 x 3 ) + t ( a 2 + b 2 x + c 2 x 2 + d 2 x 3 ) ) = L ( ( sa 1 + ta 2 ) + ( sb 1 + tb 2 ) x + ( sc 1 + tc 2 ) x 2 + ( sd 1 + td 2 ) x 3 ) = bracketleftbigg sa 1 + ta 2 + sd 1 + td 2 sb 1 + tb 2 + sc 1 + tc 2 bracketrightbigg = s bracketleftbigg a 1 + d 1 b 1 + c 1 bracketrightbigg + t bracketleftbigg a 2 + d 2 b 2 + c 2 bracketrightbigg = sL ( a 1 + b 1 x + c 1 x 2 + d 1 x 3 ) + tL ( a 2 + b 2 x + c 2 x 2 + d 2 x 3 ) Thus, L is linear. 1 2 Chapter 2 Linear Mappings EXAMPLE 2 Prove that tr : M n × n ( R ) → R defined by tr A = n summationdisplay i =1 ( A ) ii is linear (called the trace of a matrix). Solution: Let A, B ∈ M n × n ( R ) and s, t ∈ R . Then tr( sA + tB ) = n summationdisplay i =1 ( sA + tB ) ii = n summationdisplay i =1 ( s ( A ) ii + t ( B ) ii ) = s n summationdisplay i =1 ( A ) ii + t n summationdisplay i =1 ( B ) ii = s tr A + t tr B Thus, tr is linear. EXAMPLE 3 Prove that the mapping L : P 2 → M 2 × 2 ( R ) defined by L ( a + bx + cx 2 ) = bracketleftbigg a bc abc bracketrightbigg is not linear. Solution: Observe that L (1 + x + x 2 ) = bracketleftbigg 1 1 0 1 bracketrightbigg But, L ( 2(1 + x + x 2 ) ) = L (2 + 2 x + 2 x 2 ) = bracketleftbigg 2 4 0 8 bracketrightbigg negationslash = 2 L (1 + x + x 2 ) We now begin to show that many of the results we had for linear mappings L : R n → R m also hold for linear mappings L : V → W . THEOREM 1 Let V and W be vector spaces and let L : V → W be a linear mapping. Then, L ( vector 0) = vector Proof: The proof is left as an exercise. DEFINITION Addition Scalar Multiplication Let L : V → W and M : V → W be linear mappings. Then we define L + M by ( L + M )( vectorv ) = L ( vectorv ) + M ( vectorv ) and for any t ∈ R we define tL by ( tL )( vectorv ) = tL ( vectorv ) Section 2.1 General Linear Mappings 3 THEOREM 2 Let V and W be vector spaces. The set L of all linear mappings L : V → W with standard addition and scalar multiplication of mappings is a vector space....
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235_course_notes_chapter2 - Chapter 2 Linear Mappings 2.1...

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