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Unformatted text preview: Chapter 4. Linear Transformations Math1111 Matrix Representations Further Example 2 Example . Verify that the dimension formula holds true for the linear transformation L : V → W defined by L ( v ) = . Definition We call such an identically zero linear transformation the zero map . Example 2 . Let L : V → R be a linear transformation, not identically zero. Show that there is a vector v ∈ V such that L ( v ) = 1 and V = ker ( L ) ⊕ Span ( v ) . Chapter 4. Linear Transformations Math1111 Matrix Representations Further Example 2 Example . Verify that the dimension formula holds true for the linear transformation L : V → W defined by L ( v ) = . Definition We call such an identically zero linear transformation the zero map . Example 2 . Let L : V → R be a linear transformation, not identically zero. Show that there is a vector v ∈ V such that L ( v ) = 1 and V = ker ( L ) ⊕ Span ( v ) . Let U and V be subspaces of a vector space. The notation U ⊕ V means the subspace U + V and U ∩ V = { } . We call U ⊕ V the direct sum of U and V . Chapter 4. Linear Transformations Math1111 Matrix Representations Further Example 2 Example . Verify that the dimension formula holds true for the linear transformation L : V → W defined by L ( v ) = . Definition We call such an identically zero linear transformation the zero map . Example 2 . Let L : V → R be a linear transformation, not identically zero. Show that there is a vector v ∈ V such that L ( v ) = 1 and V = ker ( L ) ⊕ Span ( v ) . Definition A linear tranformation L : V → R is called a linear functional . Chapter 4. Linear Transformations Math1111 Matrix Representations Further Example 2 (Cont’d) Ans . As L is not identically zero, there is x ∈ V such that L ( x ) 6 = . Let v = α 1 x where α = L ( x ) ∈ R is nonzero. Chapter 4. Linear Transformations Math1111 Matrix Representations Further Example 2 (Cont’d) Ans . As L is not identically zero, there is x ∈ V such that L ( x ) 6 = . Let v = α 1 x where α = L ( x ) ∈ R is nonzero. Then L ( v ) = 1 . Claim 1: L ( V ) = R . Chapter 4. Linear Transformations Math1111 Matrix Representations Further Example 2 (Cont’d) Ans . As L is not identically zero, there is x ∈ V such that L ( x ) 6 = . Let v = α 1 x where α = L ( x ) ∈ R is nonzero. Then L ( v ) = 1 . Claim 1: L ( V ) = R . Let dim V = n . Then, dimker ( L ) = n dim L ( V ) = n 1 . Let u 1 , ··· , u n 1 form a basis for ker ( L ) . Claim 2: v , u 1 , ··· , u n 1 form a basis for V . Chapter 4. Linear Transformations Math1111 Matrix Representations Further Example 2 (Cont’d) Ans . As L is not identically zero, there is x ∈ V such that L ( x ) 6 = . Let v = α 1 x where α = L ( x ) ∈ R is nonzero....
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 Spring '11

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