lina-27-11-09-p1 - independent and v 1 , v 2 ∈ R 3 are...

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Linear Algebra Geometry: Sheet 8 Set on Friday, November 27: Questions 1, 2, 3 and 5 1. Show that (a) if T : R n R m and R,S : R m R k are linear maps, then ( R + S ) T = R T + S T (b) if T : R n R m , S : R m R l and R : R l R k are linear maps then ( R S ) T = R ( S T ) . 2. Let T : R n R m be a linear map, show that (a) ker T := { x R n ; T ( x ) = 0 } is a linear subspace of R n , (b) if V R n is a linear subspace, then T ( V ) := { y R m ; there exist a x R n with T ( x ) = y } is a linear subspace of R m , and with T ( R n ) = Im T conclude that Im T is a linear subspace of R m . 3. Compute the kernel and image of the following linear maps: (a) T : R 3 R , T ( x ) = x · u , where u = (1 , 2 , - 1). (b) T : R R 3 , T ( x ) = x u where u = (1 , 2 , - 1). (c) T : R 2 R 2 , T ( x,y ) = ( x - y, 5 x ). (d) T : R 3 R 3 , T ( x ) = ( u 1 · x ) v 1 + ( u 2 · x ) v 2 , where u 1 , u 2 R 3 are linearly
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Unformatted text preview: independent and v 1 , v 2 ∈ R 3 are linearly independent, too. Relate the image and kernel of T to the subspaces V u := span { u 1 , u 2 } and V v := span { v 1 , v 2 } . 4. Show that if T : R n → R m and S : R m → R l are linear maps with Im T ⊂ ker S , then S ◦ T = 0. 5. Assume T : R n → R n and S : R n → R n are bijective linear maps, i.e., ker T = ker S = { } and Im T = Im S = R n . Show that R := S ◦ T is bijective and that R-1 = T-1 ◦ S-1 . 1...
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This note was uploaded on 02/25/2010 for the course MATHEMATIC 11007 taught by Professor Spyros during the Spring '10 term at Bristol Community College.

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