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MATH1231 ASSIGNMENT TEST QUESTIONS AND ANSWERS

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The function T : R 2→R3 is defined by: x ( 4 x2 ) x Show that T islinear. A proof of linearity begins by assigning v1 and v2 to be two arbitrary vectorsin the domain of T , and α to be a ... scalar to prove the conditions: i. T ( v1+ v2 )=T (v1 )+T (v2 ) and ii. T ( α v1 )=αT (v1 ) for all scalars α (usually α∈ R ) and vectors v1 ,v2∈T . Let v , v ∈ R 2 . Then, v = ( x1 ) and v = ( y1 ) . Therefore, x + y ( 4 ( x2+ y2 ) 4 x2+4 y2 Now, to prove linearity we must compare the values of T ( v1+v2 ) with T ( v1 )+T(v2 ) . x y 4 x2 4 y2 4 x2+4 y2 T ( v1)+T ( v2)=T ( 1 ) +T ( 1 ) = ( 4 x +5 x ) +( 4 y +5 y ) =( 4 x +4 y +5 x +5 y ) . Note that the first condition T ( v1+v2 )=T (v1 )+T (v2 ) , is satisfied. Now, for α ∈ R T ( α v1)=T ( α x1 ) = ( 4 α x2 4α x +5α x . α x2 1 2 x1 To satisfy the second condition we compare the values of T ( α v1 ) and αT (v1) . x ( 4 x2 ) ( 4α x2 1+4 y +5 x 2+5 y 2 . x 1 1+5 α x2 . x1 T ( 1 ) ∈R 2 . 2 1 T ( v1+ v2 )=T ( + y + y 1 2 1 αT (v1)=αT ( ) ( 1 2 3 ( ) ( ) x3 α x3 Therefore, T (α v1 )=αT (v1 ) . Hence, by proving these two conditions are true, we have shown T is a linear transformation. Question 3 Show that S={x∈R 3 : 6 x +4 x +x =0 } is a subspace of R 3 . To verify that S is a subspace we need to prove: I. Closure under vector addition: v1 , v2∈ S then v1+ v2∈ S . II. Closure under scalar multiplication: v1 ∈ S and α a scalar, then α v1∈ S . III. The existence of a zero vector 0∈S . The remaining seven vector space axioms are inherited from these three. First to prove closure under addition, we let v1 and v2 to be two arbitrary vectors,so v1 ,v2∈S . x1 y1 Then, v1= ( x2) and v2= ( y2 ) .This gives us: x3 y3 6 x1+4 x2+x3 ¿ 0 6 y1+4 y2+ y3 ¿ 0 x1+ y1 Now, v1+ v2= x2+ y2 =6 ( x1+ y1)+4 ( x2+ y2)+( x3+ y3) . x3+ y3 Simplifying this we get, (6 x1+ 4 x2+ x3 )+(6 y1+ 4 y2+ 4 y3 )=0+ 0=0. Therefore, S is closed under vector addition. Proving closure under scalar multiplication requires usto let α∈ R . Then: x1 α x1 α v1=α ( x2) = ( α x2) . This becomes, 6 (α x1 )+4 (α x2 )+α x3=α (6 x1+4 x2+x3 )=α (0)=0. Therefore, S is closed under scalar multiplication. 0 To prove the existence of a zero vector, such that 0∈S , we consider 0= 0 . Since: 0 6 (0)+ 4 (0)+0=0 [Show More]

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