solution13

# solution13 - M341 Homework Solution 13 Section 5.1 1(c Yes...

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Unformatted text preview: M341 Homework Solution 13 Section 5.1: 1(c): Yes. If x = [ x 1 , x 2 , x 3 ] and y = [ y 1 , y 2 , y 3 ] then k ( x + y ) = k ([ x 1 , x 2 , x 3 ] + [ y 1 , y 2 , y 3 ]) = k ([ x 1 + y 1 , x 2 + y 2 , x 3 + y 3 ]) = [ x 2 + y 2 , x 3 + y 3 , x 1 + y 1 ] = [ x 2 , x 3 , x 1 ] + [ y 2 , y 3 , y 1 ] = k ( x ) + k ( y ) Furthermore if c ∈ R then k ( cx ) = k ([ cx 1 , cx 2 , cx 3 ]) = [ cx 2 , cx 3 , cx 1 ] = c [ x 2 , x 3 , x 1 ] = ck ( x ) So k preserves addition and scalar multiplication so it is a linear transformation. Since its domain and codomain are both R 3 it is also a linear operator. 1(e): No, n is neither a linear transformation nor a linear operator. It does not preserve addition nor scalar multiplication. For example if A = I 2 then n (2 A ) = 4 ̸ = 2 = 2 n ( A ) . 1(g): No, s is neither a linear transformation nor a linear operator. It does not preserve addi- tion nor scalar multiplication. For example 2 s ([ π/ 2 , π, 1] = 2[0 , , e ] = [0 , , 2 e ] but s (2[ π/ 2 , π, 1]) = s ([ π, 2 π, 2]) = [ − 1 , , e 2 ] which are clearly not equal. 4(a): f is a linear transformation because if [ x 1 , y 1 , z 1 ] , [ x 2 , y 2 , z 2 ] and [ x, y, z ] are in R 3 and c is in R then f ([ x 1 , y 1 , z 1 ] + [ x 2 , y 2 , z 2 ]) = f ([ x 1 + x 2 , y 1 + y 2 , z 1 + z 2 ]) = [ − x 1 − x 2 , y 1 + y 2 , z 1 + z 2 ] = [ − x 1 , y 1 , z 1 ] + [ − x 2 , y 2 , z 2 ] = f ([ x 1 , y 1 , z 1 ]) + f ([ x 2 , y 2 , z 2 ]) and cf ([ x, y, z ]) = c [ − x, y, z ] = [ − cx, cy, cz ] = f ([ cx, cy, cz ]) = f ( c [ x, y, z ]) ....
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solution13 - M341 Homework Solution 13 Section 5.1 1(c Yes...

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