## Practice Midterm #4 |
shellyhuo |
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How to prove whether the functions are linear? |

aleksanderjanczewski: May 5, 2015, 7:06 p.m. |
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Do you mean polynomials or regular vectors ? |

dannystapleton: May 5, 2015, 7:32 p.m. |
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Show Closure under addition and closure under scalar multiplication. In other words, T(x + y) = T(x) + T(y) for addtion, and respectively for multiplication: T(kx) = kT(x) |

dannystapleton: May 5, 2015, 7:35 p.m. |
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I found that 4a and 4d were linear functions, based on these two above ^ properties. 4b is non linear because it fails the closure under addition property, and 4c is non linear because it fails the closure under scalar multiplication property. Hope that helps! |

yulduzkhonbruin: May 5, 2015, 8:34 p.m. |
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You have to check if it preserves addition and scalar multiplication. For example, take problem 4b: T(a,b,c)=(a+b, b, 1+a) First, show if it preserves addition. we add (d, e, f) to T(a,b,c)=(a+b, b, 1+a): LHS=>T((a, b, c) + (d, e, f))=T(a, b, c) + T(d, e, f) <=RHS by linear transformation definition. we check the left hand side of the equation first: T(a+d, b+e, c+f)=(a+d+b+e, b+e, 1+a+d) Now, we check the right hand side of the equation: T(a, b, c)+T(d, e, f)=((a+b, b, 1+a))+((d+e, e, 1+d))=(a+d+b+e, b+e, 2+a+d) Notice that LHS is not equal to RHS which means it does not preserve addition. Since it does not preserve addition, it is not linear. |