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anonymous
 4 years ago
Can someone help me prove a property of vectors
anonymous
 4 years ago
Can someone help me prove a property of vectors

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Prov ethat cu is a vector in R^n

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0c is a scaler and u is a vector

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0imagine c is\[ai+bj\] if we multiply it in c in the end we have a vector

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ummm my prof wldnt accept that

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hence vector have dirction & magnitude so cu it has too characteristic get it? could i explain well?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1As \( u \in \mathbb{R}^n \) we can write \( u \) as \[ u = (x_1, x_2, ..., x_n) \] where each of the \( x_i \) are real numbers. Now by definition of scalar multiplication, \[ cu = (cx_1, cx_2, ..., cx_n) \] As each \( x_i \) is a real number as is \( c \), each component \( cx_i \) is also a real number. Hence \( cu \) is an \( n\)tuple of real numbers and therefore a member of \( \mathbb{R}^n \).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yup that is what i was looking for :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks guys :D I still need to prove 4 more properties so I may be back

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0jamsj answer is better than mine Pippa

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1Imitate the method here then. Show explicitly that the resulting quantity meets exactly the definition required. good luck.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks james I finished all the proving :D On my own which is a big feat for me. I think I am getting the hang of it
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