(a) (6 points) It is very important that you become use to the concept of span a
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(a) (6 points) It is very important that you become use to the concept of span and linear inde- pendence as it comes up in other courses such as ordinary differential equations and numer- ical methods for solving differential equations. Let u be in the vector space Span( . 2. i. Using the definition of span, write out what it means for u to be in Spanfvi, v2) ii. Show that u u is also in Span|vi, v2). show that 3u is in Span{vi, v2, va]. linear combinations of w, 2, and w3, show that u v can also be written as a linear (b) (4 points) Let fvi, v2, v3] be a set of three vectors in R. If u is in Spanfvi, v2, va), (c) (4 points) Let u, v, , 2 and ws be vectors in R. If u and v can both be written as combination of wi, w2 and w.Explanation / Answer
2.(a)
i. If the vector u is in Span{v1,v2}, it means that u is a linear combination of the vectors v1,v2, i.e. there exist scalars c1 , c2 (say), not both zero, such that u = c1v1 + c2v2.
ii. If u Span{v1,v2} and u = c1v1 + c2v2, then u+u = c1v1 + c2v2+ c1v1 + c2v2 = 2c1v1 + 2c2v2 . This implies that u+u , being a linear combination of v1,v2 , Span{v1,v2}.
(b). If the vector u is in Span{v1,v2,v3}, it means that u is a linear combination of the vectors v1,v2, v3 i.e. there exist scalars c1 , c2,c3 (say), not all zero, such that u = c1v1 + c2v2+c3v3.Then 3u = 3(c1v1 + c2v2+c3v3) = 3c1v1 + 3c2v2+3c3v3. This implies that 3u, being a linear combination of v1,v2, v3 , Span{v1,v2,v3}.
(c ). Let u = a1w1+a2w2+a3w3 and v = b1w1+b2w2+b3w3 where ai s (1i 3)are scalars , not all zero and bj s (1j 3) are scalars , not all zero. Then u+v = a1w1+a2w2+a3w3 + b1w1+b2w2+b3w3 = (a1+b1)w1+(a2+b2)w2+(a3+b3)w3. This implies that u+v, being a linear combination of w1,w2, w3 , Span{w1,w2,w3}.
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