Feb 9 - xv. But uEW by close under scalar mult. xvi.-uEW so...

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W is a subspace of a vector space V If W is a subset of V With the same ops + and scalar multiples as V Which is also a vector space under these ops. Quick Efficient test that WcV is a subspace of V i. 0ϵW where 0 is zero vector of V ii. + : , closure under in W u vϵW kuϵW iii. Closure under scalar mult: , kϵR uϵR kuϵW iv. Show WcV with i,ii,iii W is a VS i) Axiom 5 is satisfied ii) 1 is satisfied iii) 2 is satisfied v. vi. 3 is satisfied: , → + = + → u vϵV u v V u 3 is satisfied vii. Since V vs u+(v+w)=(u+v)+w so axiom 4 is satisfied viii. ix. (7) ku, vEw, kER u, v, E V k(u+v)=ku+kv (7) is satisfied x. xi. Only (6) is not obvious: xii. Show there is –uEW so that u+(-u)=0 xiii. Since uEV -uEV so that u+(-u)=0 xiv.So theoretically we could have –uEV, but –u not in W
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Unformatted text preview: xv. But uEW by close under scalar mult. xvi.-uEW so (6) xvii. xviii. xix. xx. Quick subspace test enables us to easily show that many obkects are VS easily by exploiting well-known VS: xxi.R 2 ,R 3 ,R n M mxn xxii. Ex. Det6ermine if W= xxiii. xxiv. xxv. Is a VS under usual + and scalar mult. Of vectors. xxvi. xxvii. i)0eW xxviii. 0= zero voctor in R 3 xxix. xxx. xxxi. xxxii. xxxiii. xxxiv. ii)u, vEW xxxv. xxxvi. xxxvii. xxxviii. xxxix. xl. xli. xlii. iii)ku= k xliii. xliv. xlv. xlvi. xlvii. xlviii. xlix. W is a subspace of R 3 W is a VS l. li. lii. liii. liv. Ex. W= lv. lvi. lvii. lviii. lix. Is it a vector space? i) 0eW lx. Not possible as x and y can’t equal 0 lxi. lxii. lxiii. Ex. lxiv. lxv. W= lxvi....
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Feb 9 - xv. But uEW by close under scalar mult. xvi.-uEW so...

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