Let W be the set of all vectors of the form shown on the right, where a and b re
ID: 3123217 • Letter: L
Question
Let W be the set of all vectors of the form shown on the right, where a and b represent arbitrary real numbers. Find a set S of vectors that spans W, or give an example or an explanation showing why W is not a vector space. [-a + 1 a - 2b 2b + a] Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The set W is a vector space and a spanning set is S = _____. (Use a comma to separate vectors as needed.) B. The set W is not a vector space because W is not closed under vector addition. C. The set w is not a vector space because W is not closed under scalar multiplication. D. The set W is not a vector space because the zero vector is not in W.Explanation / Answer
We are considering for arbitrary real numbers a,b the set W of vectors of the form [ -a+1 , a-2b , 2b+a].
For a subset W to be a subspace we need three conditions, namely including the zero vector,being closed under addition and closed under scalar multiplication.
Note that if we have two vectors [-a+1, a - 2b, 2b - a], [-a' +1, a´ - 2b' , 2b' - a' ], in W, then making a''= a+a' -1, and making b''= b + b' + (1/2). we have that
[-a+1,a -2b, 2b -a]+[-a' +1, a´ -2b' , 2b' -a' ]=[-a+1 + (-a' +1), a - 2b + (a´-2b'), 2b - a + (2b'-a') ]=[a''-1, a''-2b'', 2b''+a'' ]
Which means that W it is closed under addition.
Note however that for the vector [-a+1, a - 2b, 2b - a] to be equal to the zero vector [0,0,0 ] we need to satisfy the three equations,
-a + 1 = 0 -------(1);
a - 2b = 0 -------(2);
2b + a = 0 -----(3).
From the first equation a=-1. Using this value for a we get from the second equation that b= (1/2), and from the third equation we get b=-1/2. This is a contradiction.
Hence there are no values of a, b such that [-a+1, a - 2b, 2b - a] is equal to [0,0,0]. This means W can not be a subspace since it does not contain the zero vector.
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As a final remark note that W is not closed under scalar multiplication. Since this will mean that the vector k* [-a+1, a - 2b, 2b - a] can be written as a vector of the form [-a''+1, a'' - 2b'', 2b'' - a'']. This would mean that the following equations hold.
-ka + k = -a'' +1 ---------(1);
ka - 2kb = a'' - 2b'' ------(2);
2kb - ka = 2b'' + a'' -----(3);
This from equations (2) and (3) we get that a'' = k a. However from (1) we get a'' = ka -k+1. This can only be true simultaneously if k=1.
This implies that W is not closed under scalar multiplication. This also means that W is not a subspace.
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