Basis for a Vector Space in R^3Date: 11/25/98 at 13:49:55 From: Noura Subject: Basis for a vector space. Hi Dr. Math: I have several questions about bases. First, which of the following are bases for R^3? a) (2,-3,1), (4,1,1), (0,-7,1) b) (1,6,4), (2,4,-1), (-1,2,5) I know that both of them are not bases because it is written in my book. The second one is because it is linearly dependent, and so it is not a basis. I don't know why the first one is not a basis. Second, I need to determine the dimension of and a basis for the solution space of the system: X1 - 3X2 + X3 = 0 2X1 - 6X2 + 2X3 = 0 3X1 - 9X2 + 3X3 = 0 I know the dimension is 2 and the basis is: {(3,1,0), (-1,0,1)} How do you find it? Third, I need to find a standard basis vector that can be added to the set {v1,v2} to produce a basis for R^3, where: v1 = (1,-1,0), v2 = (3,1,-2) Thanks! Date: 11/25/98 at 14:56:21 From: Doctor Anthony Subject: Re: Basis for a vector space. For the first question, part (a), the vectors are not bases if you can express one of them in terms of the other two, i.e. they are linearly dependent. Note: p[ 2] + q[4] + r[ 0] = 0 [-3] [1] [-7] [ 1] [1] [ 1] This gives 3 homogeneous equations in 3 unknowns, p, q and r, and if they are not consistent, the vectors are independent. If they are consistent one vector is a linear combination of the other two and so the vectors could not form a basis for R^3. The condition for valid solutions of p, q, r other than the trivial ones p = q = r = 0 is that the determinant shown below equals zero. |2 4 0| M = |-3 1 -7| = 0 (if true then vectors are linearly dependent) |1 1 1| Since, Det(M) = 2 - 28 + 0 - 0 + 14 + 12 = 0, the vectors are linearly dependent and DO NOT form a basis for R^3. For part (b), by the same reasoning as before, we must check the value of the determinant shown below: |1 6 4| M = |2 4 -1| = 20 + 6 + 16 + 16 + 2 - 60 = 0 |-1 2 5| again the vectors are linearly dependent and DO NOT form a basis for R^3. For your second question, notice the 3 equations are all the same equation. If these are coordinates then X1 - 2X2 + X3 = 0 is the equation of a plane and so the dimension is 2. Any two lines chosen at random in this plane can be used as base vectors. For example X1 = 2, X2 = 1 and X3 = 1 will satisfy the equation, so one base vector is (2, 1, 1). A second is (1, 1, 2), but there are an infinity of pairs (as long as they are not parallel) that can be used as base vectors. Finally, for the third problem, any vector not in the plane of v1 and v2 could form the third vector of the basis. For example the vector perpendicular to both these is: |i j k| |1 -1 0| = i(2) - j(-2) + k(4) = i + j + 2k |3 1 -2| So the third vector could be (1, 1, 2). - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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