May 28th, 2015, 10:17 AM  #1 
Newbie Joined: May 2015 From: sweden Posts: 1 Thanks: 0  Vector and spaces
I have been stuck for hours trying to solve this problem. But I just dont understand how to solve it. Question 1 Which of the following vectors belong to the column space of vector A A: 0 1 2 1 1 1 1 2 5 a) (0,1,1) b) (1,0,3) c) (1, 0, 1) d) (1,1,2) d) (0,0,0) f) (1,1,1) After i rref(A) I'm stuck. The last row contains only zero's. Does that mean that matrix A is linearly dependent and how does that help me? 
May 29th, 2015, 05:45 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 895 
First, A is not a vector, it is a matrix. Second, the "column space" of a matrix is the space spanned by its columns, thought of as individual vectors. Now, why did you row reduce the matrix? I'm not saying you shouldn't have, but it looks like you are doing rote calculation without knowing why or thinking about the basic definitions and properties here. Yes, when you rowreduce you get a row of all 0s which tells you, not that the columns are not independent so the column space is not three dimensional. (If it were three dimensional, you could immediately answer this question "all of them".) It is also easy to see that the first two vectors are not multiples of each other. They are independent so the column space of this matrix is two dimensional and those first two vectors, (0, 1, 1) and (1, 1, 2) form a basis for the column space. Now we can determine if each of those given vectors is in that columns space by determining if each can be written as a linear combination of the first two column vectors. Any linear combination of (0, 1, 1) and (1, 1, 2) is of the form A(0, 1, 1)+ B(1, 1, 2)= (B, A+ B, A+ 2B). Setting those equal to the three components of each vector will give three equations for A and B. If they are solvable, then that vector is in the column space. The first given vector is (a) (0, 1, 1). The three equations are B= 0, A+ B= 1, A+ 2B= 1. Since B= 0, from the second equation, A= 1. Then A+ 2B= 1+ 2(0)= 1, satisfying the third equation. Yes, this vector is in the column space of A. The second given vector is (b) (1, 0, 3). The three equations are B= 1, A+ B= 0, A+ 2B= 3. Since B= 1, from the second equation, A= 1. Then A+ 2B= 1+ 2(1)= 3, satisfying the third equation. Yes, this vector is in the column space of A. The third given vector is (c) (1, 0, 1). The three equations are B= 1, A+ B= 0, A+ 2B= 1. Since B= 1, from the second equation, A= 1. Then A+ 2B= 1+ 2(1)= 3, which does NOT satisfy the third equation. No, this vector is NOT in the column space of A. Last edited by Country Boy; May 29th, 2015 at 05:58 AM. 

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