
#1
Jul2112, 05:56 PM

P: 38

1. The problem statement, all variables and given/known data
Determine if the vectors v_{1}=(3,1,4), v_{2}=(2,3,5), v_{3}=(5,2,9), v_{4}=(1,4,1) span ℝ^{3} 2. Relevant equations 3. The attempt at a solution So I first arranged it as a matrix, \begin{bmatrix} \begin{array}{ccccc} 3&2&5&1&b_1\\ 1&3&2&4&b_2\\ 4&5&9&1&b_3 \end{array} \end{bmatrix} Now I know what to do if it's a square matrix. I just have to see if the coefficient matrix is invertible (det ≠0). If yes that would mean that any vector b can be expressed as a linear combination. Since this is not a square matrix, I thought I'd have to row reduce it. Row reduced: \begin{bmatrix} \begin{array}{cccc} 1&0&1&1\\ 0&1&1&1\\ 0&0&0&0 \end{array} \end{bmatrix} Now what do I do? It seems to me that the system has infinitely many solutions and therefore the vectors span ℝ^{3}. But the solution manual says that it doesn't. What am I doing wrong? Thanks. 



#2
Jul2112, 09:17 PM

P: 4,570

Hey jey1234 and welcome to the forums.
Think about the rank of your matrix: how many nonzero rows do you have? If you have one nonzero row, then everything is a multiple of the first row which means that everything is a multiple of some vector. Building on this idea, what is the difference between having two nonzero rows and no nonzero rows in this matrix in terms of the dimension of the space (for one nonzero row we have one dimension since everything is a scalar multiple of that vector)? 



#3
Jul2212, 08:59 AM

Math
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Thanks
PF Gold
P: 38,881

The question was whether the vector span the space, not whether or not the form a basis. The fact that the system "has infinitely many solutions" means it has solutions and so the vectors do span the space. The fact there there is not a unique solution means they are not independent and do not form a basis for R^{3}. But you already knew that no set of four vectors can be a basis for a three dimensional vector space.




#4
Jul2212, 03:29 PM

HW Helper
Thanks
PF Gold
P: 7,197

Test if these 4 vectors span R^3 



#5
Jul2212, 04:01 PM

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P: 7,396

jey1234,
Have you noticed that v_{3} = v_{1} + v_{2} and v_{4} = v_{1}  v_{2} ? 



#7
Jul2212, 05:24 PM

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P: 7,396

The OP, jey1234 . (I will edit that !) 



#8
Jul2212, 06:05 PM

P: 200

Do you understand the consequences of your line of zeros in your row reduced matrix? That means one of the 3 vectors can be obtained by a linear combination of the other 2. What does that tell you? Well, the 3 vectors lie in a plane.
Can you see why they do not span ℝ^{3}? EDIT: Sorry, I thought you had 3 vectors in R4, but you actually have 4 vectors in R3. Since you wrote the vectors vertically, then you actually have to columnreduce them. If I were you, I would transpose the matrix and complete the row reduction. You will see that 2 of your vectors can be obtained by a linear combination of the other 2. In other words, you will have 2 lines of zeros. 



#9
Jul2212, 06:12 PM

P: 200




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