Test if these 4 vectors span R^3

by jey1234
Tags: span, test, vectors
 P: 38 1. The problem statement, all variables and given/known data Determine if the vectors v1=(3,1,4), v2=(2,-3,5), v3=(5,-2,9), v4=(1,4,-1) span ℝ3 2. Relevant equations 3. The attempt at a solution So I first arranged it as a matrix, \begin{bmatrix} \begin{array}{cccc|c} 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 ℝ3. But the solution manual says that it doesn't. What am I doing wrong? Thanks.
 P: 4,578 Hey jey1234 and welcome to the forums. Think about the rank of your matrix: how many non-zero rows do you have? If you have one non-zero 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 non-zero rows and no non-zero rows in this matrix in terms of the dimension of the space (for one non-zero row we have one dimension since everything is a scalar multiple of that vector)?
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,682 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 R3. But you already knew that- no set of four vectors can be a basis for a three dimensional vector space.
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Test if these 4 vectors span R^3

 Quote by HallsofIvy 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 R3. But you already knew that- no set of four vectors can be a basis for a three dimensional vector space.
But the OP's row reduction just shows the homogeneous system has infinitely many solutions. If you try to write [1,1,1] as a linear combination of the given vectors you get an inconsistent system.
 Emeritus Sci Advisor HW Helper PF Gold P: 7,819 jey1234, Have you noticed that v3 = v1 + v2 and v4 = v1 - v2 ?
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 Quote by SammyS Have you noticed that v3 = v1 + v2 and v4 = v1 - v2 ?
Who, me? Halls??
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 Quote by SammyS Have you noticed that v3 = v1 + v2 and v4 = v1 - v2 ?
 Quote by LCKurtz Who, me? Halls??
Neither you nor Halls ...

The OP, jey1234 .

(I will edit that !)
 P: 202 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 ℝ3? 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 column-reduce 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.
P: 202
 Quote by HallsofIvy 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 R3. But you already knew that- no set of four vectors can be a basis for a three dimensional vector space.
OP's matrix is confusing because he wrote his vectors as columns instead of rows. 2 of the vectors are actually linear combinations of the other 2, which means they do not span R3.

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