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Test if these 4 vectors span R^3

by jey1234
Tags: span, test, vectors
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jey1234
#1
Jul21-12, 05:56 PM
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.
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chiro
#2
Jul21-12, 09:17 PM
P: 4,573
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)?
HallsofIvy
#3
Jul22-12, 08:59 AM
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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.

LCKurtz
#4
Jul22-12, 03:29 PM
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Test if these 4 vectors span R^3

Quote Quote by HallsofIvy View Post
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.
SammyS
#5
Jul22-12, 04:01 PM
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jey1234,

Have you noticed that v3 = v1 + v2

and v4 = v1 - v2 ?
LCKurtz
#6
Jul22-12, 05:19 PM
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Quote Quote by SammyS View Post
Have you noticed that v3 = v1 + v2

and v4 = v1 - v2 ?
Who, me? Halls??
SammyS
#7
Jul22-12, 05:24 PM
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Quote Quote by SammyS View Post
Have you noticed that v3 = v1 + v2

and v4 = v1 - v2 ?
Quote Quote by LCKurtz View Post
Who, me? Halls??
Neither you nor Halls ...

The OP, jey1234 .

(I will edit that !)
tamtam402
#8
Jul22-12, 06:05 PM
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.
tamtam402
#9
Jul22-12, 06:12 PM
P: 202
Quote Quote by HallsofIvy View Post
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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