Test if these 4 vectors span R^3

1. Jul 21, 2012

jey1234

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.

2. Jul 21, 2012

chiro

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)?

3. Jul 22, 2012

HallsofIvy

Staff Emeritus
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.

4. Jul 22, 2012

LCKurtz

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.

5. Jul 22, 2012

SammyS

Staff Emeritus
jey1234,

Have you noticed that v3 = v1 + v2

and v4 = v1 - v2 ?

Last edited: Jul 22, 2012
6. Jul 22, 2012

LCKurtz

Who, me? Halls??

7. Jul 22, 2012

SammyS

Staff Emeritus
Neither you nor Halls ...

The OP, jey1234 .

(I will edit that !)

8. Jul 22, 2012

tamtam402

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.

Last edited: Jul 22, 2012
9. Jul 22, 2012

tamtam402

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.