Test if these 4 vectors span R^3

In summary, the given vectors v1=(3,1,4), v2=(2,-3,5), v3=(5,-2,9), v4=(1,4,-1) do not span ℝ3 as two of the vectors can be obtained by a linear combination of the other two, indicating that they are not independent. This can be seen through row reducing the matrix or by noticing that v3 and v4 are linear combinations of v1 and v2. The fact that the system has infinitely many solutions does not change the fact that the vectors do not span ℝ3.
  • #1
jey1234
38
0

Homework Statement


Determine if the vectors v1=(3,1,4), v2=(2,-3,5), v3=(5,-2,9), v4=(1,4,-1) span ℝ3

Homework Equations



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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  • #2
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
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
HallsofIvy said:
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.
 
  • #5
jey1234,

Have you noticed that v3 = v1 + v2

and v4 = v1 - v2 ?
 
Last edited:
  • #6
SammyS said:
Have you noticed that v3 = v1 + v2

and v4 = v1 - v2 ?

Who, me? Halls??
 
  • #7
SammyS said:
Have you noticed that v3 = v1 + v2

and v4 = v1 - v2 ?

LCKurtz said:
Who, me? Halls??
Neither you nor Halls ...

The OP, jey1234 .

(I will edit that !)
 
  • #8
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:
  • #9
HallsofIvy said:
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.
 

1. What does it mean for vectors to span R^3?

For a set of vectors to span R^3, it means that they are able to create any point in three-dimensional space. This means that any vector in R^3 can be written as a linear combination of these vectors.

2. How can I test if 4 vectors span R^3?

To test if 4 vectors span R^3, you can use the method of Gaussian elimination or row reduction. This involves creating a matrix with the 4 vectors as columns, and reducing it to row-echelon form. If the matrix has a pivot in every row, then the vectors span R^3.

3. Can a set of 4 linearly dependent vectors span R^3?

No, a set of 4 linearly dependent vectors cannot span R^3. This is because linearly dependent vectors are essentially multiples of each other, so they cannot create all possible points in three-dimensional space.

4. What is the minimum number of vectors needed to span R^3?

The minimum number of vectors needed to span R^3 is 3. This is because R^3 is a three-dimensional space, and three independent vectors are able to create any point in this space.

5. Can a set of 4 vectors span a space other than R^3?

Yes, a set of 4 vectors can span a different space other than R^3. The number of vectors needed to span a space depends on the dimension of that space. For example, to span R^2, you would need 2 linearly independent vectors, and to span R^4, you would need 4 linearly independent vectors.

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