Checking for Vector Subspace Equality in 4-Dimensional Space

In summary, the conversation is about checking if two sets of vectors generate the same subspace for \mathbb{R}^4. The suggested method is to use row transformations to reduce the matrices representing the two sets of vectors and see if they end up with the same reduced row-echelon form. The process is slightly more complicated for a matrix with 3 rows but can still be done by matching the first numbers in each row.
  • #1
Physicsissuef
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0

Homework Statement



Check if this sets of vectors generate same subspace for [itex]\mathbb{R}^4[/itex].

{ (1,2,0,-1), (-2,0,1,1) } and { (-1,2,1,0),(-3,-2,1,2) , (-1,6,2,-1) }

Homework Equations


The Attempt at a Solution



Here is the one matrix.
[tex]
\begin{bmatrix}
1 & 2 & 0 & -1 \\
-2 & 0 & 1 & 1
\end{bmatrix}
[/tex]
and the other
[tex]
\begin{bmatrix}
-1 & 2 & 1 & 0\\
-3 & -2 & 1 & 2\\
-1 & 6 & 2 & -1
\end{bmatrix}
[/tex]

I use row transformations to find out if this 2 matrices are equal. But how will I know which row transformations to implement?
 
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  • #2
You can shoot for reduced Row-Echelon form on both matrices, but this may not be the fastest approach.
 
  • #3
I think it is. Just row reduce both matrices and see if you get the same thing. (Since the first matrix has only two rows, to be the same, the second matrix will have to wind up with a row of zeros.)
 
Last edited by a moderator:
  • #4
This two matrices are easy to find, because of the first 2x4 matrix. And what if I have two 3x3 matrices? is it easier?
 
  • #5
It's slightly harder to row reduce a matrix with 3 rows than one with 2 rows, but not much.
 
  • #6
The problem is to "match" the process of doing row reduction of the 2 matrices...
 
  • #7
What do you mean by "matching"? You row reduce one matrix, row reduce the other and see if the results are the same.
 
  • #8
By doing row reduction to the 1-nd matrix

[tex]\begin{bmatrix}
2 & 3 & -1 & 1\\
1 & 1 & 0 & -2\\
1 & 2 & -1 & 3
\end{bmatrix}[/tex]

I get:

[tex]
\begin{bmatrix}
0 & 1 & -1 & 5\\
1 & 1 & 0 & -2\\
0 & 0 & 0 & 0
\end{bmatrix}
[/tex]

and I have the 2-nd matrix.

[tex]
\begin{bmatrix}
0 & 1 & -1 & 5\\
4 & 5 & -1 & -3
\end{bmatrix}
[/tex]

If I make [itex]-5*R_1+R_2[/itex] I will get

[tex]
\begin{bmatrix}
0 & 1 & -1 & 5\\
4 & 0 & 4 & -27
\end{bmatrix}
[/tex]

And if I make [itex]-R_1+R_2[/itex] and after that dividing [itex]R_2[/itex] by 2, I will get:

[tex]
\begin{bmatrix}
0 & 1 & -1 & 5\\
1 & 1 & 0 & -2
\end{bmatrix}
[/tex]

Do you understand me what I am talking about?
 
Last edited:
  • #9
HallsofIvy?
 
  • #10
No, I do not see what you are talking about. First, you haven't "row reduced" the first matrix: you should not have a 0 above the 1 in the first column. Also, although it is not strictly speaking part of the definition of "row reduced", you do not want a non-zero number above the first non-zero number in the second row. You should have
[tex]\left[\begin{array}{cccc}1 & 0 & 1 & -7 \\ 0 & 1 & -1 & 5\\0 & 0 & 0 & 0\end{array}\right][/tex]

For the second, you again should not have a 0 above a non-zero number in the first column: the first thing you should do is swap the two rows:
[tex]\left[\begin{array}{cccc}4 & 5 & -1 & -3 \\ 0 & 1 & -1 & -3\end{array}\right][/tex]

Now, to get rid of the non-zero, 5, above the first 1 in the second column, subtract 5 times the second row from the first:
[tex]\left[\begin{array}{cccc}4 & 0 & 4 & -28 \\ 0 & 1 & -1 & 5\end{array}\right][/tex].

Finally, dividing the first row by 4, we have
[tex]\left[\begin{array}{cccc}1 & 0 & 1 & -7 \\ 0 & 1 & -1 & 5\end{array}\right][/tex].
Which, dropping the all 0 third rwo in the first matrix, is exactly the same as the first matrix.

Do what you can to make the first numbers in each row match (the point being you can always get 1 0 ... and 0 1 ..., and see if the other numbers match.
 
  • #11
Aaah... I understand now.. Thank you very much...
 

1. What is a vector subspace in 4-dimensional space?

A vector subspace in 4-dimensional space is a subset of 4-dimensional space that satisfies the three conditions of closure under vector addition, closure under scalar multiplication, and contains the zero vector. This means that any vector in the subspace can be added to another vector in the subspace and the result will also be in the subspace, and any vector in the subspace can be multiplied by a scalar and the result will also be in the subspace.

2. How do you check for vector subspace equality in 4-dimensional space?

To check for vector subspace equality in 4-dimensional space, you must first determine if the two subspaces have the same dimension. Then, you must verify if the two subspaces have the same basis vectors. Lastly, you must check if all vectors in one subspace can be written as a linear combination of the basis vectors of the other subspace. If all three conditions are met, the subspaces are equal.

3. What is the significance of checking for vector subspace equality in 4-dimensional space?

Checking for vector subspace equality in 4-dimensional space is important because it allows us to determine if two subspaces are equivalent. This can help in various applications such as linear algebra, physics, and computer science, where understanding the properties and relationships of vectors is essential.

4. Can a vector subspace in 4-dimensional space be equal to the entire 4-dimensional space?

No, a vector subspace in 4-dimensional space cannot be equal to the entire 4-dimensional space. A subspace must contain the zero vector, which is not equivalent to the entire space. Additionally, the entire space is not a proper subset of itself, therefore, cannot be considered a subspace.

5. Are there any shortcuts for checking vector subspace equality in 4-dimensional space?

Yes, there are a few shortcuts for checking vector subspace equality in 4-dimensional space. One is checking if the subspaces have the same dimension and the same number of basis vectors. If this is true, then checking if any vector in one subspace can be written as a linear combination of the basis vectors of the other subspace is enough to determine equality. Another shortcut is checking if the subspaces have the same span, meaning all vectors in one subspace can be written as a linear combination of the vectors in the other subspace.

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