Checking for Vector Subspace Equality in 4-Dimensional Space

[Physicsissuef]

Homework Statement

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

{ (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.
<br /> \begin{bmatrix}<br /> 1 &amp; 2 &amp; 0 &amp; -1 \\ <br /> -2 &amp; 0 &amp; 1 &amp; 1<br /> \end{bmatrix}<br />
and the other
<br /> \begin{bmatrix}<br /> -1 &amp; 2 &amp; 1 &amp; 0\\ <br /> -3 &amp; -2 &amp; 1 &amp; 2\\ <br /> -1 &amp; 6 &amp; 2 &amp; -1<br /> \end{bmatrix}<br />

I use row transformations to find out if this 2 matrices are equal. But how will I know which row transformations to implement?

 

[benorin]

You can shoot for reduced Row-Echelon form on both matrices, but this may not be the fastest approach.

 

[HallsofIvy]

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.)

 

[Physicsissuef]

This two matrices are easy to find, because of the first 2x4 matrix. And what if I have two 3x3 matrices? is it easier?

 

[HallsofIvy]

It's slightly harder to row reduce a matrix with 3 rows than one with 2 rows, but not much.

 

[Physicsissuef]

The problem is to "match" the process of doing row reduction of the 2 matrices...

 

[HallsofIvy]

What do you mean by "matching"? You row reduce one matrix, row reduce the other and see if the results are the same.

 

[Physicsissuef]

By doing row reduction to the 1-nd matrix

\begin{bmatrix}<br /> 2 &amp; 3 &amp; -1 &amp; 1\\ <br /> 1 &amp; 1 &amp; 0 &amp; -2\\ <br /> 1 &amp; 2 &amp; -1 &amp; 3<br /> \end{bmatrix}

I get:

<br /> \begin{bmatrix}<br /> 0 &amp; 1 &amp; -1 &amp; 5\\ <br /> 1 &amp; 1 &amp; 0 &amp; -2\\ <br /> 0 &amp; 0 &amp; 0 &amp; 0<br /> \end{bmatrix}<br />

and I have the 2-nd matrix.

<br /> \begin{bmatrix}<br /> 0 &amp; 1 &amp; -1 &amp; 5\\ <br /> 4 &amp; 5 &amp; -1 &amp; -3<br /> \end{bmatrix}<br />

If I make -5*R_1+R_2 I will get

<br /> \begin{bmatrix}<br /> 0 &amp; 1 &amp; -1 &amp; 5\\ <br /> 4 &amp; 0 &amp; 4 &amp; -27<br /> \end{bmatrix}<br />

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

<br /> \begin{bmatrix}<br /> 0 &amp; 1 &amp; -1 &amp; 5\\ <br /> 1 &amp; 1 &amp; 0 &amp; -2<br /> \end{bmatrix}<br />

Do you understand me what I am talking about?

 

[Physicsissuef]

HallsofIvy?

 

[HallsofIvy]

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
\left[\begin{array}{cccc}1 &amp; 0 &amp; 1 &amp; -7 \\ 0 &amp; 1 &amp; -1 &amp; 5\\0 &amp; 0 &amp; 0 &amp; 0\end{array}\right]

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:
\left[\begin{array}{cccc}4 &amp; 5 &amp; -1 &amp; -3 \\ 0 &amp; 1 &amp; -1 &amp; -3\end{array}\right]

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:
\left[\begin{array}{cccc}4 &amp; 0 &amp; 4 &amp; -28 \\ 0 &amp; 1 &amp; -1 &amp; 5\end{array}\right].

Finally, dividing the first row by 4, we have
\left[\begin{array}{cccc}1 &amp; 0 &amp; 1 &amp; -7 \\ 0 &amp; 1 &amp; -1 &amp; 5\end{array}\right].
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.

 

[Physicsissuef]

Aaah... I understand now.. Thank you very much...