# Matrix vector subspace

1. Mar 31, 2008

### Physicsissuef

1. The problem statement, all variables and given/known data

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) }

2. Relevant equations

3. The attempt at a solution

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

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

2. Apr 3, 2008

### benorin

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

3. Apr 4, 2008

### HallsofIvy

Staff Emeritus
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: Apr 4, 2008
4. Apr 5, 2008

### 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?

5. Apr 5, 2008

### HallsofIvy

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

6. Apr 5, 2008

### Physicsissuef

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

7. Apr 8, 2008

### HallsofIvy

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

8. Apr 9, 2008

### Physicsissuef

By doing row reduction to the 1-nd matrix

$$\begin{bmatrix} 2 & 3 & -1 & 1\\ 1 & 1 & 0 & -2\\ 1 & 2 & -1 & 3 \end{bmatrix}$$

I get:

$$\begin{bmatrix} 0 & 1 & -1 & 5\\ 1 & 1 & 0 & -2\\ 0 & 0 & 0 & 0 \end{bmatrix}$$

and I have the 2-nd matrix.

$$\begin{bmatrix} 0 & 1 & -1 & 5\\ 4 & 5 & -1 & -3 \end{bmatrix}$$

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

$$\begin{bmatrix} 0 & 1 & -1 & 5\\ 4 & 0 & 4 & -27 \end{bmatrix}$$

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

$$\begin{bmatrix} 0 & 1 & -1 & 5\\ 1 & 1 & 0 & -2 \end{bmatrix}$$

Do you understand me what I am talking about?

Last edited: Apr 9, 2008
9. Apr 11, 2008

### Physicsissuef

HallsofIvy?

10. Apr 11, 2008

### HallsofIvy

Staff Emeritus
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
$$\left[\begin{array}{cccc}1 & 0 & 1 & -7 \\ 0 & 1 & -1 & 5\\0 & 0 & 0 & 0\end{array}\right]$$

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

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

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

11. Apr 11, 2008

### Physicsissuef

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