# Find a basis of a linear system

1. Apr 1, 2015

### Kavorka

Find a basis for the solution space of the linear system

x1-x2-2x3+x4 = 0
-3x1+3x2+x3-x4 = 0
2x1-2x2+x3 = 0

I created a matrix (not augmented, will be 0 on right side no matter what row operations) and brought it to reduced echelon form. x2 and x4 were free variables and I set them to the variables t and s respectively. I found that:

x1 = t - (1/5)s
x3 = (-2/5)s

All of this I'm fairly certain I did right, I just wanted to make sure I am stating the basis correctly because this is the first problem I've done where I've found a basis for a linear system. Stating all the x's in terms of s and t I get (forgive weird formatting not sure how to do matrices on here):

[x1]\\\\\\[1]\\\\\\\[-1/5]
[x2]\\=\t\[1]\+\s\[0]
[x3]\\\\\\[0]\\\\\\\[-2/5]
[x4]\\\\\\[0]\\\\\\\[1]

Forming the 2 dimensional basis:
[1] [-1]
[1] [0]
[0] [-2]
[0], [5]

2. Apr 1, 2015

### Staff: Mentor

You can check this for yourself. If A is the matrix of coefficients of your system above, and x1 and x2 are your two basis vectors, verify that Ax1 = 0 and that Ax2 = 0. If either of these doesn't come out to 0, then definitely you have made a mistake. If they both come out to zero, then A times any linear combination of the two vectors will also be zero.

Edit: you have a sign error in your second basis vector.

Last edited: Apr 1, 2015
3. Apr 1, 2015

### HallsofIvy

Staff Emeritus
You have four equation in three unknowns so one would initially expect the solution space to be 4- 3= 1 dimensional. Of course, if the equations are not independent that may not be true. However, I believe these are independent so your answer cannot be correct.

4. Apr 1, 2015

### Staff: Mentor

The three equations are dependent. The OP has a sign error in one of his basis vectors.

5. Apr 1, 2015

### HallsofIvy

Staff Emeritus
Oops. Thanks.

6. Apr 1, 2015

### Staff: Mentor

Happens to us all...

7. Apr 2, 2015

### BvU

From latexhelp:

Last edited by a moderator: May 7, 2017
8. Apr 2, 2015

### Staff: Mentor

I like the bmatrix environment, which uses brackets around the matrix rather than big parentheses.

https://www.physicsforums.com/file:////begin%7Bpmatrix [Broken]}
1 & 2 & 3 & 4\\
a & b & c & d\\
x & y & z & w
\end{bmatrix}

This yields
$$\begin{bmatrix} 1 & 2 & 3 & 4\\ a & b & c & d\\ x & y & z & w \end{bmatrix}$$

The vmatrix environment is useful for rendering determinants.
\begin{vmatrix]
1 & 2 & 3\\
a & b & c\\
x & y & z
\end{vmatrix}
which renders as
$$\begin{vmatrix}1 & 2 & 3\\ a & b & c\\ x & y & z \end{vmatrix}$$

Last edited by a moderator: May 7, 2017
9. Apr 2, 2015

### Kavorka

Thanks everyone! I noticed the sign error, and when multiplying my basis vectors by the original matrix I get a null matrix in both cases. I am just wondering, in back-checking your answer to say positively that your basis is correct, is this the only step you need to perform?

10. Apr 3, 2015

### BvU

I seem to remember you also need to show the basis vectors are independent, which in this case is a piece of cake.