# Basis for the homogeneous system

1. Apr 26, 2010

### thushanthan

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

Find a basis for the solution space of the homogeneous systems of linear equations AX=0

2. Relevant equations

Let A=1 2 3 4 5 6
6 6 5 4 3 3
1 2 3 4 5 6

and X= x
y
z
u
v
w

3. The attempt at a solution

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

2. Relevant equations

3. The attempt at a solution

2. Apr 27, 2010

### HallsofIvy

In Latex, your matrix problem is
$$\begin{bmatrix}1 & 2 & 3 & 4 & 5 & 6 \\ 6 & 6 & 6 & 4 & 3 & 3 \\ 1 & 2 & 3 & 4 & 5 & 6 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \\ u \\ v \\ w\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0\end{bmatrix}$$

That is the same as the three equations, x+ 2y+ 3z+ 4u+ 5v+ 6w= 0, 6x+ 6y+ 4z+ 4u+ 3v+ 3w= 0, and x+ 2y+ 3z+ 4u+ 5v+ 6w= 0.

Of course, the first and third are exactly the same so we only have two equations. We can solve those two equations for two of the variables in terms of the other 4. Replace those two in <x, y, z, u, v, w> with their (linear) expressions in the other 4.

For example, suppose the solution were u= 2x- 3y+ 4z- w, v= x+ y- 3z+ 4w (I just made those up. Solve the two equations yourself.)

Then we could write <x, y, z, u, v, w>= <x, y, z, 2x- 3y+ 4z- w, x+ y-3 z+ 4w, w>.

Now separate variables: <x, 0, 0, 2x, x, 0>+ <0, y, 0, -3y, y, 0>+ <0, 0, z, 4z, -3z, 0>+ <0, 0, 0, -w, 4w, w>.

Finally, take each variable out of its vector:
x<1, 0, 0, 2, 1, 0>+ y<0, 1, 0, -3, 1, 0>+ z<0, 0, 0, 1, 4, -3, 0>+ w<0, 0, 0, -1, 4, 1>.

Since any vector in the nullspace can be written as a linear combination of those 4 vectors, they form a basis for the null space.

(Again, that is NOT the solution to YOUR problem. You will have to solve those two equations for two of the variables your self.)

3. Apr 27, 2010

### thushanthan

Thank you!! Now I got it