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Describe all solutions of Ax = 0 as Span(v1,v2...vp) for suitable vectors v1,v2,...,vp

A = \begin{bmatrix}1&-2&3&-6&5&0\\0&0&1&2&4&-6\\0&0&0&0&1&3\\0&0&0&0&0&0\end{bmatrix}

So, I know I'm looking for the null space here. Normally, I would use Gaussian Elimination and reduce to reduced row echelon form. However, I'm not sure what to do when the matrix looks like this. The last row is complete 0's and so the only thing I can come up with is

x_5 = -3x_6

and of course, I just have the other variables to build my solution. Does this mean that x_4,x_6 are free variables? How would I find the solution with that last row of 0's

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    Yes (and another one as well). – 2017-02-20
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    I'm sorry, I believe I am lost when it comes to this question. Would x_5 be the last free variable? – 2017-02-20

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Using elementary row operations, you obtain \begin{bmatrix} 1&-2&0&-12&0&39\\0&0&1&2&0&-18\\ 0&0&0&0&1&3\\0&0&0&0&0&0\end{bmatrix}
Your free variables are $x_2,x_4,x_6$ and the bound variables are $x_1,x_3,x_5$. In turn, set one free variable equal to 1 and the others to 0 to obtain a basis vector for the solution space. A basis is then found to be $(2,1,0,0,0,0), (12,0,-2,1,0,0), (-39,0,18,0,-3,1)$.

You can also view this null space as the perp space to the row space of the given matrix (which is obvious if you think about it).

The row of zeroes is not a problem. It says $0=0$ which simply does not provide any information so it can be ignored.

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    Can you explain how you achieved that basis? I am lost to that part. – 2017-02-20
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    Set free variable $x_2=1$ and the other free variables $x_4=0$ and $x_6=0$. Then solve for the bound variables $x_1$, $x_3$, and $x_5$. That gives you one of the three basis vectors. Repeat this process with $x_4=1$,$x_2=0$, $x_6=0$ to get another basis vector. Then get the third basis vector using $x_6=1,x_2=0,x_4=0$. Note that the dimension of the null space is the number of free variables. Also note that if $N$ is the null space and $R$ is the row space then $N=R^{\perp}$ so $\text{dim} N+\text{dim} R = \text{dim domain}$. – 2017-02-20
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    So each basis is a separate solution that is in the Span? – 2017-02-20
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    @LordHoneydew See [here](http://math.stackexchange.com/a/1521354/265466) for a detailed description of how to read a kernel basis from a rref matrix. – 2017-02-20
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    As I'm already confused, I think that link sort of confuses even more. Sorry. – 2017-02-20