Finding the kernel of a linear map Our exercise is to find all solutions to the equation $Ax = 0$, among others for the following matrix  
$$A =\begin{pmatrix} 6 & 3 & -9 \\ 2 & 1 & -3 \\ -4 & -2 & 6 \end{pmatrix}.$$
This amounts to finding the kernel, and obviously, the rows of the matrix are multiples of each other, so we can reduce the equations to:
$$6x_1 + 3x_2 - 9x_3 = 0.$$
Choosing $x_3 = 0$, $x_1 = 1$ and $x_2 = -2$ would fulfill the equation in my opinion.
However, the official solution is described as the following set:
$$\lambda \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} 0 \\ 3 \\ 1 \end{pmatrix}$$
with $\mu, \lambda \in$  R.
My questions now:
Where does the vector (and its multiples) $\lambda \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \end{smallmatrix}\right]$ come from?
Also, since after the reduction we are left with only one row, the dimension of the kernel should accordingly be $1$ - however $\left[\begin{smallmatrix} 0 \\ 3 \\ 1  \end{smallmatrix}\right]$ and $\left[\begin{smallmatrix} 1 \\ -2 \\ 0  \end{smallmatrix}\right]$ (from $x_3 = 0$, $x_1 = 1$ and $x_2 = -2$) seem to be two different basis vectors - what have I done or understood wrongly?
Many thanks
 A: The dimension of the subspace given by $6x_1+3x_2-9x_3=0$ is $2$. This is the equation of a plane in $\mathbb{R}^3$.
The number of independant equations defining a subspace is not the dimension of this subspace. It is $n-$ dimension of the subspace. If you had $2$ independant equations left, it would be a line, because $3-2=1$ and the dimension of a line is $1$.
Then, once you know that the dimension of the subspace is $2$, all you have ti do to find a basis is finding two linearly independant vectors satisfying its equation. Just looking at it, it is obvious that $(1, 1, 1)$ is a solution. In order to find the other, you can just fix two coordinates (e.g. $x=1$ and $z=1$ and deduce the third one using the equation.
A: You correctly found that $$\ker  A= \{ (x_1,x_2,x_3) \in \mathbb R^3 :6x_1 +3x_2 - 9x_3 = 0\},$$
which is equivalent to $$\ker A = \left\{\left( x_1, x_2, \frac {2x_1}3 +\frac {x_2}3\right): x_1, x_2 \in \mathbb R\right\}.$$
If we move forward one step we have that $$\ker A  = \left\{x_1 \cdot \left(1,0,\frac 23\right) +x_2 \left(0,1,\frac 13\right) : x_1,x_2 \in \mathbb R\right\}.$$
That means that the kernel is spanned by the $2$ linearly independent vectors  $(1,0,\frac 23),\,(0,1,\frac 13)$ (thus, $\dim\ker A = 2)$, or equivalently the kernel of $A$ is spanned by any $2$ linearly independent vectors which can be written as a linear combination of the 2 above vectors.


*

*First case: Consider $x_1 = 0, x_2 = 3\implies e_1 = (0,3,1).$

*Second case: Consider $x_1 =1 , x_2 =1 \implies e_2 =(1,1,1). $
We can easily check that $e_1,e_2$ are linearly independent vectors.
A: First, note that 
$$\begin{pmatrix} 1 \\ -2 \\ 0  \end{pmatrix} =
\begin{pmatrix} 1\\1\\1\end{pmatrix} - \begin{pmatrix} 0 \\ 3 \\ 1 \end{pmatrix}, $$
So your missing solution 
$\left[\begin{smallmatrix} 1 \\ -2 \\ 0  \end{smallmatrix}\right] $ 
is, in fact, a special case of the general solution 
$ \lambda \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \end{smallmatrix}\right] + \mu \left[\begin{smallmatrix} 0 \\ 3 \\ 1 \end{smallmatrix}\right] $ with $\lambda =1$ and $\mu = -1$.
Second, the general solution can be obtained from reduced equation $6x_1 + 3x_2 - 9x_3 = 0$ by imposing certain assumptions:


*

*For example, if we assume $x_1 = 0$, we will get the following
$$
\begin{cases} x_1 = 0 \\ 6x_1 + 3x_2 - 9x_3 = 0\end{cases} 
\implies \begin{cases}x_1 = 0 \\ 3x_2 - 9x_3 = 0 \end{cases} 
\implies \begin{cases}x_1 = 0 \\ x_2 = 3 x_3\end{cases} 
$$
Assuming parametrization $x_3 = \mu$, we get 
$$
\begin{cases} x_1 = 0 \\ 6x_1 + 3x_2 - 9x_3 = 0\end{cases} 
\implies 
\begin{cases}x_1=0 \\ x_2 = 3 x_3 \\ x_3 = \mu & - \operatorname{parameter}\end{cases} 
\iff 
\begin{cases}x_1 = 0 \\ x_2 = 3\mu \\ x_3 = \mu \end{cases} 
$$
We can rewrite this solution in the vector form:
$$
\begin{cases}x_1 = 0 \\ x_2 = 3\mu \\ x_3 = \mu \end{cases} 
\iff
x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} 
= \begin{bmatrix} 0 \\ 3\mu \\ \mu \end{bmatrix} 
\iff 
x
= \begin{bmatrix} 0 \\ 3  \\ 1\end{bmatrix}  \mu
$$

*Similarly, imposing condition $x_1 = x_2$, we  write
$$
\begin{cases} x_1 = x_2 \\ 6x_1 + 3x_2 - 9x_3 = 0\end{cases} 
\implies \begin{cases}x_1 = x_2 \\ 6x_1 + 3x_2 - 9x_3 = 0 \end{cases}  
\implies \begin{cases}x_1 = x_2 \\ x_1 = x_3\end{cases} 
$$
Assuming parametrization $x_1 = \lambda$, we get 
$$
\begin{cases} 
x_1 = \lambda  & -\operatorname{parameter}\\ x_2 = x_1 \\ x_3 = x_1
\end{cases} 
\implies 
\begin{cases} x_1 = \lambda \\ x_2 = \lambda \\ x_3 = \lambda \end{cases} 
$$
The vector form of the solution then looks like 
$$
\begin{cases} x_1 = \lambda \\ x_2 = \lambda \\ x_3 = \lambda \end{cases} 
\implies 
x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} 
= \begin{bmatrix} \lambda \\ \lambda \\ \lambda \end{bmatrix} 
\iff 
x= \begin{bmatrix} 1 \\ 1  \\ 1\end{bmatrix}  \lambda
$$
Third, Since parametrized solutions $x = x(\lambda)$ and $x = x(\mu)$ are linearly independent, and since the equation  $ 6x_1 + 3x_2 - 9x_3 = 0$ is linear, the general solution can be written as the sum of two independent parametrized particular solutions, i.e.
$$
x = x_\lambda + x_\mu = \begin{bmatrix} 1 \\ 1  \\ 1\end{bmatrix}  \lambda + \begin{bmatrix} 0 \\ 3  \\ 1\end{bmatrix}  \mu
$$

Finally, Note that we could have imposed different assumptions on $x$, and that would result in different parametrized solutions. 
However, as long as these solutions will be linearly independent, the will still span the same two-dimentional space of solutions which we have now. 
Thus, the kernel of $A$, which is also a space of all solutions of the system $Ax = 0$, is the linear span of vectors $\left[\begin{smallmatrix} 1 \\ 1 \\ 1 \end{smallmatrix}\right]$ and $\left[\begin{smallmatrix} 0 \\ 3 \\ 1 \end{smallmatrix}\right]$:
$$
\ker A 
= \operatorname{span} \left(\; \begin{bmatrix} 1 \\ 1  \\ 1\end{bmatrix}, \begin{bmatrix} 0 \\ 3  \\ 1\end{bmatrix}  \;\right) 
= \left\{\; \begin{bmatrix} 1 \\ 1  \\ 1\end{bmatrix}, \begin{bmatrix} 0 \\ 3  \\ 1\end{bmatrix} \; \right\}
= \left\{\ \vec v = \begin{bmatrix} 1 \\ 1  \\ 1\end{bmatrix}  \lambda + \begin{bmatrix} 0 \\ 3  \\ 1\end{bmatrix}  \mu 
\mathrel{\bigg|}
\lambda, \mu \in \mathbb R\ \right\}
$$
