linear algebra kernel problem The question is to find matrix $A$ where $\text{ker}(A)$ is span of two vectors, the first is $[1 \ 0 \  0 \  1 \  1]$ and the second is $[3  \ 1  \ 2  \  0  \  0]$.
I tried by taking different $3$ by $5$ matrices with rank $3$ ... but does not get the proper matrix... 
 A: If $K$ is a matrix with those two vectors as columns, you need to find a matrix $A$ that first of all satisfies $A\cdot K=0$ (that is a zero matrix with two columns and as many rows as $A$ has). This relation alone will guarantee that $\ker(A)$ contains the span of your two vectors (which is also the column space or image of$~K$), but the kernel might be larger. Indeed this will happen if $A$ is taken to simply (for instance a zero matrix); indeed by rank-nullity you need a matrix of rank $3$ to get a kernel that (only) has dimension$~2$.
The equation $A\cdot K=0$ just means that each row $R$ of $A$ satisfies $R\cdot K=0$; one can independently choose as many such rows as one likes; one of course needs at least $3$ such rows (and well chosen) in order to make the rank$~3$. The problem at hand is a well known one, but slightly disguised: your unknown $R$ is a row vector written to the left of a given matrix, while you are probably more used to solving linear systems of equations in the form $C\cdot X=b$ where $C$ is a coefficient matrix and $X$ a column vector of unknowns (and $b$ is a given column vector, which in our case will be zero). One can simply transform into this form by taking transposes: $R\cdot K=0$ is equivalent to $K^\top\cdot R^\top=0^\top$ (the transpose of the original zero row gives our zero column as right hand side). Get the general solution to this linear homogeneous system of $2$ equation in the $5$ unknown entries of $R^\top$; the solution will involve $3$ independent parameters, which give you (by setting the parameters successively to the columns of the $3\times3$ identity matrix) a total of $3$ independent solutions for $R^\top$. Transpose these solutions to give you $3$ independent rows for your $3\times5$ matrix $A$.
A: In order to find a matrix such that its kernel is given by the span of $2$ linearly independent vectors, then you must find $3$ vectors ($3$ because your space if $5$ and you have $2$ vectors), which are orthogonal to the given $2$ vectors. 
Then, you can create $A$ using these $3$ vectors as rows. 
Gram–Schmidt process can help you in this task.
