About linear map and null space Find a linear map $T:\mathbb{Q}^{6}\rightarrow \mathbb{Q}^{4}$ such that the kernel of T be the subspace of $\mathbb{Q}^{6}$consisting of all the vectors $(a_{1},...,a_{6})$ such that $\displaystyle \sum_{i=1}^{6}a_{i}=0$
Well, I need the set $$\lbrace a\in \mathbb{Q}^{6},\sum_{i=1}^{6}a_{i}=0| \hspace{0.1cm}T(a)=0\rbrace$$ so $$T(a_{1}e_{1}+...+a_{6}e_{6})=0$$
Where $\lbrace e_{1},...,e_{6}\rbrace$ is the canonical basis. I don't know what more can I do, if anyone have an idea I'll really appreciate it.
 A: Let $\mathbf u=(1,1,1,1,1,1)^\top$ and pick any nonzero vector $\mathbf v\in\mathbb Q^4$. Consider $T(\mathbf a)=(\mathbf u^\top \mathbf a)\mathbf v$.
A: Let $W$ be the subspace of $\Bbb Q^6$ given by
$$
W=\left\{v\in\Bbb Q^6:\sum v_k=0\right\}
$$
The vectors in $W$ are exactly the vectors of the form
\begin{align*}
(v_1,v_2,v_3,v_4,v_5,v_6)
&= (v_1,v_2,v_3,v_4,v_5,-v_1-v_2-v_3-v_4-v_5) \\
&= v_1\cdot(1,0,0,0,0,-1)+v_2\cdot(0,1,0,0,0,-1)+v_3\cdot(0,0,1,0,-1) \\
&  \qquad+v_4\cdot(0,0,0,1,0,-1)+v_5\cdot(0,0,0,0,1,-1)
\end{align*}
This implies that the vectors
\begin{align*}
q_1 &= (1,0,0,0,0,-1) & q_4 &= (0,0,0,1,0,-1) \\
q_2 &= (0,1,0,0,0,-1) & q_5 &= (0,0,0,0,1,-1) \\
q_3 &= (0,0,1,0,0,-1)
\end{align*}
form a basis for $W$. In particular, note that $W$ has codimension one in $\Bbb Q^6$ (that is, $\dim W=5=6-1$).
Now, let $p\in\Bbb Q^6$ be any vector not in $W$. Then $\beta=\{q_1,q_2,q_3,q_4,q_5,p\}$ is a basis for $\Bbb Q^6$. Defining a linear map $T:\Bbb Q^6\to\Bbb Q^4$ amounts to defining $T$ on $\beta$. To ensure that $\ker T=W$ we can use the formulas
$$
T(q_k)=\mathbf 0\qquad T(p)=(1,0,0,0)
$$
