Prove that an element of the basis is an element of the Kernel after linear transformation Let $T:R^4\rightarrow R^4$ and basis $B=(v_1,v_2,v_3,v_4)$.
$$T(v_1)+T(v_2)=T(v_3)\; \text{ and } \; T(v_1)+T(v_3)=T(v_2)$$
Prove that $v_1\in Ker(T)$
What I wrote is:
$$T(v_1)=T(v_3)-T(v_2)\; \text{ and } \; T(v_1)=T(v_2)-T(v_3)$$
due to the properties of linear transformation 
$$T(v_3-v_2)=T(v_2-v_3)$$ 
$$v_3-v_2=v_2-v_3\Rightarrow 2v_3=2v_2\Rightarrow v_3=v_2$$
Therefore $T(v_1)=0$ and $v_1\in Ker(T)$.
Is this proof valid?
 A: In fact if the characteristic of the background field be not 2 Then you should do the following after arriving $T(v_3-v_2)=T(v_2-v_3)$;
$$T(v_3-v_2)-T(v_2-v_3)=0\Longrightarrow T(v_3-v_2-(v_2-v_3))=0\Longrightarrow T(2(v_3-v_2))=0$$
$$\Longrightarrow 2T(v_3-v_2)=0\Longrightarrow T(v_3-v_2)=0$$
which you said yourself $T(v_1)=T(v_3-v_2)$, Then $v_1\in\ker T$.
In the case characteristic is 2, there is counterexample for the main argument itself. But in your question the field I think is $\mathbb{R}$ as you used the symbol $R$, so your solution is correct.
For a moment assume $k=\bar{\mathbb{Z}}_2$ (instead of $\mathbb{R}$) then the counterexample is $v_1=\left[\begin{array}{c}
1\\
1\\
0\\
0
\end{array}\right]$, $v_2=\left[\begin{array}{c}
1\\
0\\
0\\
0
\end{array}\right]$ and $v_3=\left[\begin{array}{c}
0\\
1\\
0\\
0
\end{array}\right]$, and take your transformation be the identity. None of $v_i$s are in kernel but we do have those two equations.
A: By linearity, $T(v_1+v_2-v_3)=0$ and $T(v_1+v_3-v_2)=0$ yield $0=T(v_1+v_2-v_3+v_1+v_3-v_2)=T(2v_1)=2T(v_1)$, thus $T(v_1)=0$, meaning $v_1\in \ker(T)$.
