Let left{v_{1}, v_{2}, dots, v_{n}right} be a basis for a vector space V.

Step 1
Let $\{v_1,v_2,\dots ,v_n\}$ be the basis for a vector space.
Define the map, ${\displaystyle T:V\to V\text{such that}T\left(v_i\right)=0\text{for}i=1,2,\dot{s},n.}$
To prove, the above defined map is a linear transformation.
Step 2
$\text{Let}{v}_1,v_2\in V\text{and}r\in \mathbb{F}\text{a scalar.}$
${\displaystyle \Rightarrow T\left(v_1\right)=0=T\left(v_2\right)}$
${\displaystyle \text{As}{v}_1{v}_2\in V\Rightarrow v_1+v_2\in V,r{v}_1\in V}$
${\displaystyle \Rightarrow T(v_1+v_2)=0}$
${\displaystyle =0+0}$
${\displaystyle =T\left(v_1\right)+T\left(v_2\right)}$
${\displaystyle \text{Also}}$
${\displaystyle T\left(r{v}_1\right)=0=r.0}$
${\displaystyle \Rightarrow T\left(r{v}_1\right)=rT\left(v_1\right)}$
Therefore, the above defined map is a linear transformation and this is true for every vector in V.
Step 3
Let v be an arbitrary vector in V such that ${\displaystyle v=c_1{v}_1+c_2{v}_2+\dot{s}+c_n{v}_n}$
Where, ${\displaystyle C_i}$ 's are scalars.
The vector v can be written as such linear combination since $\{v_1,v_2,\dots ,v_n\}$ is the basis for a vector space V.
Apply the transformation T on both sides and use the fact that T is linear transformation.
${\displaystyle \Rightarrow T\left(v\right)=T(c_1{v}_1+c_2{v}_2+\dot{s}+c_n{v}_n)}$
${\displaystyle =T\left(c_1{v}_1\right)+T\left(c_2{v}_2\right)+\dot{s}+T\left(c_n{v}_n\right)}$
${\displaystyle =c_{1}T\left(v_1\right)+c_{2}T\left(v_2\right)+\dot{s}+c_{n}T\left(v_n\right)}$