Question

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

Transformation properties

Let $$\left\{v_{1},\ v_{2}, \dots,\ v_{n}\right\}$$ be a basis for a vector space V. Prove that if a linear transformation $$\displaystyle{T}\ :\ {V}\rightarrow\ {V}$$ satisfies $$\displaystyle{T}{\left({v}_{{{1}}}\right)}={0}\ \text{for}\ {i}={1},\ {2},\dot{{s}},\ {n},$$ then T is the zero transformation.
Getting Started: To prove that T is the zero transformation, you need to show that $$\displaystyle{T}{\left({v}\right)}={0}$$ for every vector v in V.
(i) Let v be an arbitrary vector in V such that $$\displaystyle{v}={c}_{{{1}}}\ {v}_{{{1}}}\ +\ {c}_{{{2}}}\ {v}_{{{2}}}\ +\ \dot{{s}}\ +\ {c}_{{{n}}}\ {v}_{{{n}}}.$$
(ii) Use the definition and properties of linear transformations to rewrite $$\displaystyle{T}\ {\left({v}\right)}$$ as a linear combination of $$\displaystyle{T}\ {\left({v}_{{{1}}}\right)}$$.
(iii) Use the fact that $$\displaystyle{T}\ {\left({v}_{{i}}\right)}={0}$$ to conclude that $$\displaystyle{T}\ {\left({v}\right)}={0}$$, making T the zero tranformation.

2021-03-08

Step 1
Let $$\left\{v_{1},\ v_{2}, \dots,\ v_{n}\right\}$$ be the basis for a vector space.
Define the map, $$\displaystyle{T}\ :\ {V}\rightarrow{V}\ \text{such that}\ {T}\ {\left({v}_{{{i}}}\right)}={0}\ \text{for}\ {i}={1},\ {2},\dot{{s}},\ {n}.$$
To prove, the above defined map is a linear transformation.
Step 2
$$\text{Let}\ v_{1},\ v_{2}\ \in\ V\ \text{and}\ r\ \in \mathbb{F}\ \text{a scalar.}$$
$$\displaystyle\Rightarrow\ {T}\ {\left({v}_{{{1}}}\right)}={0}={T}{\left({v}_{{{2}}}\right)}$$
$$\displaystyle\text{As}\ {v}_{{{1}}}\ {v}_{{{2}}}\in\ {V}\Rightarrow\ {v}_{{{1}}}\ +\ {v}_{{{2}}}\ \in\ {V},\ {r}{v}_{{{1}}}\ \in\ {V}$$
$$\displaystyle\Rightarrow\ {T}\ {\left({v}_{{{1}}}\ +\ {v}_{{{2}}}\right)}={0}$$
$$\displaystyle={0}\ +\ {0}$$
$$\displaystyle={T}\ {\left({v}_{{{1}}}\right)}\ +\ {T}\ {\left({v}_{{{2}}}\right)}$$
$$\displaystyle\text{Also}$$
$$\displaystyle{T}\ {\left({r}{v}_{{{1}}}\right)}={0}={r}{.0}$$
$$\displaystyle\Rightarrow\ {T}\ {\left({r}{v}_{{{1}}}\right)}={r}{T}\ {\left({v}_{{{1}}}\right)}$$
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 $$\left\{v_{1},\ v_{2}, \dots,\ v_{n}\right\}$$ is the basis for a vector space V.
Apply the transformation T on both sides and use the fact that T is linear transformation.
$$\displaystyle\Rightarrow\ {T}\ {\left({v}\right)}={T}\ {\left({c}_{{{1}}}{v}_{{{1}}}\ +\ {c}_{{{2}}}{v}_{{{2}}}\ +\ \dot{{s}}\ +\ {c}_{{{n}}}{v}_{{{n}}}\right)}$$
$$\displaystyle={T}\ {\left({c}_{{{1}}}{v}_{{{1}}}\right)}\ +\ {T}{\left({c}_{{{2}}}{v}_{{{2}}}\right)}\ +\ \dot{{s}}\ +\ {T}{\left({c}_{{{n}}}{v}_{{{n}}}\right)}$$
$$\displaystyle={c}_{{{1}}}{T}\ {\left({v}_{{{1}}}\right)}\ +\ {c}_{{{2}}}{T}\ {\left({v}_{{{2}}}\right)}\ +\ \dot{{s}}\ +\ {c}_{{{n}}}{T}\ {\left({v}_{{{n}}}\right)}$$
$$\displaystyle\text{Also,}\ {T}{\left({v}_{{{1}}}\right)}={T}{\left({v}_{{{2}}}\right)}=\dot{{s}}\ ={T}{\left({v}_{{{n}}}\right)}={0}$$
$$\displaystyle\Rightarrow\ {T}{\left({v}\right)}={c}_{{{1}}}\ {\left({0}\right)}\ +\ {c}_{{{2}}}\ {\left({0}\right)}\ +\ \dot{{s}}\ +\ {c}_{{{n}}}\ {\left({0}\right)}$$
$$\displaystyle\Rightarrow\ {T}{\left({v}\right)}={0}\ +\ {0}\ +\ \dot{{s}}\ +\ {0}$$
$$\displaystyle\Rightarrow\ {T}{\left({v}\right)}={0}$$
Which gives, the transformation T is zero tranformation.