Question

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

Transformation properties
ANSWERED
asked 2021-03-07

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.

Answers (1)

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.

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