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.

To show:
\(\displaystyle{a}\ {<}\ {\frac{{{a}+{b}}}{{{2}}}}\ {<}\ {b}\)
Given information:
a and b are real numbers.

Angles A and B are vertical angles.
If \(\displaystyle{m}\angle{A}={104}∘\) what is the \(m\angle B?\)
\(\displaystyle{m}\angle{B}=∘{d}{e}{g}{r}{e}{e}{s}\)

Angles A and B are complementary. If \(\displaystyle{m}\angle{A}={27}∘\), what is the \(m\angle B?\) \(\displaystyle{m}\angle{B}= \circ {d}{e}{g}{r}{e}{e}{s}\)

For the V vector space contains all \(\displaystyle{2}\times{2}\) matrices. Determine whether the \(\displaystyle{T}:{V}\rightarrow{R}^{{1}}\) is the linear transformation over the \(\displaystyle{T}{\left({A}\right)}={a}\ +\ {2}{b}\ -\ {c}\ +\ {d},\) where \(A=\begin{bmatrix}a & b \\c & d \end{bmatrix}\)

Angles A and B are supplementary.
If \(\displaystyle{m}\angle{A}={78}°\) what is \({m}\angle{B}\)