Question

Prove that V is a vector space. \(\displaystyle{V}={T}{\left({m},\ {n}\right)}\), the set of linear transformations \(\displaystyle{T}\ :\ {R}^{{{m}}}\ \rightarrow\ {R}^{{{n}}}\), together with the usual addition and scalar multiplication of functions.

Answer 1

Step 1
\(\displaystyle{V}={T}{\left({m},\ {n}\right)},\) the set of linear transformations \(\displaystyle{T}\ :\ {R}^{{{m}}}\ \rightarrow\ {R}^{{{n}}},\) together with the usual addition and scalar multiplication of functions are clearly defined.
next we need to determine if the five conditions of Definion 7.1 are all met.
If \(\displaystyle{T}{\left({u}\right)}\) and \(\displaystyle{T}{\left({v}\right)}\) are a linear transformation if for all vectors u and v and \(\displaystyle{R}^{{{m}}},\) then their sum will also be a linear transformation in \(\displaystyle{R}^{{{m}}}\), that is, \(\displaystyle{V}={T}{\left({m},\ {n}\right)}\) is closed under addition.
If \(\displaystyle{T}{\left({u}\right)}\) is a linear transformation if for all vectors u in \(\displaystyle{R}^{{{m}}}\) and c is a scalar then i and \(\displaystyle{c}{T}{\left({u}\right)}\) will also be a linear transformation thus \(\displaystyle{V}={T}{\left({m},\ {n}\right)}\) is also closed under scalar multiplication.
M If \(\displaystyle{T}{\left({0}\right)}={0}\) is an identical zero transformation, then if \(\displaystyle{T}{\left({u}\right)}\) is in \(\displaystyle{R}^{{{m}}}\) linear transformation, then T(0)\ +\ T(u)=T)ZSK for all \(\displaystyle{T}{\left({u}\right)}\) in \(\displaystyle{R}^{{{m}}}\), so that \(\displaystyle{T}{\left({0}\right)}\) is the zero transformation.
If \(\displaystyle{T}{\left({u}\right)}\) linear transformation, it has its inverse linear transformation \(\displaystyle-{T}{\left({u}\right)}\) so that
\(\displaystyle{T}{\left({u}\right)}\ +\ {\left(-{T}{\left({u}\right)}\right)}={T}{\left({u}\right)}\ +\ {T}{\left(-{u}\right)}={0}\)
Step 2
a) We are checking whether you are
\(\displaystyle{v}_{{{1}}}\ +\ {v}_{{{2}}}={v}_{{{2}}}\ +\ {v}_{{{1}}}\)
PSKv_{1}\ +\ v_{2}=T(v_{1})=T(x_{1}\ +\ x_{2}\ +\ \cdots\ x_{m})\ +\ T(y_{1}\ +\ y_{2}\ +\ \cdots\ y_{n})
\(\displaystyle={T}{\left({y}_{{{1}}}\ +\ {y}_{{{2}}}\ +\ \cdots\ {y}_{{{n}}}\right)}\ +\ {T}{\left({x}_{{{1}}}\ +\ {x}_{{{2}}}\ +\ \cdots\ {x}_{{{m}}}\right)}={v}_{{{2}}}\ +\ {v}_{{{1}}}\)
Therefore, commutativity is valid.
b) Check
\(\displaystyle{\left({v}_{{{1}}}\ +\ {v}_{{{2}}}\right)}\ +\ {v}_{{{3}}}={v}_{{{1}}}\ +\ {\left({v}_{{{2}}}\ +\ {v}_{{{3}}}\right)}\)
\(\displaystyle{\left({v}_{{{1}}}\ +\ {v}_{{{2}}}\right)}\ +\ {v}_{{{3}}}={\left({\left({T}{\left({v}_{{{1}}}\right)}\ +\ {T}{\left({v}_{{{1}}}\right)}\right)}\ +\ {T}{\left({v}_{{{3}}}\right)}\right.}\)
\(\displaystyle={\left({T}{\left({x}_{{{1}}}\ +\ {x}_{{{2}}}\ +\ \cdots\ {x}_{{{m}}}\right)}\ +\ {T}{\left({y}_{{{1}}}\ +\ {y}_{{{2}}}\ +\ \cdots\ {y}_{{{n}}}\right)}\right)}\ +\ {T}{\left({z}_{{{1}}}\ +\ {z}_{{{2}}}\ \cdots\ +\ {z}_{{{m}}}\right)}\)
\(\displaystyle={T}{\left({x}_{{{1}}}\ +\ {x}_{{{2}}}\ +\ \cdots\ {x}_{{{m}}}\right)}\ +\ {T}{\left({y}_{{{1}}}\ +\ {y}_{{{2}}}\ +\ \cdots\ {y}_{{{n}}}\right)}\ +\ {T}{\left({z}_{{{1}}}\ +\ {z}_{{{2}}}\ +\ \cdots\ +\ {z}_{{{m}}}\right)}={v}_{{{1}}}\ +\ {\left({v}_{{{2}}}\ +\ {v}_{{{3}}}\right)}\)
The law of associativity also applies.
c) Let c be scalar, check
\(\displaystyle{c}{\left({v}_{{{1}}}\ +\ {v}_{{{2}}}\right)}={x}{v}_{{{1}}}\ +\ {c}{v}_{{{2}}}\)
\(\displaystyle{c}{\left({T}{\left({x}_{{{1}}}\ +\ {x}_{{{2}}}\ +\ \cdots\ {x}_{{{m}}}\right)}\ +\ {T}{\left({y}_{{{1}}}\ +\ {y}_{{{2}}}\ +\ \cdots\ {y}_{{{n}}}\right)}\right)}={c}{\left({T}{\left({x}_{{{1}}}\right)}\ +\ {T}{\left({x}_{{{2}}}\ +\ \cdots\ {T}{\left({x}_{{{m}}}\right)}\ +\ {T}{\left({y}_{{{1}}}\right)}\ +\ {T}{\left({y}_{{{2}}}\right)}\ +\ \cdots\ {T}{\left({y}_{{{n}}}\right)}\right)}\right.}\)
\(\displaystyle={c}{T}{\left({x}_{{{1}}}\right)}\ +\ {c}{T}{\left({x}_{{{2}}}\right)}\ +\ \cdots\ +\ {c}{T}{\left({x}_{{{m}}}\right)}\ +\ {c}{T}{\left({y}_{{{1}}}\right)}\ +\ {c}{T}{\left({y}_{{{2}}}\right)}\ +\ \cdots\ {c}{T}{\left({y}_{{{n}}}\right)}\)
\(\displaystyle={c}{\left({T}{\left({x}_{{{1}}}\right)}\ +\ {T}{\left({x}_{{{2}}}\right)}\ +\ \cdots\ {T}{\left({x}_{{{m}}}\right)}\right)}\ +\ {c}{\left({T}{\left({y}_{{{1}}}\right)}\ +\ {T}{\left({y}_{{{2}}}\right)}\ +\ \cdots\ {T}{\left({y}_{{{n}}}\right)}\right)}\)
\(\displaystyle\Rightarrow\ {c}{\left({v}_{{{1}}}\ +\ {v}_{{{2}}}\right)}={c}{v}_{{{1}}}\ +\ {c}{v}_{{{2}}}\)
Step 3
d) Let \(\displaystyle{c}_{{{1}}},\ {c}_{{{2}}}\) be scalars, check
\(\displaystyle{\left({c}_{{{1}}}\ +\ {c}_{{{2}}}\right)}{v}_{{{1}}}={c}_{{{1}}}{v}_{{{1}}}\ +\ {c}_{{{2}}}{v}_{{{1}}}\)
\(\displaystyle{\left({c}_{{{1}}}\ +\ {c}_{{{2}}}\right)}{\left({T}{\left({x}_{{{1}}}\ +\ {x}_{{{2}}}\ +\ \cdots\ {x}_{{{m}}}\right)}={\left({c}_{{{1}}}\ +\ {c}_{{{2}}}\right)}{\left({T}{\left({x}_{{{1}}}\right)}\ +\ {\left({c}_{{{1}}}\ +\ {c}_{{{2}}}\right)}{\left({T}{\left({x}_{{{2}}}\right)}\ +\ \cdots\ +\ {\left({c}_{{{1}}}\ +\ {c}_{{{2}}}\right)}{\left({T}{\left({x}_{{{m}}}\right)}\right)}\right.}\right.}\right.}\)
\(\displaystyle={c}_{{{1}}}{\left({T}{\left({x}_{{{1}}}\right\rbrace}\ +\ {c}_{{{2}}}{\left({T}{\left({x}_{{{1}}}\right)}\right)}\ +\ \cdots\ +\ {\left({c}_{{{1}}}{T}{\left({x}_{{{m}}}\right)}\right)}\ +\ {c}_{{{2}}}{\left({T}{\left({x}_{{{m}}}\right)}\right)}\right.}\) \(\displaystyle={c}_{{{1}}}{\left({T}{\left({x}_{{{1}}}\ +\ {x}_{{{2}}}\ +\ \cdots\ {x}_{{{m}}}\right)}\ +\ {c}_{{{2}}}{\left({T}{\left({x}_{{{1}}}\ +\ {x}_{{{2}}}\ +\ \cdots\ {x}_{{{m}}}\right)}\right.}\right.}\)
\(\displaystyle{R}{i}{>}{a}{r}{r}{o}{w}\ {\left({c}_{{{1}}}\ +\ {c}_{{{2}}}\right)}{v}_{{{1}}}={c}_{{{1}}}{v}_{{{1}}}\ +\ {c}_{{{2}}}{v}_{{{1}}}\)
e) Check if it is
\(\displaystyle{1}{v}_{{{1}}}={v}_{{{1}}}\)
1v_{1}=1(T(x_{1}\ +\ x_{2}\ +\ \cdots\ x_{m})=1\ \cdot\ (T(x_{1}\ +\ x_{2}\ +\ \cdots\ x_{m})=(T(x_{1}\ +\ x_{2}\ +\ \cdots\ x_{m})=v_{1}ZSK
All six terms of the 7.1 definion are satisfied therefore \(\displaystyle{V}={T}{\left({m},\ {n}\right)}\) is a vector space.