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

Step 1
${\displaystyle V=T(m,n),}$ the set of linear transformations ${\displaystyle T:R^m\to R^n,}$ 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, ${\displaystyle V=T(m,n)}$ 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 ${\displaystyle cT\left(u\right)}$ will also be a linear transformation thus ${\displaystyle V=T(m,n)}$ 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 $T(0)+T(u)=T)$ 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
${\displaystyle T\left(u\right)+(-T\left(u\right))=T\left(u\right)+T(-u)=0}$
Step 2
a) We are checking whether you are
${\displaystyle v_1+v_2=v_2+v_1}$
$v_1+v_2=T(v_1)=T(x_1+x_2+\cdots x_m)+T(y_1+y_2+\cdots y_n)$
${\displaystyle =T(y_1+y_2+\cdots y_n)+T(x_1+x_2+\cdots x_m)=v_2+v_1}$
Therefore, commutativity is valid.
b) Check
${\displaystyle (v_1+v_2)+v_3=v_1+(v_2+v_3)}$
${\displaystyle (v_1+v_2)+v_3=((T\left(v_1\right)+T\left(v_1\right))+T\left(v_3\right)}$

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