# 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.

Question
Transformations of functions
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.

2020-11-25
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.

### Relevant Questions

Consider $$\displaystyle{V}={s}{p}{a}{n}{\left\lbrace{\cos{{\left({x}\right)}}},{\sin{{\left({x}\right)}}}\right\rbrace}$$ a subspace of the vector space of continuous functions and a linear transformation $$\displaystyle{T}:{V}\rightarrow{V}$$ where $$\displaystyle{T}{\left({f}\right)}={f{{\left({0}\right)}}}\times{\cos{{\left({x}\right)}}}−{f{{\left(π{2}\right)}}}\times{\sin{{\left({x}\right)}}}.$$ Find the matrix of T with respect to the basis $$\displaystyle{\left\lbrace{\cos{{\left({x}\right)}}}+{\sin{{\left({x}\right)}}},{\cos{{\left({x}\right)}}}−{\sin{{\left({x}\right)}}}\right\rbrace}$$ and determine if T is an isomorphism.

Give the correct answer and solve the given equation $$V = T(m,\ n),$$
The set of lineartrans formations
$$T\ :\ R^{m}\ \rightarrow\ R^{n},$$
together with the usualadditionand scalar multiplication of functions.
Let U and W be vector spaces over a field K. Let V be the set of ordered pairs (u,w) where u ∈ U and w ∈ W. Show that V is a vector space over K with addition in V and scalar multiplication on V defined by
(u,w)+(u',w')=(u+u',w+w') and k(u,w)=(ku,kw)
(This space V is called the external direct product of U and W.)
For each of the following functions f (x) and g(x), express g(x) in the form a: f (x + b) + c for some values a,b and c, and hence describe a sequence of horizontal and vertical transformations which map f(x) to g(x).
$$\displaystyle{f{{\left({x}\right)}}}={x}^{{2}}-{2},{g{{\left({x}\right)}}}={2}+{8}{x}-{4}{x}^{{2}}$$
For each of the following functions f (x) and g(x), express g(x) in the form a: f (x + b) + c for some values a,b and c, and hence describe a sequence of horizontal and vertical transformations which map f(x) to g(x).
$$\displaystyle{f{{\left({x}\right)}}}={x}^{{2}}+{3},{g{{\left({x}\right)}}}={x}^{{2}}-{6}{x}+{8}$$
For each of the following functions f (x) and g(x), express g(x) in the form a: f (x + b) + c for some values a,b and c, and hence describe a sequence of horizontal and vertical transformations which map f(x) to g(x).
$$\displaystyle{f{{\left({x}\right)}}}={x}^{{2}},{g{{\left({x}\right)}}}={3}{x}^{{2}}-{24}{x}+{8}$$
For each of the following functions f (x) and g(x), express g(x) in the form a: f (x + b) + c for some values a,b and c, and hence describe a sequence of horizontal and vertical transformations which map f(x) to g(x).
$$\displaystyle{f{{\left({x}\right)}}}={x}^{{2}},{g{{\left({x}\right)}}}={2}{x}^{{2}}+{4}{x}$$
Guided Proof Let $${v_{1}, v_{2}, .... V_{n}}$$ be a basis for a vector space V.
Prove that if a linear transformation $$T : V \rightarrow V$$ satisfies
$$T (v_{i}) = 0\ for\ i = 1, 2,..., n,$$ then T is the zero transformation.
To prove that T is the zero transformation, you need to show that $$T(v) = 0$$ for every vector v in V.
(i) Let v be the arbitrary vector in V such that $$v = c_{1} v_{1} + c_{2} v_{2} +\cdots + c_{n} V_{n}$$
(ii) Use the definition and properties of linear transformations to rewrite T(v) as a linear combination of $$T(v_{j})$$ .
(iii) Use the fact that $$T (v_{j}) = 0$$
to conclude that $$T (v) = 0,$$ making T the zero transformation.
$$A 0 = 0$$
$$A(cv) = cAv$$
$$A(v\ +\ w) = Av\ +\ Aw$$