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I've been doing some independent study on vector spaces and have moved on to looking at linear operators, in particular those of the form [itex] T:V \rightarrow V[/itex]. I know that the set of linear transformations [itex]\mathcal{L}\left( V,V\right) =\lbrace T:V \rightarrow V \vert \text{T is linear} \rbrace [/itex] form a vector space over [itex] \mathbb{F}[/itex], under the operations of point-wise addition and scalar multiplication, but I've been trying to show this explicitly by showing that these operations satisfy the vector space axioms. Here's my attempt so far:

Let [itex]V[/itex] be a vector space over [itex]\mathbb{F}[/itex] and let [itex]\mathcal{L}\left( V,V\right) =\lbrace T:V \rightarrow V \vert \text{T is linear} \rbrace [/itex] be the set of all linear transformations from [itex]V[/itex] into itself. As the operators [itex]T \in \mathcal{L}\left(V,V\right)[/itex] are linear, they satisfy the following conditions [tex]T\left(\mathbf{v} + \mathbf{w}\right)= T\left(\mathbf{v}\right) + T\left(\mathbf{w}\right) \quad\forall \mathbf{v},\mathbf{w} \in V,\; T\in \mathcal{L}\left(V,V\right)[/tex] and [tex] T\left(c\mathbf{v}\right) = cT\left(\mathbf{v}\right) [/tex]

And as such, we define point-wise addition of two operators [itex]S, T\in \mathcal{L}\left(V,V\right)[/itex] as [tex]\left(T+S\right) \left(\mathbf{v}\right) = T\left(\mathbf{v}\right) + S\left(\mathbf{v}\right)[/tex] and scalar multiplication of and operator [itex]T\in \mathcal{L}\left(V,V\right) [/itex] by some scalar from the underlying field [itex]\mathbb{F}[/itex] as [tex]T\left(c\mathbf{v}\right) = \left(cT\right) \left(\mathbf{v}\right) = cT\left( \mathbf{v}\right)[/tex]

Given this, we can now show that [itex]\mathcal{L}\left(V,V\right)[/itex] forms a vector space. First, we note that as both [itex]S, T\in \mathcal{L}\left(V,V\right)[/itex] clearly their "operator sum" [itex]T+S \in\mathcal{L}\left(V,V\right)[/itex] and similarly [itex]cT \in\mathcal{L}\left(V,V\right)[/itex] and hence the set [itex]\mathcal{L}\left(V,V\right)[/itex] is closed under these binary operations. Now, we shall check that [itex]\mathcal{L}\left(V,V\right)[/itex] satisfies the vector space axioms.

[itex]1.[/itex] (Commutativity of vector addition). Given [itex]S, T\in \mathcal{L}\left(V,V\right)[/itex] and [itex]\mathbf{v} \in V[/itex], and noting that as both [itex]S[/itex] and [itex]T[/itex] maps [itex]V[/itex] into itself, such that [itex] S\left(\mathbf{v}\right), T\left(\mathbf{v}\right) \in V [/itex] and therefore satisfy the vector space axioms, we have that [tex]\left(T+S\right) \left(\mathbf{v}\right) = T\left(\mathbf{v}\right) + S\left(\mathbf{v}\right)= S\left(\mathbf{v}\right) +T\left(\mathbf{v}\right)= \left(S+T\right) \left(\mathbf{v}\right).[/tex]

[itex]2.[/itex] (Associativity of vector addition). Given [itex]S, T, U\in \mathcal{L}\left(V,V\right)[/itex] and [itex]\mathbf{v} \in V[/itex], we have that [tex] \left(T +\left(S+U\right)\right) \left(\mathbf{v}\right) = T\left(\mathbf{v}\right) + \left(S+U\right)\left( \mathbf{v}\right) = T\left(\mathbf{v}\right) + \left(S\left(\mathbf{v}\right) + U\left(\mathbf{v}\right) \right)[/tex] [tex]\qquad\qquad\qquad\quad\quad = \left(T\left(\mathbf{v}\right) + S\left(\mathbf{v}\right) \right) + U\left(\mathbf{v}\right) = \left(T+S\right) \left(\mathbf{v}\right) + U\left(\mathbf{v}\right)[/tex] [tex]\qquad\qquad\qquad\quad\quad =\left(\left(T +S \right) +U\right) \left(\mathbf{v}\right).[/tex]

[itex]3.[/itex] (Identity element of vector addition). Let us define an operator [itex]\tilde{T}[/itex] such that [itex]\tilde{T}\left( \mathbf{v}\right) = \mathbf{0}[/itex]. Hence, given [itex]T\in \mathcal{L}\left(V,V\right)[/itex] and [itex]\mathbf{v} \in V[/itex], we have that [tex]\left( T+ \tilde{T} \right)\left(\mathbf{v} \right) = T\left(\mathbf{v}\right) + \tilde{T}\left( \mathbf{v}\right) = T\left(\mathbf{v} \right) + \mathbf{0} = T\left(\mathbf{v} \right) [/tex]

where we have noted that [itex]T\left( \mathbf{v}\right) \in V[/itex]

hence an identity exists for [itex]\mathcal{L}\left(V,V\right)[/itex]. (*I'm really not sure I've done this part correctly, it doesn't seem right?!*).

[itex]4.[/itex] (Inverse elements of addition) Noting that each [itex] \mathbf{v} \in V[/itex] has a unique inverse, [itex] -\mathbf{v} \in V[/itex], such that [itex] \mathbf{v} + \left(-\mathbf{v}\right) = \mathbf{0} [/itex], and that [itex] \left(-1\right)\mathbf{v}= - \mathbf{v} \;\;\forall\;\mathbf{v} \in V[/itex], we have [tex] T\left(\mathbf{0}\right) = T\left( \mathbf{v} + \left(-\mathbf{v}\right) \right) = T\left(\mathbf{v}\right) + T\left(-\mathbf{v}\right)= T\left(\mathbf{v}\right) + T\left(\left(-1\right)\mathbf{v}\right) = T\left(\mathbf{v}\right)+ \left(-1\right)T\left(\mathbf{v}\right) = T\left(\mathbf{v}\right) + \left(-T\left(\mathbf{v}\right)\right)= \left(T + \left(-T\right)\right)\left(\mathbf {v}\right) = \mathbf{0} [/tex] where we have noted that [itex]T\left(\mathbf{0}\right) = T\left(0\mathbf{v}\right) = 0T\left(\mathbf{v}\right) = \mathbf{0} [/itex] (as [itex]0\mathbf{v}=\mathbf{0} \;\; \forall \mathbf{v} \in V[/itex] and [itex]T\left(\mathbf{v}\right) \in \mathcal{L}\left(V,V\right)[/itex].

Hence, each element [itex]T\in \mathcal{L}\left(V,V\right)[/itex] has an inverse [itex]-T\in \mathcal{L}\left(V,V\right)[/itex]. (*Again, I'm not fully sure I'm correct on this one?!*).

[itex]5.[/itex] (Compatibility of scalar multiplication with field multiplication). Let [itex]T\in \mathcal{L}\left(V,V\right)[/itex], [itex]\mathbf{v} \in V[/itex] and [itex]c_{1},c_{2} \in \mathbb{F}[/itex], noting that [itex]\left(c_{1}c_{2}\right) \mathbf{v} = c_{1}\left( c_{2}\mathbf{v}\right) \;\;\forall\, \mathbf{v} \in V, \; c_{1},c_{2} \in \mathbb{F} [/itex], we have that [tex] T\left(\left( c_{1}c_{2}\right) \mathbf{v}\right) = \left( c_{1}c_{2}\right)T\left( \mathbf{v}\right)[/tex] and also [tex] T\left(\left( c_{1}c_{2}\right) \mathbf{v}\right)= T\left(c_{1}\left( c_{2}\mathbf{v}\right)\right)= c_{1}T\left(c_{2}\mathbf{v}\right) = c_{1}\left( c_{2} T\left(\mathbf{v}\right) \right)[/tex] Hence, we have that, [tex] \left( c_{1}c_{2}\right) T\left( \mathbf{v}\right) = c_{1}\left( c_{2} T \right) \left(\mathbf{v}\right) [/tex]

[itex]6.[/itex] (Identity element of scalar multiplication). Noting that [itex]1\mathbf{v} = \mathbf{v} \;\; \forall \mathbf{v} \in V [/itex], then given [itex]T\in \mathcal{L}\left(V,V\right)[/itex] and [itex]\mathbf{v} \in V[/itex] we have [tex]T\left(\mathbf{v} \right)= T\left(1\mathbf{v} \right) = 1T\left(\mathbf{v} \right) = \left(1 T\right)\left(\mathbf{v} \right) [/tex] as required.

[itex]7.[/itex] (Distributivity of scalar multiplication with respect to vector addition). Let [itex]S,T\in \mathcal{L}\left(V,V\right)[/itex], [itex]\mathbf{v} \in V[/itex] and [itex]c \in \mathbb{F}[/itex]. Then, [tex] \left(S+T\right)\left(c\mathbf{v}\right)= c\left(S+T\right)\left(\mathbf{v}\right) [/tex] and also [tex] \left(S+T\right)\left(c\mathbf{v}\right) = S\left(c\mathbf{v}\right) + T\left(c\mathbf{v}\right) = cS\left(\mathbf{v}\right) + cT\left(\mathbf{v}\right) = \left( cS+ cT\right)\left(\mathbf{v}\right) [/tex] and hence, [tex] c\left(S+T\right)\left(\mathbf{v}\right) =\left( cS+ cT\right)\left(\mathbf{v}\right) [/tex]

[itex]8.[/itex] (Distributivity of scalar multiplication with respect to field addition). Let [itex]T\in \mathcal{L}\left(V,V\right)[/itex], [itex]\mathbf{v} \in V[/itex] and [itex]c_{1},c_{2} \in \mathbb{F}[/itex]. Then, noting that [itex] \left(c_{1} +c_{2}\right)\mathbf{v} = c_{1}\mathbf{v} + c_{2}\mathbf{v} \;\;\forall\; \mathbf{v} \in V, c_{1},c_{2} \in \mathbb{F} [/itex], we have that [tex] T\left(\left(c_{1} +c_{2}\right)\mathbf{v}\right) = \left(c_{1} +c_{2}\right) T\left(\mathbf{v}\right)[/tex] and also [tex] T\left(\left(c_{1} +c_{2}\right)\mathbf{v}\right)= T\left(c_{1} \mathbf{v}\right) + T\left(c_{2}\mathbf{v}\right) = c_{1}T\left( \mathbf{v}\right)+ c_{2}T\left( \mathbf{v}\right) = \left(c_{1}T + c_{2}T\right)\left( \mathbf{v}\right) [/tex] and hence, [tex]\left(c_{1} +c_{2}\right) T\left(\mathbf{v}\right)= \left(c_{1}T + c_{2}T\right)\left( \mathbf{v}\right)[/tex] Therefore, the set [itex]\mathcal{L}\left( V,V\right) [/itex] forms a vector space over [itex]\mathbb{F} [/itex] under point-wise addition and scalar multiplication.

Is this a correct analysis?

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# Linear operators and vector spaces

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