Isomorphic?

Algebra Level 2

Let E , F \mathbb{E}, \mathbb{F} be vectorial spaces with finite dimension over a field K \mathbb{K} and V \mathbb{V} a subspace of E \mathbb{E} . In these conditions, let's denote M \mathscr{M} the vectorial subspace of L ( E , F ) \mathcal{L} \left(\mathbb{E}, \mathbb{F} \right) whose elements are linear applications from E \mathbb{E} to F \mathbb{F} such that for every vector of V \mathbb{V} its image is 0 F 0_{\mathbb{F}} . Let W \mathbb{W} be a vectorial subspace of E \mathbb{E} such that E = V W \mathbb{E}=\mathbb{V} \oplus \mathbb{W} .

Then, M \mathscr{M} and L ( W , F ) \mathcal{L} \left(\mathbb{W}, \mathbb{F} \right) are isomorphic spaces.

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1 solution

We have that M = { f L ( E , F ) : v V , f ( v ) = 0 F } \mathscr{M}= \{ f \in \mathcal{L} \left( \mathbb{E}, \mathbb{F} \right) : \forall v \in \mathbb{V}, f(v)=0_{\mathbb{F}} \} .

Let

ϕ : M f f W ( W , F ) \phi: \mathscr{M} \stackrel{f \mapsto {f|}_{\mathbb{W}}}{\longrightarrow} \left( \mathbb{W}, \mathbb{F} \right)

In these conditions, let's prove that ϕ \phi is an isomorphism.

Let's prove that ϕ \phi is injective.

Let x , y M x,y \in \mathscr{M} such that ϕ ( x ) = ϕ ( y ) \phi(x)= \phi (y) . We have that x , y L ( E , F ) x,y \in \mathcal{L} \left(\mathbb{E}, \mathbb{F} \right) .

It is clear that

t W , x ( t ) = y ( t ) \forall t\in \mathbb{W}, x(t)=y(t)

and

t V , x ( t ) = y ( t ) = 0 F \forall t\in \mathbb{V}, x(t)=y(t)=0_{\mathbb{F}} .

Let t 1 E t_1 \in \mathbb{E} . As E = V W \mathbb{E}=\mathbb{V} \oplus \mathbb{W} , we have that there are well determined a 1 V a_1 \in \mathbb{V} and b 1 W b_1 \in \mathbb{W} such that

t 1 = a 1 + b 1 t_1=a_1+b_1 .

So,

x ( t 1 ) = x ( a 1 + b 1 ) = x ( a 1 ) + x ( b 1 ) = x ( b 1 ) = x(t_1)=x(a_1+b_1)=x(a_1)+x(b_1)=x(b_1)=

y ( b 1 ) = y ( b 1 ) + 0 F = y ( b 1 ) + y ( a 1 ) = y ( a 1 + b 1 ) y(b_1)=y(b_1)+0_{\mathbb{F}}=y(b_1)+y(a_1)=y(a_1+b_1) .

So, x = y x=y , which means that ϕ \phi is injective, as wanted.

Let's prove, now, that ϕ \phi is surjective.

Let g L ( W , F ) g \in \mathcal{L} \left(\mathbb{W}, \mathbb{F} \right) . Let's consider g L ( E , F ) g' \in \mathcal{L} \left(\mathbb{E}, \mathbb{F} \right) such that

g ( x ) = { g ( x ) , x W 0 F , x V g'(x)= \begin{cases} g(x), x \in \mathbb{W}\\ 0_{\mathbb{F}}, x \in \mathbb{V} \end{cases}

It's clear that ϕ ( g ) = g \phi (g')=g , so ϕ \phi is surjective.

Let's prove, in final place, that ϕ \phi is a linear application.

Let x , y M x,y \in \mathscr{M} and α , β K \alpha, \beta \in \mathbb{K} . It is clear that ϕ ( α x + β y ) , α ϕ ( x ) , β ϕ ( y ) L ( W , F ) \phi( \alpha x+\beta y), \alpha \phi(x), \beta \phi(y) \in \mathcal{L} \left(\mathbb{W}, \mathbb{F} \right) .

Let t W t \in \mathbb{W} . We have that

ϕ ( α x + β y ) ( t ) = ( α x + β y ) ( t ) = α x ( t ) + β y ( t ) = \phi( \alpha x+\beta y) (t)=( \alpha x+\beta y) (t)=\alpha x(t)+\beta y(t)=

α ϕ ( x ) ( t ) + β ϕ ( y ) ( t ) \alpha \phi(x)(t)+ \beta \phi(y)(t) .

So, as ϕ ( α x + β y ) , α ϕ ( x ) , β ϕ ( y ) L ( W , F ) \phi( \alpha x+\beta y), \alpha \phi(x), \beta \phi(y) \in \mathcal{L} \left(\mathbb{W}, \mathbb{F} \right) and

t W , ϕ ( α x + β y ) ( t ) = α ϕ ( x ) ( t ) + β ϕ ( y ) ( t ) \forall t \in \mathbb{W}, \phi( \alpha x+\beta y)(t)=\alpha \phi(x)(t)+ \beta \phi(y)(t)

We can conclude that

ϕ ( α x + β y ) = α ϕ ( x ) + β ϕ ( y ) \phi(\alpha x+\beta y)=\alpha \phi(x)+ \beta \phi(y) .

In conclusion, ϕ \phi is an isomorphism, which means that M \mathscr{M} and L ( W , F ) \mathcal{L} \left(\mathbb{W}, \mathbb{F} \right) are isomorphic spaces.

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