Let be vectorial spaces with finite dimension over a field and a subspace of . In these conditions, let's denote the vectorial subspace of whose elements are linear applications from to such that for every vector of its image is . Let be a vectorial subspace of such that .
Then, and are isomorphic spaces.
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We have that M = { f ∈ L ( E , F ) : ∀ v ∈ V , f ( v ) = 0 F } .
Let
ϕ : M ⟶ f ↦ f ∣ W ( W , F )
In these conditions, let's prove that ϕ is an isomorphism.
Let's prove that ϕ is injective.
Let x , y ∈ M such that ϕ ( x ) = ϕ ( y ) . We have that x , y ∈ L ( E , F ) .
It is clear that
∀ t ∈ W , x ( t ) = y ( t )
and
∀ t ∈ V , x ( t ) = y ( t ) = 0 F .
Let t 1 ∈ E . As E = V ⊕ W , we have that there are well determined a 1 ∈ V and b 1 ∈ W such that
t 1 = a 1 + b 1 .
So,
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 ) .
So, x = y , which means that ϕ is injective, as wanted.
Let's prove, now, that ϕ is surjective.
Let g ∈ L ( W , F ) . Let's consider g ′ ∈ L ( E , F ) such that
g ′ ( x ) = { g ( x ) , x ∈ W 0 F , x ∈ V
It's clear that ϕ ( g ′ ) = g , so ϕ is surjective.
Let's prove, in final place, that ϕ is a linear application.
Let x , y ∈ M and α , β ∈ K . It is clear that ϕ ( α x + β y ) , α ϕ ( x ) , β ϕ ( y ) ∈ L ( W , F ) .
Let t ∈ W . We have that
ϕ ( α x + β y ) ( t ) = ( α x + β y ) ( t ) = α x ( t ) + β y ( t ) =
α ϕ ( x ) ( t ) + β ϕ ( y ) ( t ) .
So, as ϕ ( α x + β y ) , α ϕ ( x ) , β ϕ ( y ) ∈ L ( W , F ) and
∀ t ∈ W , ϕ ( α x + β y ) ( t ) = α ϕ ( x ) ( t ) + β ϕ ( y ) ( t )
We can conclude that
ϕ ( α x + β y ) = α ϕ ( x ) + β ϕ ( y ) .
In conclusion, ϕ is an isomorphism, which means that M and L ( W , F ) are isomorphic spaces.