Answer to Question #207038 in Linear Algebra for Simphiwe Dlamini

Question

Show that for any g element of L(V;C) and u element of V with g(u) not equal 0: V =null g operation { xu: x element of C}

Answer 1

f V is finite dimensional and W is a subspace of V then we can

find a subspace U of V for which V = W ⊕ U. 

Also null T is a subspace of V . Setting W = null T, to get a subspace U of V for which

V = null T ⊕ U

Now we want to prove any subspace U for which V = null T ⊕ U satisfies the desired property.

Since V = null T ⊕ U, we already have null T ∩ U={0}. So we just need to show that range T =

{T u | u ∈ U}. First we show that range T ⊂ {T u | u ∈ U}. So let w ∈ range T. That means there

is some v ∈ V for which T(v) = w. Since v ∈ V and we have that V = null T ⊕ U, we can find

vectors n ∈ null T and u ∈ U for which v = n + u. Thus,

T(v) = T(n) + T(u)

= 0 + T(u) since n ∈ null T

We had that w = T(v). So, w = T(u) for some u ∈ U. That means that w ∈ {T u | u ∈ U}. Thus

range T ⊂ {T u | u ∈ U}.

Now we show that {T u | u ∈ U} ⊂ range T. But for any element u ∈ U, u is also in V as

U ⊂ V . Thus T u is in the image of T by definition. Therefore {T u | u ∈ U} ⊂ range T.

So we have shown that range T = {T u | u ∈ U}. Thus, there exists a subspace U of V s.t.

V = null T ⊕ U and range T = {T u | u ∈ U}.