Blow all vector spaces are defined over a field F with characteristic zero. Let

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Blow all vector spaces are defined over a field F with characteristic zero.
Let W be a subspace of a finite dimensional vector space V.

Blow all vector spaces are defined over a field F with characteristic zero. Let W be a subspace of a finite dimensional vector space V. (a) Show that there exists a subspace X of V such that V = W oplus X. (b)Define the projection(p 76) P:V rightarrow V on W along X by P(w+x)=w, forall w in W, xin X . Then prove that (i)P is linear (ii)R(P)=W (iii)N(P)=X (iv) P^{2} = P (p 76) Definition. : Let V be a vector space and W{1} and W{2} be subspaces of V such that V=W{1}oplus W{2} .(Recall the definition of direct sum : Definition: A vector space V is called the direct sum of W{1} and W{2} if W{1} and W{2} are subspaces of V such that W{1}cap W{2}= {0} and W{1}+W{2}=V . We denote that V is the direct sum of W{1} and W{2} by writing V= W{1}oplus W{2} .) A function T : Vrightarrow V is called the projection on W{1} along W{2} if, for x=x{1}+x{2} with x{1}in W{1} and x{2}in W{2} , we have T(x)=x{1}.

Explanation / Answer

Let V a vector space of finite dimension and W a subspace of V with respective dimension n and m.

1)

Let B=(e1,...,em) a basis of W.

We know by a well-know theorem that we can extend B into a basis (e1,...,en) of V using elements of VW. Let X=(e_{m+1},...,e_n} the extended elements.

Then let's prove that V = W(+)X

First W n X is empty by this theorem because X contains elements of VW.

Since (e1,...,en) is a basis of V then:

v in V => we can write v = (a1e1 + ... + am em)+(a_{m+1}e_{m+1} + ... + a_n e_n)

Which is in W(+)X, so V = W(+)X

Hence proved.

2)

Let P : V -> V s.t. P(w+x) = w

i) Linearity is evident : P(v1+k v2) = P(w1+kw2 + x1+k x2) = w1+kw2 = P(v1) + k P(v2)

ii) Because for all w in W P(w) = w in W then W is a subset of R(P)

And clearly if y in R(P) then y = P(u)= w in W for some y = w+x in V so R(P) = W

iii) First by construction for all x in X : P(x) = 0 so X is a subset of N(P)

Now if u = w+x in N(P) then P(u) = w = 0 so u = x in X

Thus N(P) = X

iv) Let u = w+x then P^2 (u) = P^2 ( w + x ) = P ( w ) = w = P(w+x) = P(u)

So P^2 = P

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