Re: Linear Algebra, C3. E4. Request for help

On Sun, 07 Jan 2007 22:52:01 +0100, G. Frege <nomail@invalid> wrote:

Well, I tried to construct some sort of proof myself...

F is a field. V is a _finite-dimensional_ vector space over F.
Suppose that T is a linear map from V to F. Prove that if u in V
is not in null T, then

V = null T + {a*u : a in F}.

Ok.

I'll use the definition:

L(x) := {a*x : a in F}

Let u e V, such that u !e null T. null T is a vector space over F.
Then null T n L(u) = {0}. (u is not in null T. If there were a vector
v other than 0 such that v e null T & v e L(u), than v would be of the
form b*x for some b =/= 0, b e F. But then we would have (1/b)*v = u e
null T, since T is a vector space. Contradiction!)

Now I make use of two "high level" results:

1. A theorem concerning the sum of (finite-dimensional) subspaces:

dim M + dim N = dim(M + N) + dim(M n N),

where M and N are finitely dimensional subspaces of some vector space
V.

2. A theorem concerning linear maps:

dim null f + dim img f = n,

where V is a n-dimensional vector space and f is a linear map from V
in some vector space W.

Now the rest of the proof is straightforward. We have that both null T
and L(u) are (finitely dimensional) subspaces of V, since from 2. we
get dim null T = n - dim img T = dim V - 1. And (easy) dim L(u) = 1.
Moreover dim(null T n L(u)) = 0 (since null T n L(u) = {0}, and dim
{0} = 0).

Hence from 1. we get:

dim null T + dim L(u) = dim(null T + L(u)) +
dim(null T n L(u)),

(dim V - 1) + 1 = dim(null T + L(u)) + 0,

and hence we get:

dim(null T + L(u)) = dim V.

But null T + L(u) is a subspace of V. Hence null T + L(u) = V. qed.


F.

--

E-mail: info<at>simple-line<dot>de
.

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