MHB Linear Transformation (Fredholm Alternative Theorem)

[Swati]

Let T:V->V be a linear operator on an n-dimensional vector space. Prove that exactly one of the following statements holds:

(i) the equation T(x)=b has a solution for all vectors b in V.

(ii) Nullity of T>0

 

[Sudharaka]

Let T:V->V be a linear operator on an n-dimensional vector space. Prove that exactly one of the following statements holds:

(i) the equation T(x)=b has a solution for all vectors b in V.

(ii) Nullity of T>0

Hi Swati, :)

Suppose that the first statement is true. That is \(T\) is surjective. Then,

\[\mbox{dim }(V)=\mbox{dim }(\mbox{Im }T)=n\]

Then by the Rank-Nullity Theorem,

\[\mbox{Nullity }T=\mbox{dim }(\mbox{Ker }T)=0\]

Conversely you can show that if the second statement is true the first statement cannot be true.

Kind Regards,
Sudharaka.

 

[Swati]

how to proof if second statement is true then first statement is false.

 

[Opalg]

how to proof if second statement is true then first statement is false.

Use the rank-nullity theorem again, to show that T is surjective.

 

[Deveno]

if the SECOND statement is true, T CANNOT be surjective:

by the rank-nullity theorem:

dim(V) = rank(T) + nullity(T).

if nullity(T) > 0, then rank(T) < dim(V), so that:

dim(im(T)) < dim(V).

thus there is some b in V not in im(T).

(i only posted this because Opalg's post answers the wrong question).

 

[Opalg]

if the SECOND statement is true, T CANNOT be surjective:

by the rank-nullity theorem:

dim(V) = rank(T) + nullity(T).

if nullity(T) > 0, then rank(T) < dim(V), so that:

dim(im(T)) < dim(V).

thus there is some b in V not in im(T).

(i only posted this because Opalg's post answers the wrong question).

Yes of course. I should have said: Use the rank-nullity theorem again, to show that T is NOT surjective.