# Linear Transformation (Fredholm Alternative Theorem)

#### Swati

##### New member
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

##### Well-known member
MHB Math Helper
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

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

#### Opalg

##### MHB Oldtimer
Staff member
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

##### Well-known member
MHB Math Scholar
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

##### MHB Oldtimer
Staff member
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.