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Linear Transformation (Fredholm Alternative Theorem)

Swati

New member
Oct 30, 2012
16
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
Feb 5, 2012
1,621
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
Oct 30, 2012
16
how to proof if second statement is true then first statement is false.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
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
Feb 15, 2012
1,967
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
Feb 7, 2012
2,725
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.