Solving Tx=y: One Solution & R(T) Condition

[juaninf]

Help for prove this please

Let $$T\in{L(V,W)}$$
The equation $$Tx=y$$ have one solution iff $$y\in{R(T)}$$

 

[Mark44]

What have you tried?

 

[HallsofIvy]

A couple of questions: when you say "has one solution" do you mean exactly one solution or at least one solution?

Also, do you know the definition of "R(T)"?

It looks to me like this problem is either impossible or trivial!

 

[arkajad]

Read carefully and try to understand without any doubt whatsoever the definition of R(T).

 

[blue_raver22]

.

To prove this statement, we need to show that if y\in{R(T)}, then the equation Tx=y has one solution, and if the equation Tx=y has one solution, then y\in{R(T)}.

First, let y\in{R(T)}. This means that y is in the range of T, or in other words, there exists an x\in{V} such that Tx=y. This x is the solution to the equation Tx=y, and it is unique because if there were another x' such that Tx'=y, then T(x-x')=0, which contradicts the assumption that T is a linear transformation and x\neq x'.

On the other hand, if the equation Tx=y has one solution, let's say x, then y=Tx\in{R(T)}. This is because T is a linear transformation, so T(ax+by)=aT(x)+bT(y), and since Tx=y, we have T(ax+by)=aTx+bTy=aTx+by=a(y)+by=(a+b)y. Therefore, y is a linear combination of the vectors in the range of T, which means y\in{R(T)}.

Therefore, we have proven that the equation Tx=y has one solution iff y\in{R(T)}.