Proof of composite linear transformations

[Flyboy27]

Prove that if [tex]T:R^{m} \rightarrow R^{n}[/tex] and [tex]U:R^{n} \rightarrow R^{p}[/tex] are linear transformations that are both onto, then [tex]UT:R^{n} \rightarrow R^{p}[/tex] is also onto.

Can anyone point me in the right direction? Is there a theorem that I can pull out of the def'n of onto that I can begin this proof?

 

[mathwonk]

this is trivial, direct from definition of onto.

 

[M98Ranger]

To prove that UT is onto, we need to show that for every vector y in R^p, there exists a vector x in R^m such that UT(x) = y.

Since U is onto, for every vector y in R^p, there exists a vector z in R^n such that U(z) = y.

And since T is onto, for every vector z in R^n, there exists a vector x in R^m such that T(x) = z.

Therefore, for every vector y in R^p, there exists a vector x in R^m such that UT(x) = U(T(x)) = U(z) = y.

Hence, UT is onto, and the proof is complete.

This proof uses the definition of onto, which states that a linear transformation is onto if every element in the codomain (R^p in this case) has at least one preimage in the domain (R^m in this case). By using the onto property of both T and U, we can show that UT also satisfies this property and is therefore onto.