Answer
See below
Work Step by Step
From example 6.4.4, we have:
$(T_2T_1)(A)=2tr(A)\\
T_2T_1:M_n (R) \rightarrow R$
Obtain: $Ker(T)=\{A \in M_n(R):T_2T_1(A)=0\}\\
=\{ A \in M_n(R):2tr (A)=0\}
=\{ A \in M_n(R): tr(A)=0\}\\
=\{[a_{ij}]\in M_n(R): \sum a_{ij}=0\}\\
=\{ [a_{ij}] \in M_n(R):a_{11}+a_{22}+...+a_{nn}=0\}\\
\rightarrow a_{11}+a_{22}+...+a_{nn}=0\\
\rightarrow a_{11}=-(a_{22}+...+a_{nn})\\
\dim [Ker (T)]=n^2-1$
Since $T$ is one-to-one $\rightarrow Ker (T)=0$
Hence, $T_2T_1$ is not one-to-one.
Apply Rank Nullity Theorem:
$\dim [Ker(T_2T_1)]+\dim [Rng(T)=\dim M_n(R)\\
(n^2-1)+\dim [Rng(T_2T_1)]=n^2\\
\dim [Rng(T_2T_1)]=1$
Since, $Rng(T_2T_1) \subset R\\
\dim R=1=\dim [Rng(T_2T_1)]\\
\rightarrow RngT_2T_1=R\\
\rightarrow T_2T_1$ is onto.
