Let $M_n(K)$ denote the space of all $n\times n$ matrices with entries in a field $K$. Fix a non-singular matrix $A=(A_{ij})\in M_n(K)$, and consider the linear map $T:M_n(K)\to M_n(K)$ given by: $$T(X)=AX$$ Then:
trace $(T)$ = $n\times\sum_{i=1}^nA_{ii}$
trace $(T)$ = $\sum_{i=1}^n$$\sum_{j=1}^nA_{ij}$
I have proved that option 1 is true when we take $A$ and $X$ as $2\times 2$ matrix, but for a matrix of order $n$, I cannot understand how to prove it, can anyone help me to solve it out? Thanks.
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$\begingroup$The trace of $T$ is the trace of the Kronecker product $I\otimes A$, therefore it should be $\sum_{i=1}^n\sum_{j=1}^na_{jj}=n\operatorname{trace}(A)$.
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