A linear map $T$ in $M_n(K)$ Let $M_n(K)$ denote the space of all $n×n$ matrices with entries in a field $K$. Fix a non-singular matrix $A=(A_{ij})\in M_n(K)$ and consider the linear map $T:M_n(K)→M_n(K)$ given by:
$T(X)=AX$.
Then:


*

*$trace(T)=n\sum_{i=1}^nA_{ii}$

*$trace(T)=\sum_{i=1}^n\sum_{j=1}^nA_{ij}$

*$Rank(T)=n^2$
I am confused that in what sense options 1 and 3 are given correct? And then why not option 2 is correct in that sense? Please help.

 A: For (c), since $A$ is non-singular hence $T$ is also non-singular, so nullity$(T)=0$. By rank-nullity theorem rank$(T)=n^2$ which is the dimension of $M_n(K)$.
(a) can be proved by considering a $2\times2$ case and then generalized.
(b) is false because trace involves only diagonal entries. 
Another view for (c)
Suppose $$T(X)=T(Y)$$
$$AX=AY$$
$$A^{-1}AX=A^{-1}AY$$
$$X=Y$$
therefore $T$ is one-one, hence rank$(T)$=dim$(M_n(K))$
A: Consider the matrices $E_{ij}$, having coefficients always vanishing except for coefficient at row $i$ and column $j$ which has value equal to $1$.
$$\mathcal{E}=(E_{1,1}, \dots , E_{1,n},E_{2,1} \dots, E_{2,n},\dots,E_{n,1}, \dots, E_{n,n})$$ is a basis of $M_n(K)$.
You'll verify that for $1 \le i \le n , 1 \le j \le n$, you have $$T(E_{i,j})=\sum_{k=1}^n A_{k,i} E_{k,j}.$$ This is no more no less the way to get the matrix of $T$ in the basis $\mathcal{E}$ where $A=(A_{i,j})$.
To get (1), you can now apply the definition of the trace. The diagonal of the matrix of $T$ in the basis $\mathcal{E}$ is $$(A_{1,1},A_{1,1}, \dots,A_{1,1},A_{2,2},\dots,A_{2,2}, \dots, A_{n,n},\dots, A_{n,n})$$ proving that $\text{tr}(T)=n \sum_{i=1}^n A_{i,i}=n \text{tr}(A)$.
Obviously (2) is conflicting with (1)...
Regarding (3), what is the kernel of $T$? A matrix $X$ is in the kernel of $T$ if and only if $$T(X)=AX=0$$ which means that $A^{-1} A X=X=0$ as $A$ is supposed to be invertible. That means that $T$ is a non-singular linear map of $M_n(K)$ which is of dimension $n^2$. Hence $\text{rank}(T)=n^2$.
