# 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:

1. $trace(T)=n\sum_{i=1}^nA_{ii}$
2. $trace(T)=\sum_{i=1}^n\sum_{j=1}^nA_{ij}$
3. $Rank(T)=n^2$ 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.

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$(M_n(K))$

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 $\text{rank}(T)=n^2$.