Vtyjkyuk

I really cant understand what could be tr(T) actually. Also how to show T is non singular with rank n^2, please need a help with elaboration. Actually I found from somewhere that tr(T)= n*tr(A) regarding this l. t I have mentioned, but how? w. r. t std basis of M(n, R), m(T) gives a diagonal n^2*n^2 diagonal which does not satisfy that trace property. I'm wrong somewhere, please help to find that

If $X$ is a finite dimensional linear space, with basis $x_1,ldots,x_m$, and $T:Xto X$ a linear transformation, defined as
$$
Tx_i=sum_i=1^m t_i,jx_j,
$$
then its trace is defined to be Tr$(T)=sum_i=1^n t_ii$.

In the question, consider as a basis, $E_ij$, $i,j=1,ldots,n$, the matrix with $1$ in the $(i,j)$ place, and zeroes everywhere else.

Then
$$
T(E_i,j)=sum_k=1^n a_k,iE_k,j
$$
and hence
$$
textTr(T)=n(a_1,1+cdots+a_n,n)=ntextTr(A).
$$

Also note that if $X=(x_1,x_2cdots x_n)inmathbb R^ntimes n$, where $x_1,ldots, x_n$ re the columns of $X$, then $T(X)=(Ax_1,Ax_2cdots Ax_n)$. So if $T(X)=0$, then $Ax_1=cdots=Ax_n=0$, and since $A$ is nonsingular, so is $T$.

In fact, $p_T(lambda)=big(p_A(lambda)big)^n$, where $p_A$ and $p_T$ are the characteristic polynomials of $A$ and $T$, respectively.

Hint:

Let $E_ij : 1 le i, jle n$ be the canonical basis of the space $M_n(mathbbR)$. The trace $operatornameTr T$ is the sum of diagonal elements of the matrix of $T$ w.r.t. this basis.

If we denote the columns of $A = (a_ij)$ by $A = beginbmatrix A_1 & A_2 & cdots & A_nendbmatrix$, verify that
$$T(E_ij) = AE_ij = beginbmatrix 0 & cdots & 0 & A_i & 0 & cdots & 0 endbmatrix = sum_k=1^n a_kiE_kj$$
where the $i$-th column of $A$ appears as the $j$-th column here.

Therefore the diagonal elements of the matrix of $T$ are $$underbracea_11, ldots, a_11_n, underbracea_22, ldots, a_22_n, ldots, underbracea_nn, ldots, a_nn_n$$ so $$operatornameTr T = nsum_i=1^n a_ii =n operatornameTr A$$