Trace of the transformation $T$ is a linear transformation from $M(n,\Bbb R)$ to $M(n,\Bbb R)$ and $A$ be a non zero nilpotent matrix of order $n\times n$, $T(X)=(A-I)X$, then what will be the trace of $T$?
 A: For $(i, j) \in \{1, \dots, n\}^2 =: [n]^2$, let $E_{i,j}$ be the $n\times n$ matrix with a $1$ in the $i,j$ entry and zeros elsewhere. The set $\{ E_{i,j} : (i,j) \in [n]^2\}$ forms a basis of $M(n, \mathbb R)$. Note $M(n, \mathbb R) \simeq \mathbb R^{n^2}$, and hence $E_{i,j}$ may be identified with the standard orthonormal basis of the inner-product space $\mathbb R^{n^2}$ (once we enumerate $[n]^2$). This turns $M(n, \mathbb R)$ into an inner-product space, and hence $\operatorname{tr}(T) = \sum_{i, j} \langle T E_{i,j}, E_{i,j} \rangle$, where the inner-product on $M(n, \mathbb R)$ is computed by sending matrices in $M(n, \mathbb R)$ to vectors in $\mathbb R^{n^2}$ and computing the standard inner-product there.

Let $e_1, \dots, e_n$ be the standard basis of $\mathbb R^n$. Since $A$ is nilpotent, it's traceless, i.e.,  $\sum_{i=1}^n \langle A e_i, e_i \rangle = 0$, from which we deduce
$
\sum_{i,j} \langle A E_{i,j}, E_{i,j} \rangle = 0.
$
It's not hard to verify that $\sum_{i,j} \langle I E_{i,j}, E_{i,j} \rangle = n^2$. Using the above formula for $\operatorname{tr}(T)$, we find
$$
\operatorname{tr}(T) = \sum_{i,j} \langle (A - I) E_{i,j}, E_{i,j} \rangle = \sum_{i,j} \langle A E_{i,j}, E_{i,j} \rangle - \sum_{i,j} \langle I E_{i,j}, E_{i,j} \rangle = -n^2.
$$ 

An alternative way to see this: if the isomorphism $M(n, \mathbb R) \to \mathbb R^{n^2}$ is given by vertically stacking column $1$, $\dots$, column $n$ of a matrix, then the matrix representation of "$T = A-I$" as an $M(n, \mathbb R)$ operator is $A \otimes I - I \otimes I$ (where $\otimes$ is the tensor or Kronecker product), and using the property $\operatorname{tr}(U \otimes V) = \operatorname{tr}(U) \operatorname{tr}(V)$, we have
$
\operatorname{tr}(T) = \operatorname{tr}(A \otimes I - I \otimes I) = 0 \cdot n - n \cdot n = -n^2.
$
