# Question #f4e98

Mar 15, 2017

$t r \left(2 {A}^{7} + 3 {A}^{5} + 4 {A}^{2} + A + I\right) = 6$

#### Explanation:

If $A$ is nilpotent with order $2$; then

${A}^{2} = 0$

And more importantly;

${A}^{m} = 0 \forall m \in \mathbb{N} , m \ge 2$

From which we can deduce that:

$t r \left({A}^{m}\right) = 0 \forall m \in \mathbb{N} , m \ge 2$

And we can use the trace properties:

$t r \left(A + B\right) = t r \left(A\right) + t r \left(B\right)$
$t r \left(m A\right) \setminus \setminus \setminus \setminus = m \setminus t r \left(A\right)$
$t r \left({I}_{n}\right) \setminus \setminus \setminus \setminus \setminus \setminus \setminus = n$ where ${I}_{n}$ is the $n \times n$ identity matrix

And so:

$t r \left(2 {A}^{7} + 3 {A}^{5} + 4 {A}^{2} + A + I\right)$
$\text{ } = t r \left(2 {A}^{7}\right) + t r \left(3 {A}^{5}\right) + t r \left(4 {A}^{2}\right) + t r \left(A\right) + t r \left(I\right)$
$\text{ } = 2 t r \left({A}^{7}\right) + 3 t r \left({A}^{5}\right) + 4 t r \left({A}^{2}\right) + t r \left(A\right) + t r \left(I\right)$
$\text{ } = 0 + 0 + 0 + t r \left(A\right) + t r \left(I\right)$
$\text{ } = t r \left(A\right) + t r \left(I\right)$
$\text{ } = 3 + 3$
$\text{ } = 6$

Mar 15, 2017

If ${A}^{2} = 0$ then

$B = 2 {A}^{7} + 3 {A}^{5} + 4 {A}^{2} + A + I = A + I$

and $\text{tr"(B)="tr"(A)+"tr} \left(I\right) = 3 + 3 = 6$