Note that
$$
A=\boldsymbol{n}\boldsymbol{1}^t+i\boldsymbol{1}\boldsymbol{n}^t+(1+i)\mathrm{diag}(\boldsymbol{n}),
$$
where $\boldsymbol{n}=(1,2,\ldots,n)$, $\boldsymbol{1}=(1,1,\ldots,1)$.
We need to show that $A\boldsymbol{u}=0$, implies that $\boldsymbol{u}=0$.
We have
$$
A\boldsymbol{u}=\boldsymbol{n}\boldsymbol{1}^t\boldsymbol{u}+i\boldsymbol{1}\boldsymbol{n}^t
\boldsymbol{u}+(1+i)\mathrm{diag}(\boldsymbol{n})\boldsymbol{u}=
(\boldsymbol{1},\boldsymbol{u})\boldsymbol{n}+i(\boldsymbol{n},
\boldsymbol{u})\boldsymbol{1}+(1+i)\mathrm{diag}(\boldsymbol{n})\boldsymbol{u}. \tag{1}
$$
If $\boldsymbol{u}=(u_1,\ldots,u_n)$, then $A\boldsymbol{u}=0$, implies that
$$
\mathrm{diag}(\boldsymbol{n})\boldsymbol{u}=(u_1,2u_2,\ldots,nu_n),
$$
is a linear combination of $\boldsymbol{1}$ and $\boldsymbol{n}$, i.e.,
$$
(u_1,2u_2,\ldots,nu_n)=c_1(1,1,\ldots,1)+c_2(1,2,\ldots,n), \tag{2}
$$
with $c_1$ and $c_2$ not both vanishing. Plugging $(2)$ to $(1)$ we get
$$
0=A\boldsymbol{u}=(\boldsymbol{1},c_1\boldsymbol{1}+c_2\boldsymbol{n})\boldsymbol{n}
+i(\boldsymbol{n},
c_1\boldsymbol{1}+c_2\boldsymbol{n})\boldsymbol{1}+(1+i)\big(c_1\boldsymbol{1}+c_2\boldsymbol{n}\big).
$$
Equivalently
$$
i(\boldsymbol{n},c_1\boldsymbol{1}+c_2\boldsymbol{n})+c_1(1+i)=0 \tag{$A_1$}
$$
and
$$
(\boldsymbol{1},c_1\boldsymbol{1}+c_2\boldsymbol{n})+c_2(1+i)=0. \tag{$A_2$}
$$
It is not hard to calculate the $2\times 2$ determinant of the system $(A_1)-(A_2)$ and obtain that the only solution is $c_1=c_2=0$.