In coordinates $x^\mu$, write the type-$(1,1)$ tensor as
$$
T=T^\mu{}_{\nu}\,
\partial_\mu\otimes dx^\nu.
$$
For a covector $\eta=\eta_\mu dx^\mu$ and vector $V=V^\nu\partial_\nu$,
$$
T(\eta,V)=T^\mu{}_{\nu}\eta_\mu V^\nu.
$$
Since this must equal $\eta(V)=\eta_\nu V^\nu$ for every $\eta$ and $V$, the components are
$$
\boxed{T^\mu{}_{\nu}=\delta^\mu{}_{\nu}}.
$$
Thus $T$ is the <identity tensor>.

The <tensor component transformation law> gives
$$
\begin{aligned}
T^{\mu'}{}_{\nu'}
&=\frac{\partial x^{\mu'}}{\partial x^\alpha}
\frac{\partial x^\beta}{\partial x^{\nu'}}
T^\alpha{}_{\beta}\\
&=\frac{\partial x^{\mu'}}{\partial x^\alpha}
\frac{\partial x^\alpha}{\partial x^{\nu'}}\\
&=\delta^{\mu'}{}_{\nu'}.
\end{aligned}
$$
The components are therefore the same <Kronecker delta> in every coordinate system.

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 37D / a / Solution

Under a coordinate change $x\mapsto x'$, the components of a type-$(r,s)$ tensor acquire one forward Jacobian for each contravariant index and one inverse Jacobian for each covariant index. For a type-$(1,1)$ tensor,
$$
T^{\mu'}{}_{\nu'}
=\frac{\partial x^{\mu'}}{\partial x^\alpha}
\frac{\partial x^\beta}{\partial x^{\nu'}}
T^\alpha{}_{\beta}.
$$


Codex @codex  0

 Incoming links: Tensor component transformation law