On tensor manipulation and algebra I am reading Quantum Field Theory in a Nutshell by Anthony Zee. On page 33, I can't figure out how he got equation 3.
The initial equation is $$[-(k^2 - m^2)g^{\mu\nu} + k^{\mu}k^{\nu}]D_{\nu\lambda}(k)=\delta^{\mu}_{\lambda}$$
which he manipulated to get equation (I.5.3) which is
$$D_{\nu\lambda}(k)=\frac{-g_{\nu\lambda}+k_{\mu}k_{\nu}/m^2}{k^2-m^2}.\tag{I.5.3}$$
Can anyone suggest me how to do it?
 A: Introducing a complete set of orthogonal projectors
$$\begin{align}P^{\mu\nu}_\perp=&g^{\mu\nu}-\frac{k^\mu k^\nu}{k^2}\\P^{\mu\nu}_\parallel=&\frac{k^\mu k^\nu}{k^2},\end{align}\tag{1}$$
which satisfy
$$P_\perp^{\mu\nu}+P_\parallel^{\mu\nu}=g^{\mu\nu}\tag{2.1}$$
$$(P_{(i)})^{\mu\nu}(P_{(j)})_{\nu\lambda}=(P_{(i)})^\mu_\lambda~\delta^{(j)}_{(i)}\tag{2.2}$$
where $i,j=\perp,\parallel$, you can write the tensor
$$A^{\mu\nu}=-(k^2-m^2)g^{\mu\nu}+k^\mu k^\nu\tag{3}$$
as a linear combination of them. You can see it's
$$A^{\mu\nu}=-(k^2-m^2)P_\perp^{\mu\nu}+m^2P_\parallel^{\mu\nu}.\tag{4}$$
We can also expand its inverse $D_{\nu\lambda}$ in terms of $(1)$
$$D_{\nu\lambda}=aP^{\perp}_{\nu\lambda}+bP^{\parallel}_{\nu\lambda}.\tag{5}$$
Imposing $A^{\mu\nu}D_{\nu\lambda}=\delta^{\mu}_\lambda$ and using $(2)$ you get $a=-1/(k^2-m^2)$ and $b=1/m^2$, i.e. the coeficients of each projector in $(5)$ are the inverse of the coeficients in $(4)$, so
$$D_{\nu\lambda}=\frac{-1}{k^2-m^2}P^{\perp}_{\nu\lambda}+\frac{1}{m^2}P^{\parallel}_{\nu\lambda},\tag{5}$$
