Is there a special meaning for this operator? Does the operator
$$\prod_{j\neq i} \frac{\mathbf{\Lambda}-\lambda_i}{\lambda_j-\lambda_i}$$
have any special meaning that makes it useful in quantum mecahnics, given that $\{\lambda_i\}$ are the eigenvalues of the Hermitian non-degenerate operator $\mathbf\Lambda$?
 A: It is the projection operator onto the eigenspace to $\lambda_j$.


*

*It evaluates to $1$ on $\mathrm{Eig}(\lambda_j)$.

*Every other eigenspace is annihilated because of the nominator. Note that the product is over all but one eigenvalue.

*It is self-adjoint (every factor is self-adjoint and they mutually commute) and

*squares to itself.


Due to the product it is useful/practical to use in finite dimensions only I believe.
I have somewhere seen it utilized in the context of spin chains, where one would project composite spins onto a definite $\vec{S}^2$ subspace. It was a seminar talk, but maybe I can find reference.
A: Let us assume that $|\lambda\rangle$ is an arbitary state ket of $\Lambda$ i.e $\Lambda|\lambda\rangle=\lambda|\lambda\rangle$. 
Then
\begin{eqnarray}
\left[\prod_{\lambda''\neq\lambda'}\frac{\Lambda-\lambda''}{\lambda'-\lambda''}\right]|\lambda\rangle & = & \left[\prod_{\lambda''\neq\lambda'}\frac{\Lambda-\lambda''}{\lambda'-\lambda''}\right]\sum_{\lambda'''}\overbrace{|\lambda'''\rangle \langle\lambda'''|}^{\mathbb{I}}\lambda\rangle\\
& = & \sum_{\lambda'''}\langle\lambda'''|\lambda\rangle\left[\prod_{\lambda''\neq\lambda'}\frac{\Lambda-\lambda''}{\lambda'-\lambda''}\right]|\lambda'''\rangle\\
& = & \sum_{\lambda'''}\langle\lambda'''|\lambda\rangle\overbrace{\left[\prod_{\lambda''\neq\lambda'}\frac{(\lambda'''-\lambda'')}{(\lambda'-\lambda'')}\right]}^{\delta_{\lambda'\lambda'''}}|\lambda'''\rangle\\
& = & \sum_{\lambda'''}\langle\lambda'''|\lambda\rangle\delta_{\lambda'\lambda'''}|\lambda'''\rangle\\
& = & \sum_{\lambda'}|\lambda'\rangle\langle\lambda'|\lambda\rangle\\
\left[\prod_{\lambda''\neq\lambda'}\frac{\Lambda-\lambda''}{\lambda'-\lambda''}\right] & = & \sum_{\lambda'}|\lambda'\rangle\langle\lambda'| = \mathbb{P}_{\lambda'}
\end{eqnarray}
So it projects to the eigenket $|\lambda'\rangle$.
Regrading the uselfuness of $\mathbb{P}_{\lambda'}$ in quantum mechanics,
Lets consider the spin operator $S_{z}$, which has representation in terms of $|+\rangle = \begin{pmatrix} 1 \\0 \end{pmatrix}$ and $|-\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ as, 
$$S_{z}=\frac{\hbar}{2}(|+\rangle|\langle+|- |-\rangle\langle-|)$$ 
Where, $S_{z}|+\rangle=\frac{\hbar}{2}|+\rangle$ and $S_{z}|-\rangle=-\frac{\hbar}{2}|-\rangle$.
Then, for $\lambda'=\frac{\hbar}{2}$, we have
\begin{eqnarray}
\left[\prod_{\lambda''\neq\lambda'}\frac{S_{z}-\lambda''}{\lambda'-\lambda''}\right] & = & \left[\prod_{\lambda''\neq\hbar/2}\frac{S_{z}-\lambda''}{(\hbar/2-\lambda'')}\right] \\
& = & \left[\frac{S_{z}+\frac{\hbar}{2}\mathbb{I}}{\hbar/2+\hbar/2}\right]\\
& = & \frac{1}{\hbar}\left[\frac{\hbar}{2}(|+\rangle|\langle+|- |-\rangle\langle-|)+\frac{\hbar}{2}(|+\rangle|\langle+|+ |-\rangle\langle-|)\right]\\
& = & \frac{1}{\hbar}.\hbar|+\rangle|\langle+|\\
& = & |+\rangle\langle+|
\end{eqnarray}
Likewise So, for $\lambda'=-\frac{\hbar}{2}$, it can be shown that $\prod_{\lambda''\neq\lambda'}\frac{S_{z}-\lambda''}{\lambda'-\lambda''}=|-\rangle\langle-|$.
Which in fact justify $\mathbb{P}$ as the projection operators for spin-1/2 system.
