Solving Linear Algebra Problem: Eliminating p_2 with Diagonal D

[hotvette]

I have a book that goes though a detailed development of the following:

$$\left[\begin{matrix}J^TJ & J^TV\\ VJ & D \end{matrix}\right]\left[\begin{matrix}p_1 \\ p_2\end{matrix}\right] = \left[\begin{matrix}J^Tr\\ Vr + Ds \end{matrix}\right]$$

where V and D are diagonal matrices and everything is known except $p_1$ and $p_2$.

Then it says "since the lower right submatrix D is diagonal, it is easy to eliminate $p_2$ from this system and obtain a smaller n x n system to be solved for $p_1$ alone." The implication is that it's so easy, explanation isn't needed. However, I don't see it.

Can someone explain how to eliminate $p_2$ given D is diagonal?

I've managed to come up with:

$$[VJ(J^TJ)^{-1}J^TV - D]p_2 = VJ(J^TJ)^{-1}J^Tr - Vr - Ds$$

but this seems far more complicated than what is implied and I don't see how D being diagonal simplifies anything.

 

[Rainbow Child]

If $D$ is diagonal with non zero eigenvalues, then it has an inverse $D^{-1}$. Thus from

$$V\,J\,p_1+D\,p_2=V\,r+D\,s$$

you can solve for $p_2$, i.e.

$$p_2=D^{-1}\,(V\,r+D\,s)-D^{-1}\,V\,J\,p_1$$

and use this to eliminate $p_2$ from the other equation.

 

[hotvette]

Thanks. I can't believe I didn't see it. So simple.