I was able to figure out the first part of this problem, but I have no concept of how it relates to Schur complements, so I'm not sure (no pun intended) how to proceed. The question is as follows:
Consider $2x^2 + 2xy + 2y^2 + z^2 + 2xz$. Write the symmetric matrix representing this quadratic form. Now, express this as a sum of squares by using this symmetric matrix and Schur complements.
I determined the symmetric matrix representation as:
$$\begin{bmatrix}
2 & 1 & 1 \\
1 & 2 & 1 \\
1 & 0 & 1 \\
\end{bmatrix} $$
And that's as far as I've gotten. Any help would be much appreciated.
Best Answer
I think what the question means is to apply the $L^TDL$ decomposition that produces Schur complements, rather than to use Schur complements themselves. In other words, you are expected to apply recursively the decomposition $$ \pmatrix{A&B\\ B^T&C}= \pmatrix{I&BC^{-1}\\ 0&I} \pmatrix{A-BC^{-1}B^T&0\\ 0&C} \pmatrix{I&0\\ C^{-1}B^T&I} $$ to diagonalise a symmetric matrix by matrix congruence. In your case, apply $L^TDL$ decomposition once, we get $$ \left(\begin{array}{cc|c} 2&1&1 \\ 1&2&0 \\ \hline 1&0&1 \end{array}\right) =\pmatrix{1&0&1\\ 0&1&0\\ 0&0&1} \pmatrix{1&1&0\\ 1&2&0\\ 0&0&1} \pmatrix{1&0&0\\ 0&1&0\\ 1&0&1}. $$ Apply the decomposition once more to the submatrix $ \left[\begin{array}{c|c} 1&1\\ \hline 1&2 \end{array}\right], $ we get \begin{align} \pmatrix{ 2&1&1\\ 1&2&0\\ 1&0&1} =\pmatrix{1&0&1\\ 0&1&0\\ 0&0&1} \pmatrix{1&\frac12&0\\ 0&1&0\\ 0&0&1} \pmatrix{\frac12&0&0\\ 0&2&0\\ 0&0&1} \pmatrix{1&0&0\\ \frac12&1&0\\ 0&0&1} \pmatrix{1&0&0\\ 0&1&0\\ 1&0&1}. \end{align} Therefore, your matrix can be written as $L^TDL$ for some invertible matrix $L$, where $D=\operatorname{diag}(\frac12,2,1)$. Put $(u,v,w)=(x,y,z)L^T$, your quadratic form becomes $\frac12u^2+2v^2+w^2$, which is a sum of squares.