###### tags: `one-offs` `linear algebra` # A Mnemonic for Schur Complements **Overview**: In this note, I describe a quick strategy for recalling the form of the Schur complement decomposition of a positive semidefinite matrix. ## The Technique For some time, I had difficulty remembering the exact form of the Schur complement formula. I thus began working with the following mnemonic: given a positive semidefinite matrix $M$, write it in block form as \begin{align} M=\left[\begin{array}{cc} A & B\\ B^{\top} & C \end{array}\right], \end{align} where $A$ and $C$ are positive semidefinite. Writing the state variable as $z=\left(x,y\right)$, define \begin{align} f\left(z\right) &=\frac{1}{2}z^{\top}Mz \\ &=\frac{1}{2}\left(x^{\top}Ax+x^{\top}By+y^{\top}B^{\top}x+y^{\top}Cy\right). \end{align} Now, minimise $f\left(x,y\right)$ with respect to $y$, with $x$ held fixed. The first-order stationarity condition imposes that $B^{\top}x+Cy=0$, and so we set $y=-C^{-1}B^{\top}x$ to see that \begin{align} f\left(x,y\right) &\geqslant \frac{1}{2}\left(x^{\top}Ax-x^{\top}BC^{-1}B^{\top}x-\left(C^{-1}B^{\top}x\right)^{\top}B^{\top}x+\left(C^{-1}B^{\top}x\right)^{\top}C\left(C^{-1}B^{\top}x\right)\right) \\ &= \frac{1}{2}x^{\top}\left(A-BC^{-1}B^{\top}\right)x \\ &=: \frac{1}{2}x^{\top}Sx. \end{align} By construction, S must itself be positive semidefinite. Piecing things back together, one can then write that \begin{align} f\left(x,y\right) =\frac{1}{2}x^{\top}Sx+\frac{1}{2}\left(y+C^{-1}B^{\top}x\right)^{\top}C\left(y+C^{-1}B^{\top}x\right), \end{align} which somehow gives the full story: $M$ is PSD iff both of $S$ and $C$ are, and $S$ has this specific form. I like this approach as a mnemonic because one simply has to remember “minimise $y$ given $x$”, and routine linear algebra does the rest of the work.
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