# Completion of the squares (actually of the quadratic forms)

I have this paragraph in the book I cannot understand (this is not the first time I encounter this thing but I usually move on).

$$A^\prime TA+A^\prime QB+B^\prime UA+B^\prime RB = (A+T^{-1}QB)^\prime T(A+T^{-1}QB)+B^\prime (R-UT^{-1}Q)B$$

I can do the calculus in the reverse order, but I don’t know how is someone supposed to figure out how to do it, and I don’t understand why such a procedure is called “completing the squares”.

Can someone please elaborate on that?

NOTE: prime stands for transpose and those are all matrices.

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Just curious, what book are you using? –  Edison Jun 20 '11 at 20:07
Various: System Identification (Stoica), Estimation with application to tracking and navigation (Li, Bar-Shalon)... –  gurghet Jun 20 '11 at 21:03

Hint:
Completing the square for a quadratic polynomial of the form $ax^{2}+2bxy+dy^{2}$ is given by $$a\left(x+\frac{b}{a}y\right)^{2}+\left(d-\frac{b^{2}}{a}\right)y^{2}.$$

If $A$ and $D$ are symmetric square matrices, and $B$ is $n \times m$, then

$$\left[ {\begin{array}{cc} x \\ y \\ \end{array} } \right]^T \left[ {\begin{array}{cc} A & B \\ B^T & D \\ \end{array} } \right] \left[ {\begin{array}{cc} x \\ y \\ \end{array} } \right] = \left(x+A^{-1}By\right)^{T}A\left(x+A^{-1}By\right)+y^{T}\left(D-BA^{-1}B\right)y$$
Note that the left hand side of your equation can be written as

$$\left[ {\begin{array}{cc} A^T & B^T \\ \end{array} } \right] \left[ {\begin{array}{cc} T & Q \\ U & R \\ \end{array} } \right] \left[ {\begin{array}{cc} A \\ B \\ \end{array} } \right].$$

Hope this helps.

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