Perfect square in terms of matrices

Question

If $a,b\in \mathbb R$ are constants, $x \in \mathbb R$ is the unknown variable, then I know how to perform a perfect square: $ax^2+bx=a(x^2+\frac{bx}{a}+\frac{b^2}{4a^2})-\frac{b^2}{4a}=a(x+\frac{b}{2a})^2-\frac{b^2}{4a}$.

Now my question is what if $a,b$ are $n\times n$ constant matrices and $x \in \mathbb R^n$ is a $n\times 1$ unknown vector, $c$ is a constant vector?

How to perform a perfect square for $xAx^{T}+xBc$? Can we get something like $A(x-*)^2 -B*$?

This can be analogous to central conic or quadric, see another post here for your further interest or application. — Ng Chung Tak, Aug 20 '21 at 19:46

Ben Grossmann · Accepted Answer · 2023-08-04T03:21:57.403

2

If $A$ is a symmetric matrix, we have $$ x^TAx + b^Tx = \left(x + \frac 12 A^{-1}b\right)^TA\left(x + \frac 12 A^{-1}b\right) - \frac 14 b^TA^{-1}b. $$ If $A$ is not a symmetric matrix, note that $x^TAx = x^TSx$, where $S = \frac 12 (A + A^T)$, and apply the above to the expression $x^TSx + b^Tx.$

edited Aug 04 '23 at 03:21

answered Aug 20 '21 at 19:26

Ben Grossmann

225,327

For dimensional consistency you'll need to change $,x^T!\to x;$ inside of the parentheses. – greg Aug 21 '21 at 12:51
@greg Thanks, fixed it – Ben Grossmann Aug 21 '21 at 14:36
@BenGrossmann After a second checking, it seems the LHS is not equal to RHS. On the RHS, I got terms like $b^{T}x+x^{T}b$. – Topoguy Aug 27 '21 at 16:55
@Topoguy But $b^Tx = x^Tb$; both are just the dot-product of $b$ and $x$. – Ben Grossmann Aug 27 '21 at 17:34
@BenGrossmann Yes, you are right. – Topoguy Aug 27 '21 at 17:36
Would you be able to explain how we could have seen this a priori? With scalars I can follow an algorithm involving factoring and adding and subtracting things, but here all I can see if that I can verify the RHS equals the LHS. – EE18 Aug 04 '23 at 01:48
@EE18 Consider the expression $$ \left(x + y\right)^TA\left(x + y\right) = x^TAx + 2y^TAx + y^Ty. $$ You want to find a $y$ such that the "$x$-terms" of this expression match those of $x^TAx + b^Tx$. – Ben Grossmann Aug 04 '23 at 19:43
Thank you! @BenGrossmann – EE18 Aug 04 '23 at 20:06

Perfect square in terms of matrices

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