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The question I am trying to solve is: how to find the critical points of a matrix equation

$$X^{2}-BX-C=0?$$

And then if I follow the method of http://www2.math.umd.edu/~jmr/241/crits.html, first I should find the Jacobian, and there is where i find a doubt: How is it the differentiation of a matrix with respect to a matrix, I mean: I have a function depending on $X$,

$$f(X)=X^{2}-BX-C,$$

where $X$ is the variable and is a square matrix; $B$, $C$ are constant square matrices.

How should be the Jacobian? Is that correct to find the critical points in that way for matrix equation?.

Gina Torres
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2 Answers2

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Yes, to find the zeros of the derivative is the right way to find the critical points. Note that for your function $f \colon \def\Mat{\mathop{\rm Mat}\nolimits}\def\R{\mathbb R}\Mat_n(\R)\to \Mat_n(\R)$ we have for $X, H \in \Mat_n(\R)$ (I think it is easier to compute the derivative here using the definition): \begin{align*} f(X+H) &= A(X+H)^2 - B(X +H) - C\\ &= AX^2 - BX - C + AXH + AHX - BH + AH^2\\ &= f(X) + AXH + AHX-BH + o(\|H\|). \end{align*} So $f'(X)H = AXH + AHX-BH$. To compute the zeros, it may be helpful to vectorize, we have $\def\vec{\mathop{\rm vec}\nolimits}\def\Id{\mathop{\rm Id}\nolimits}$ $$ \vec(f'(X)H) = \vec(AXH) + \vec(AHX)-\vec(BH) = (\Id \otimes (AX-B))\vec(H) + (X^t \otimes A)\vec(H) $$ So $f'(X) = 0$ iff $\Id \otimes (AX-B) = -X^t \otimes A$.

martini
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Consider the mapping $f \, : \, \mathcal{M}_{n}(\mathbb{R}) \, \rightarrow \, \mathcal{M}_{n}(\mathbb{R})$ defined by :

$$\forall X \in \mathcal{M}_{n}(\mathbb{R}), \; f(X) = AX^{2}-BX-C$$

Let $X$ be a matrix in $\mathcal{M}_{n}(\mathbb{R})$, you can determine the differential of $f$ at $X$ (it is a linear mapping from $\mathcal{M}_{n}(\mathbb{R})$ to itself whose matrix is the Jacobian matrix). To do so : take some matrix $H$ and consider $f(X+H)$.

We have:

$$ \begin{eqnarray*} f(X+H) & = & A(X+H)^{2} - B(X+H) - C \\ & = & A\big( X^{2} + XH + HX + H^{2} \big) - BX - BH - C \\ & = & f(X) + AXH + AHX - BH + H^{2} \\ \end{eqnarray*} $$

It proves that :

$$ \mathrm{D}_{X} f \cdot H = AXH + AHX - BH $$

The critical points of $f$ are such that :

$$ \forall H, \; (AX-B)H = - AHX $$

However, I don't know it these critical points can be determined in closed form. This article http://imajna.oxfordjournals.org/content/20/4/499.full.pdf+html presents numerical methods which can be used to solve quadratic matrix equations such as $AX^{2}-BX-C$.

pitchounet
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  • Which is the relationship between the critical points and the solutions found for $AX^2−BX−C=0$? – Gina Torres Oct 08 '13 at 16:10
  • I don't think there exists a particular relationship between matrices $X$ such that $AX^{2}-BX-C=0$ (the zeros of $f$) and the critical points of $f$. For example, if $A$ is invertible, $f$ has $1$ critical point which is $X = \frac{1}{2}A^{-1}B$. At this point, $f(X) = -\frac{1}{4} BA^{-1}B - C$. There is no reason for $f(X)$ to be zero. Conversely, if you take a matrix $X$ such that $AX^{2}-BX-C = 0$, I don't see why it would be a critical point of $f$. – pitchounet Oct 09 '13 at 09:27
  • Ok I understood and yes A is invertible, Thanks. And last point: Should X as a critical point make $Id⊗(AX−B)=−X^t⊗A$ in @martini 's equation, because I have proved it numerically and both sides of the equation are different I don't know why. – Gina Torres Oct 09 '13 at 09:50
  • I don't know what $\mathrm{Id} \otimes (AX-B)$ means. $X$ is a critical point of $f$ if, for all matrix $H$, we have :

    $$ AXH+AHX-BH = 0 $$

    which is equivalent to ask that

    $$ (AX-B)H = -AHX $$

    for all $H$. I guess it means the same thing.

     – pitchounet Oct 09 '13 at 10:04
  • In martini's answer appears after vectorizing $f′(X)H=AXH+AHX−BH$, he says that $f′(X)=0 iff Id⊗(AX−B)=−X^t⊗A$, $⊗$ is the kronecker product. So what I did was to test if the X critical point found with your equation demonstrate his equation. But doesn't. However, I understand your point and I guess maybe i am commiting a mistake somewhere. – Gina Torres Oct 09 '13 at 10:14
  • No, you're right. If I take $X = \frac{1}{2}A^{-1}B$, then we don't have $(AX-B)H=-AHX$ for all $H$. I'll edit my post. – pitchounet Oct 09 '13 at 11:28
  • But how are the critical points, then?. Does $X=1/2 A^{−1}B$ continue being a critical point in the way you have recommended? – Gina Torres Oct 09 '13 at 12:13
  • I don't know if the critical points of $f$ can be determined in closed form. I think it can't. Have a look at http://imajna.oxfordjournals.org/content/20/4/499.full.pdf+html (I will edit my post). If that doesn't answer your question, try asking about it on Math.SE! – pitchounet Oct 09 '13 at 13:28
  • I have written equation as AX^2-BX-C=0 but A is a identity matrix, so if it is the identity $(X−B)H=−HX$ is as well correct, right? And in thi case how should I consider H? identity for example? – Gina Torres Oct 10 '13 at 12:30
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    If $A$ is the identity matrix, you have to find $X$ such that :

    $$ (X-B)H=-HX $$

    holds for any $H$, not just the identity matrix.

     – pitchounet Oct 10 '13 at 15:25
  • But is that possible, because here I have one more matrix variable H? one equation and 2 variables (X,H), so infinite solutions, isn't it? – Gina Torres Oct 10 '13 at 15:54