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Suppose a discrete matrix equation $$X^{2}-BX-C=0$$, B and C are known square constant matrices, B is invertible and C is not, X is the variable and also is a square matrix. Which is the method to find its critical points? Are these critical points the same as the solutions of the equation?

Gina Torres
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  • What do you understand by "critical points"? Roots? – TZakrevskiy Oct 15 '13 at 13:17
  • In continuous functions I understand that critical points are points where all derivatives of the function become zeros, and I could find them by deriving the function and find its roots I mean f'(X)=0, in matrix equations is more complex but in general they are the same. Roots of a equation are solutions of the function in where it is always zero, but sometimes roots and critical points coincide for example sen(X)=X, that is why I am asking. But in the case of Discrete matrix equations how should I manage it? – Gina Torres Oct 15 '13 at 13:24
  • Well, the derivative of a matrix function with respect to a matrix is a tensor of fourth order, hence I suggest using the Einstein notation to write an expression for the derivative. On the other hand, solving a nonlinear matrix equation is a very non-evident task; consider, for example, the equation $X^2=0$. The solutions form a manifold (in particular, there's a continuum of solutions). – TZakrevskiy Oct 15 '13 at 13:44
  • Ok, but threating with a discrete matrix equation, do derivatives also work? Because as I know the derivative of a function exist if it is continuous, for discrete equations how derivatives are?, what is an Einstein Notation, could you explain a bit? – Gina Torres Oct 15 '13 at 13:54
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    Einstein notation is a way to simplify notations in large sums. As for derivatives, let $f:M_n(\Bbb C)\to M_n(\Bbb C)$ a matrix function. To find the derivative, you take $ij$-th component of $f(X)$ and derive it with respect to $X_{kl}$. Like this you obtain $ijkl$-th component of $\nabla_X f(X)$ (there're four indices, hence it's a tensor of fourth order). – TZakrevskiy Oct 15 '13 at 14:13
  • Ok but the question is: Does the same method to find critical points apply for discrete equations? Because as I said, derivatives just exist if function is continuous. – Gina Torres Oct 15 '13 at 14:20
  • Could you elaborate on derivatives just exist if function is continuous? We all know continuous functions that are nowhere differentiable... – TZakrevskiy Oct 15 '13 at 15:58
  • I don't know, that is why I am asking, could you give some reference to read about it, and could you explain me please? – Gina Torres Oct 16 '13 at 08:36
  • My previous comment was about what I knew but of course you can advice me and correct me if it is not true, that is why I am asking. As I have researched today of course it is possible to derivate a discrete function could you explain me the process more detailed about the tensor?. – Gina Torres Oct 16 '13 at 09:17
  • let us continue this discussion in chat – TZakrevskiy Oct 16 '13 at 10:03
  • I have found something that not coherent respect to matrices, for example if matrices are 2 x 2 so equation $$I⊗ (X-B)=-X^t ⊗ I$$, left side you have a tensor 4 x 4 like $$\begin{pmatrix}X11-B11& X12-B12& 0& 0\X21-B21 &X22-B22& 0 &0\0& 0 &X11-B11 & X12-B12\0& 0& X21-B21 &X22-B22\end{pmatrix}$$, and right side you have $$\begin{pmatrix}-X11& 0 &-X21& 0\ 0 &-X11& 0& -X21\-X12 &0& -X22 &0\0& -X12& 0 &-X22\end{pmatrix}$$, so for X12 and X21 by equating both sides you have two equations one is X12=0 and other is that X12-B12=0 and for X21 you have X21=0 X21-B21=0. What happen with equation? – Gina Torres Oct 16 '13 at 13:10
  • I have to add that B12 and B21 are not zero. – Gina Torres Oct 16 '13 at 13:11
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    Well, in this case we obtain an equation with no solutions, i.e. the function has no critical points. – TZakrevskiy Oct 16 '13 at 20:17
  • But, is the equation correct? – Gina Torres Oct 16 '13 at 20:33
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    Yes, the equation is correct. – TZakrevskiy Oct 17 '13 at 09:28
  • Ok so the results reflect the nature of B. – Gina Torres Oct 17 '13 at 09:44

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