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I am attempting to use Maple to solve a matrix equation of the form $aB^2 + bB + cI = B^{-2}$, where $B$ is a $3 \times 3$ matrix, and I is the $3 \times 3$ identity matrix (i.e. to find the values a, b and c which satisfy the equation).

My idea was to use a piece of code like 

A:=Matrix([[7,4,-2],[4,7,5],[2,-3,8]]);
d:=Vector([[8],[5],[2]]);

for matrix inversion to solve a system of three linear equations but that would assume that the right-hand side of the equation is a column vector when it is actually another $3 \times 3$ matrix. Is there a simple way to look at this which I am missing?

Math Lover
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Tom
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  • I don't know anything about Maple, but I do know that at worse you could translate this problem into a simultaneous system of $27$ equations with $3$ unknowns by simply computing coordinate-wise. I would hope that Maple would have something within that which would not force you to do this, but ultimately it would work. – davidlowryduda Dec 21 '17 at 22:12
  • I was hoping there might be something in the linear algebra package with Maple: could you provide a bit more information on how I would follow your approach? – Tom Dec 22 '17 at 00:00
  • Sure. An equality between 3x3 matrices is also an equality between all the entries of those matrices. So compute the matrix entries for $B^2$ and $B^{-2}$, and then look at the equality for all the $(1,1)$ coordinates, and all the $(1,2)$ coordinates, and so on, for each of the matrices. If there is a solution, then a solution will be found very quickly. – davidlowryduda Dec 22 '17 at 00:13
  • Not to worry, I've mentioned to solve it now: I used something like your suggestion, it was just a case of writing the code to get Maple to do it. – Tom Dec 24 '17 at 04:27

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using new letters, there is the characteristic polynomial for $B.$ As $B$ is 3 by 3 and invertible, we have $r \neq 0$ in $$ B^3 + p B^2 + q B + r I = 0. $$ When needed, this gives us the useful $$ B^3 = -p B^2 - q B - r I . $$ Next, we get $$ B^4 = -p B^3 - q B^2 - r B = -p (-p B^2 - q B - r I ) - q B^2 - r B , $$ $$ B^4 = (p^2 - q) B^2 + (pq -r)B + prI. $$ Looks right. For you, these coefficients will be integers. Also, you should confirm with the actual numbers that show up.

Back to $ B^3 + p B^2 + q B + r I = 0, $ we have $r \neq 0$ in $$ B^3 + p B^2 + q B = -rI, $$ $$ B^2 + p B + q I = -r B^{-1}. $$ So $$ B^{-1} = - \frac{1}{r} B^2 - \frac{p}{r} B - \frac{q}{r} I. $$ Nice rational coefficients, no guesswork. Finally, we get $B^{-2} = \left(B^{-1} \right)^2,$ which involves those coefficients, and will have $B^4$ and $B^3.$ All you do is substitute the expressions we found for $B^4$ and $B^3,$ add like terms, you get $B^{-2}$ in terms of $B^2, B,I.$

Will Jagy
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  • I have tried this but end up getting a bit of a mess, although everything is in terms of $B^2$, $B$ and $I$ and I can have 0 on the right hand side if I move $B^{-2}$ to the left-hand side, how do I then use this to find $q$, $p$ and $r$ given values for the entries of $B$, $I$ and $B^2$ and the original equation $qB^2 + pB + rI = B^{-2}$ ? – Tom Dec 22 '17 at 18:29