I have the following linear equation:
$$a X^2 - b X - cI = 0$$
Where, I is the identity matrix 5. And a,b,c will be arbitrary values to be later set, 1,2,3 is fine. How might one take that linear equation and turn it into a matrix?
Thank you
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$\begingroup$ Are you sure $I$ isn't the identity? Also this isn't, strictly speaking, a linear equation as you're solving for a quadratic term. And is $X$ diagonal or is every element a variable? $\endgroup$ – b3m2a1 Mar 20 '18 at 1:25
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$\begingroup$ @b3m2a1 Yes I correctly it, it is the identity matrix. Every element in a variable. $\endgroup$ – Jwizz Mar 20 '18 at 1:28
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$\begingroup$ If every element can change it'll take a long time for Mathematica to compute the solutions. I can give you how you could set up such a system, but not how the results. $\endgroup$ – b3m2a1 Mar 20 '18 at 1:36
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$\begingroup$ @b3m2a1 would appreciate it. Been stuck on this problem for quite some time. New to mathematica. $\endgroup$ – Jwizz Mar 20 '18 at 1:37
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2$\begingroup$ Then we have to wait until your friend wakes up. In the meantime, you might have a look at the algebraic Ricatti equation. $\endgroup$ – Henrik Schumacher Mar 20 '18 at 1:44
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Here's a way you could set this up, since you say you $X$ is all variable:
baseMX = Hold[
a*MatrixPower[X, 2] \[Minus] b*X \[Minus] c*IdentityMatrix[5]];
subbedMX =
baseMX /. {X ->
Array[Symbol["x" <> Map[ToString, {##}]] &, {5, 5}], a -> 1,
b -> 2, c -> 3} // ReleaseHold;
baseEq = subbedMX == Array[0 &, {5, 5}];
Then looking at subbedMX:
subbedMX // TeXForm
$$\tiny\begin{pmatrix} \text{x11}^2-2 \text{x11}+\text{x12} \text{x21}+\text{x13} \text{x31}+\text{x14} \text{x41}+\text{x15} \text{x51}-3 & \text{x11} \text{x12}+\text{x22} \text{x12}-2 \text{x12}+\text{x13} \text{x32}+\text{x14} \text{x42}+\text{x15} \text{x52} & \text{x11} \text{x13}+\text{x33} \text{x13}-2 \text{x13}+\text{x12} \text{x23}+\text{x14} \text{x43}+\text{x15} \text{x53} & \text{x11} \text{x14}+\text{x44} \text{x14}-2 \text{x14}+\text{x12} \text{x24}+\text{x13} \text{x34}+\text{x15} \text{x54} & \text{x11} \text{x15}+\text{x55} \text{x15}-2 \text{x15}+\text{x12} \text{x25}+\text{x13} \text{x35}+\text{x14} \text{x45} \\ \text{x11} \text{x21}+\text{x22} \text{x21}-2 \text{x21}+\text{x23} \text{x31}+\text{x24} \text{x41}+\text{x25} \text{x51} & \text{x22}^2-2 \text{x22}+\text{x12} \text{x21}+\text{x23} \text{x32}+\text{x24} \text{x42}+\text{x25} \text{x52}-3 & \text{x13} \text{x21}+\text{x22} \text{x23}-2 \text{x23}+\text{x23} \text{x33}+\text{x24} \text{x43}+\text{x25} \text{x53} & \text{x14} \text{x21}+\text{x22} \text{x24}-2 \text{x24}+\text{x23} \text{x34}+\text{x24} \text{x44}+\text{x25} \text{x54} & \text{x15} \text{x21}+\text{x22} \text{x25}-2 \text{x25}+\text{x23} \text{x35}+\text{x24} \text{x45}+\text{x25} \text{x55} \\ \text{x11} \text{x31}+\text{x33} \text{x31}-2 \text{x31}+\text{x21} \text{x32}+\text{x34} \text{x41}+\text{x35} \text{x51} & \text{x12} \text{x31}+\text{x22} \text{x32}-2 \text{x32}+\text{x32} \text{x33}+\text{x34} \text{x42}+\text{x35} \text{x52} & \text{x33}^2-2 \text{x33}+\text{x13} \text{x31}+\text{x23} \text{x32}+\text{x34} \text{x43}+\text{x35} \text{x53}-3 & \text{x14} \text{x31}+\text{x24} \text{x32}+\text{x33} \text{x34}-2 \text{x34}+\text{x34} \text{x44}+\text{x35} \text{x54} & \text{x15} \text{x31}+\text{x25} \text{x32}+\text{x33} \text{x35}-2 \text{x35}+\text{x34} \text{x45}+\text{x35} \text{x55} \\ \text{x11} \text{x41}+\text{x44} \text{x41}-2 \text{x41}+\text{x21} \text{x42}+\text{x31} \text{x43}+\text{x45} \text{x51} & \text{x12} \text{x41}+\text{x22} \text{x42}-2 \text{x42}+\text{x32} \text{x43}+\text{x42} \text{x44}+\text{x45} \text{x52} & \text{x13} \text{x41}+\text{x23} \text{x42}+\text{x33} \text{x43}-2 \text{x43}+\text{x43} \text{x44}+\text{x45} \text{x53} & \text{x44}^2-2 \text{x44}+\text{x14} \text{x41}+\text{x24} \text{x42}+\text{x34} \text{x43}+\text{x45} \text{x54}-3 & \text{x15} \text{x41}+\text{x25} \text{x42}+\text{x35} \text{x43}+\text{x44} \text{x45}-2 \text{x45}+\text{x45} \text{x55} \\ \text{x11} \text{x51}+\text{x55} \text{x51}-2 \text{x51}+\text{x21} \text{x52}+\text{x31} \text{x53}+\text{x41} \text{x54} & \text{x12} \text{x51}+\text{x22} \text{x52}-2 \text{x52}+\text{x32} \text{x53}+\text{x42} \text{x54}+\text{x52} \text{x55} & \text{x13} \text{x51}+\text{x23} \text{x52}+\text{x33} \text{x53}-2 \text{x53}+\text{x43} \text{x54}+\text{x53} \text{x55} & \text{x14} \text{x51}+\text{x24} \text{x52}+\text{x34} \text{x53}+\text{x44} \text{x54}-2 \text{x54}+\text{x54} \text{x55} & \text{x55}^2-2 \text{x55}+\text{x15} \text{x51}+\text{x25} \text{x52}+\text{x35} \text{x53}+\text{x45} \text{x54}-3 \\ \end{pmatrix}$$
This gives a sense for how unwieldy it is.
Then you can run Solve[baseEq] but I have low expectations you'll actually get a solution.
If X can be diagonal this is much, much nicer (so much so as to pretty much be trivial). It'd then look like:
subbedMX2 =
baseMX /. {X ->
DiagonalMatrix@Array[Symbol["x" <> Map[ToString, {##}]] &, 5],
a -> 1, b -> 2, c -> 3} // ReleaseHold;
baseEq2 = subbedMX2 == Array[0 &, {5, 5}];
And subbedMX2 will be:
$\left( \begin{array}{ccccc} \text{x1}^2-2 \text{x1}-3 & 0 & 0 & 0 & 0 \\ 0 & \text{x2}^2-2 \text{x2}-3 & 0 & 0 & 0 \\ 0 & 0 & \text{x3}^2-2 \text{x3}-3 & 0 & 0 \\ 0 & 0 & 0 & \text{x4}^2-2 \text{x4}-3 & 0 \\ 0 & 0 & 0 & 0 & \text{x5}^2-2 \text{x5}-3 \\ \end{array} \right)$
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$\begingroup$ Thank you so much for this! I really appreciate the insights! I hope this doesn't come off as annoying to you. But out of curiosity. How would this look differently if diagonal? $\endgroup$ – Jwizz Mar 20 '18 at 1:52
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$\begingroup$ @Jwizz it'll be significantly simpler. I'll post how to build it. $\endgroup$ – b3m2a1 Mar 20 '18 at 1:53
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$\begingroup$ @VsevolodA. yeah that made my point for me very nicely $\endgroup$ – b3m2a1 Mar 20 '18 at 1:59
