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# Optimizing a function of a given matrix

Let us consider the $4\times 4$ symmetric matrix $$A_x=\left(\begin{array}{rrrr} 0 & 1 & 1 & 1 \ 1 & 0 & 2^x & 2^x \ 1 & 2^x & 0 & 2^x \ 1 & 2^x & 2^x & 0 \end{array}\right)$$

Here I need to find $\min { x>0: det(A_x)=0 \, or \, ||A_x^{-1}||=0 } ,$ where by $||M||$ we mean the sum of all entries of the matrix $M.$ I'm looking for a general sage program where my input will be a matrix with entries as functions of an inderminant (like the matrix $A_x$ above) which will give me the unique $x$ corresponding to my matrix. If no such real value exists, it should result as $\infty$ Can anyone help me? Thank you in advance.

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## 1 Answer

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One could use Sage to explore the problem.

Define the matrix $A$:

sage: A = matrix(SR, 4, [0, 1, 1, 1, 1, 0, 2^x, 2^x, 1, 2^x, 0, 2^x, 1, 2^x, 2^x, 0])
sage: A
[  0   1   1   1]
[  1   0 2^x 2^x]
[  1 2^x   0 2^x]
[  1 2^x 2^x   0]


Compute its determinant:

sage: det(A)
-3*2^(2*x)


The sum of its entries:

sage: sum(A.list())
6*2^x + 6


Its inverse:

sage: ~A
[-2/3*2^x      1/3      1/3      1/3]
[     1/3 -2/3/2^x  1/3/2^x  1/3/2^x]
[     1/3  1/3/2^x -2/3/2^x  1/3/2^x]
[     1/3  1/3/2^x  1/3/2^x -2/3/2^x]


The sum of the entries of its inverse:

sage: sum((~A).list())
-2/3*2^x + 2


Now solve $\det(A) = 0$ and $\lVert A^{-1} \rVert = 0$, extract positive solutions, take the minimum, and compute a numerical approximation.

sage: S = solve(det(A) == 0, x); S
[]
sage: T = solve(sum((~A).list()) == 0, x); T
[x == (log(6) - log(2))/log(2)]
sage: min([e.rhs() for e in S + T if e.rhs() > 0]).n()
1.58496250072116


This can all be put in a function which, given a matrix A as above, returns this minimum positive solution.

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## Comments

Actually, I was stuck at this step. I tried the code "solve" to solve the two systems, but I don't get numerical value from this. Again I tried "find_root" code with the interval but one of the systems (First one) does not have a solution, and so it gives a long list of errors. Therefore I cannot evaluate the minimum of all solutions (min() doesn't work). For this particular matrix, I can find it by direct calculation. But what I need a general method to get my required (numerical value) whenever I input a matrix as a function of $x$. I need something like this

X=det(A); Y=sum((~A).list()); S=solve([X,x>0],x); T=solve([Y,x>0],x); min(S+T)

This doesn't give me numerical value. Note the last step, I want sage to calculate the minimum value of the ...(more)

( 2018-07-25 05:49:07 +0200 )edit

Edited question to add last step.

( 2018-07-25 06:28:46 +0200 )edit

Thank you for your solution. It works with that particular matrix. But it doesn't work for all matrices.

For example consider $$A=\left(\begin{array}{rrrrr} 0 & 2^{x} & 1 & 1 & 1 \ 2^{x} & 0 & 1 & 1 & 1 \ 1 & 1 & 0 & 2^{x} & 2^{x} \ 1 & 1 & 2^{x} & 0 & 2^{x} \ 1 & 1 & 2^{x} & 2^{x} & 0 \end{array}\right)$$

or

$$A=\left(\begin{array}{rrrr} 0 & 1 & 2^{x} & 1 \ 1 & 0 & 1 & 2^{x} \ 2^{x} & 1 & 0 & 1 \ 1 & 2^{x} & 1 & 0 \end{array}\right)$$

Here I get an error "cannot evaluate symbolic expression numerically"

Note: For all my matrices, it is known that the solution is at most equal to 2. So if something like min(find_root(X,0,2)+find_root(Y,0,2)) could be used, that is also enough for me.

( 2018-07-25 08:39:42 +0200 )edit

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Asked: 2018-07-23 16:50:52 +0200

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Last updated: Jul 25 '18