Maximal eigenvalue of a matrix as a function of three variables

Question

Assume that $A$ is a positive square real matrix of size $4\times 4$ along with three variables $x_1,x_2,x_3$ (all are positive real numbers):

$$A = \begin{bmatrix} 1 & x_1 & 2 &x_3\\ 1/x_1 & 1 & x_2 & 1/5 \\ 1/2 & 1/x_2 & 1 & 3\\ 1/x_3 & 5 & 1/3 & 1 \end{bmatrix}$$

My question is, how to find the maximal eigenvalue of $A$?

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When I write the following on a MATLAB commands, I will get functions. So, which function will be considered as the maximal eigenvalue (as a function of three variables)?

>> syms x1 x2 x3;

>> A=[1    x1   2   x3;
      1/x1 1    x2  1/5;
      1/2  1/x2 1   3;
      1/x3 5    1/3 1]; 
>> e=eig(A)

Please help? Thanks

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Perron–Frobenius Theorem: A real square matrix with positive entries has a unique largest real eigenvalue and that the corresponding eigenvector can be chosen to have strictly positive components.

Answer 1

A partial answer.

The characteristic polynomial of $A$

$\chi(x):=x^4+c_1x^3+c_2x^2+c_3x+c_4 \tag{1}$

has $c_1=-tr(A)=-4$ (notation "tr" for trace) and

$$c_2=\dfrac{1}{2!}\begin{vmatrix}tr(A)&1\\tr(A^2)&tr(A)\end{vmatrix}=\dfrac{1}{2}\begin{vmatrix}4&1\\16&4\end{vmatrix}=0 \tag{2}$$

(indeed $tr(A)=4$ and $tr(A^2)=16$ by an easy computation).

(I use in (2) a formula that can be found for example here).

As a consequence of (1) and (2), an eigenvalue $x$ of $A$ is such that :

$$\chi(x)=x^4-4x^3+c_3x+c_4=0 \tag{3}$$

Remarks :

1) Formulas for $c_3$ and $c_4$ (as obtained with Matlab symbolic toolbox) are very intricated.

2) Please note that there is no term in $x^2$.

We know (as said by the author of the question) that by Perron-Frobenius theorem, $A$ has at least a real root; therefore, we have at least 2 real roots (indeed, a polynomial equation with real coefficients has complex roots coming by pairs, like in Noah's Arch).

Extensive simulations show that there are always 2 real roots and 2 complex (conjugate) roots of (3). Moreover, Matlab is able to give symbolic formulas for the roots $x_1,x_2,x_3,x_4$ in this order. We can observe that $x_1$ and $x_2$ are the two real roots (always with $x_2$ the largest one) and the third and fourth are always (conjugate) complex ; I have no proof for these facts.

Among the real roots, $x_2$ is very dominant, compared with $x_1.$

Additional remarks : the formula for the determinant of $A$ is factorizable into simple factors :

$$ c_4=\det(A)=(15x_2 - 1)(x_1x_2 - 2)(x_1 - 5x_3)(x_3 - 6)/(30x_1x_2x_3).$$

2) The fact that there are necessarily at least two real roots can be established without Perron-Frobenius. It is due to the absence of the $x^2$ term ; indeed setting to zero the coefficient of the missing term in $x^2$ :

$$x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4=0\tag{4}$$

If on the contrary, we had two pairs of ( conjugated) complex roots, say with $x_2=\bar{x_1}$ and $x_4=\bar{x_3}$, we could transform (4) into :

$$|x_1|^2+2 \Re(x_1) \Re(x_3)+|x_3|^2=0$$

which is possible only if $x_1=x_3=0$, but in this case, we have a double real root !

3) The fact that (apparently) the second formula always gives the largest eigenvalus can be considered as exceptional. For example, the family of matrices :

$$A_x=\begin{pmatrix}0&x\\x&0\end{pmatrix}$$

has the following formulas for its eigenvalues :

$$x_1=x, \ \ x_2=-x$$

on which we can see that the largest eigenvalue is given by the second formula when $x<0$ and by the first one when $x>0$...