Proving the positive semidefiniteness of a 6X6 symbolic matrix

Specifically, I want to check the positive semidefiniteness of the following 6X6 symbolic matrix

{{-2 (-((11 m1^2 - 24 m1 m3 + 72 m3^2)/(
144 k^2 m1^2 m3^2 \[Sigma]^2)) - 2 \[Sigma]^2),
I - (m1 - 6 m3)/(6 k m1 m3 \[Sigma]^2), (
3/m1 + 1/m3 + (12 \[Sigma]^4)/m2)/(6 k \[Sigma]^2), 2 \[Sigma]^2,
1/(8 k^2 m3^2 \[Sigma]^2) + 4 \[Sigma]^2, (m1 + 12 m3)/(
12 k m1 m3 \[Sigma]^2)}, {-I - (m1 - 6 m3)/(6 k m1 m3 \[Sigma]^2),
1/\[Sigma]^2, 1/(2 \[Sigma]^2), 0, 0, 1/\[Sigma]^2}, {(
3/m1 + 1/m3 + (12 \[Sigma]^4)/m2)/(6 k \[Sigma]^2), 1/(
2 \[Sigma]^2), 1/\[Sigma]^2 + (4 \[Sigma]^2)/(k^2 m2^2),
I + (4 \[Sigma]^2)/(k m2), (1/m3 + (16 \[Sigma]^4)/m2)/(
4 k \[Sigma]^2), 1/\[Sigma]^2}, {2 \[Sigma]^2,
0, -I + (4 \[Sigma]^2)/(k m2), 4 \[Sigma]^2, 4 \[Sigma]^2,
0}, {1/(8 k^2 m3^2 \[Sigma]^2) + 4 \[Sigma]^2, 0, (
1/m3 + (16 \[Sigma]^4)/m2)/(4 k \[Sigma]^2), 4 \[Sigma]^2,
1/(8 k^2 m3^2 \[Sigma]^2) + 6 \[Sigma]^2,
I + 1/(4 k m3 \[Sigma]^2)}, {(m1 + 12 m3)/(12 k m1 m3 \[Sigma]^2),
1/\[Sigma]^2, 1/\[Sigma]^2, 0, -I + 1/(4 k m3 \[Sigma]^2), 3/(
2 \[Sigma]^2)}}


Which depends on the 5 parameters: $$m_{1}$$, $$m_{2}$$, $$m_{3}$$, $$k$$, $$\sigma$$, which are all positive, that is

$$m_{1}>0, m_{2}>0, m_{3}>0, k>0, \sigma>0 \tag{1}$$ .

I try to apply the methodology proposed in (Checking if a symbolic matrix is positive semi-definite) but Mathematica stays doing the calculation and does not yield a result.

Then, as another way to tackle the problem, I tried to numerically inspect the minimum and maximum values of the eigenvalues of the above matrix, by using NMinimize[] and NMaximize[] subject to the constraints given in (1). My code is

s=Simplify[Eigenvalues[{{-2 (-((11 m1^2-24 m1 m3+72 m3^2)/(144 k^2 m1^2 m3^2 \[Sigma]^2))-2 \[Sigma]^2),I-(m1-6 m3)/(6 k m1 m3 \[Sigma]^2),(3/m1+1/m3+(12 \[Sigma]^4)/m2)/(6 k \[Sigma]^2),2 \[Sigma]^2,1/(8 k^2 m3^2 \[Sigma]^2)+4 \[Sigma]^2,(m1+12 m3)/(12 k m1 m3 \[Sigma]^2)},{-I-(m1-6 m3)/(6 k m1 m3 \[Sigma]^2),1/\[Sigma]^2,1/(2 \[Sigma]^2),0,0,1/\[Sigma]^2},{(3/m1+1/m3+(12 \[Sigma]^4)/m2)/(6 k \[Sigma]^2),1/(2 \[Sigma]^2),1/\[Sigma]^2+(4 \[Sigma]^2)/(k^2 m2^2),I+(4 \[Sigma]^2)/(k m2),(1/m3+(16 \[Sigma]^4)/m2)/(4 k \[Sigma]^2),1/\[Sigma]^2},{2 \[Sigma]^2,0,-I+(4 \[Sigma]^2)/(k m2),4 \[Sigma]^2,4 \[Sigma]^2,0},{1/(8 k^2 m3^2 \[Sigma]^2)+4 \[Sigma]^2,0,(1/m3+(16 \[Sigma]^4)/m2)/(4 k \[Sigma]^2),4 \[Sigma]^2,1/(8 k^2 m3^2 \[Sigma]^2)+6 \[Sigma]^2,I+1/(4 k m3 \[Sigma]^2)},{(m1+12 m3)/(12 k m1 m3 \[Sigma]^2),1/\[Sigma]^2,1/\[Sigma]^2,0,-I+1/(4 k m3 \[Sigma]^2),3/(2 \[Sigma]^2)}}]];
(*The first eigenvalue is 0*)
s[[1]]

(*Maximum and Minimum of second eigenvalue*)
NMinimize[s[[2]],m1>0&&m2>0&&m3>0&&k>0&&\[Sigma]>0,{m1,m2,m3,m3,k,\[Sigma]}]
NMaximize[s[[2]],m1>0&&m2>0&&m3>0&&k>0&&\[Sigma]>0,{m1,m2,m3,m3,k,\[Sigma]}]

(*Maximum and Minimum of third eigenvalue*)
NMinimize[s[[3]],m1>0&&m2>0&&m3>0&&k>0&&\[Sigma]>0,{m1,m2,m3,m3,k,\[Sigma]}]
NMaximize[s[[3]],m1>0&&m2>0&&m3>0&&k>0&&\[Sigma]>0,{m1,m2,m3,m3,k,\[Sigma]}]

(*Maximum and Minimum of fourth eigenvalue*)
NMinimize[s[[4]],m1>0&&m2>0&&m3>0&&k>0&&\[Sigma]>0,{m1,m2,m3,m3,k,\[Sigma]}]
NMaximize[s[[4]],m1>0&&m2>0&&m3>0&&k>0&&\[Sigma]>0,{m1,m2,m3,m3,k,\[Sigma]}]

(*Maximum and Minimum of fifth eigenvalue*)
NMinimize[s[[5]],m1>0&&m2>0&&m3>0&&k>0&&\[Sigma]>0,{m1,m2,m3,m3,k,\[Sigma]}]
NMaximize[s[[5]],m1>0&&m2>0&&m3>0&&k>0&&\[Sigma]>0,{m1,m2,m3,m3,k,\[Sigma]}]

(*Maximum and Minimum of sixth eigenvalue*)
NMinimize[s[[6]],m1>0&&m2>0&&m3>0&&k>0&&\[Sigma]>0,{m1,m2,m3,m3,k,\[Sigma]}]
NMaximize[s[[6]],m1>0&&m2>0&&m3>0&&k>0&&\[Sigma]>0,{m1,m2,m3,m3,k,\[Sigma]}]


Particularly I tried this method due to the following reasoning:

If some eigenvalue becomes purely negative, their maximum and minimum values ​​will also be negative, and, in this situation, the matrix defined above will not be a semidefinite matrix

Then, in my code, I find that the second eigenvalue (s[[2]]) has a negative maximum and minimum, therefore, proving that the matrix has negative eigenvalues and therefore being a non-semidefinite matrix. But the problem is that Mathematica shows the following error at the output of the Max and Min calculation

minimize::cvmit: Failed to converge to the requested accuracy or precision within 100 iterations. >>


Thus, so I have two questions

1. How to remove the aforementioned error?

2. Do you have any other suggestions to prove the positive semidefiniteness of the aforementioned matrix?

Your matrix isn't positive semi-definite for all values of $$m1>0, m2>0, m3>0, k>0, σ>0$$.

Proof by contradiction...one of the Eigenvalues is negative for the parameter settings below.

mat = (* your matrix *)

Min@Eigenvalues@(mat /. {m1 -> 1, m2 -> 2, m3 -> 3, k -> 4, σ -> 5}) // N
(* -0.0882134  *)


Digging a little deeper for positve semidefiniteness, all of the leading principal minors must be non-negative. Testing for the 5th principal minor (the determinant of the upper left $$5 \times 5$$ submatrix)

Reduce[{Det@mat[[1 ;; 5, 1 ;; 5]] < 0,
m1 > 0, m2 > 0, m3 > 0, k > 0, σ > 0},
{m1, m2, m3, k, σ}]

(* m1 > 0 && m2 > 0 && m3 > 0 && k > 0 && σ > 0 *)


So the 5th principal minor is always less than zero when your parameters are greater than zero.

• I specifically want to prove the non-semidefiniteness for any value such that $m_{1}>0, m_{2}>0, m_{3}>0, k>0, \sigma>0$ (you have using specific values!). In other words, I want to find a purely negative eigenvalue for any positive parameter: $m_{1}>0, m_{2}>0, m_{3}>0, k>0, \sigma>0$ Commented Dec 5, 2019 at 15:41
• See my edit. I am showing a purely negative principal minor, which implies non positive semi-def. Commented Dec 5, 2019 at 15:51
• You're right!. this is because of Silvester´s Criterion, which says: if every principal minor (determinant of a principal matrix) of the Matrix, is non-negative, then, the matrix is positive semidefinite (Matrix Analisis, R. Horn, 2ed. pp. 439). Effectively, the 5X5 is a leading principal matrix, whose determinant is negative, thus the matrix is not a positive semidefinite. Thanks a lot! Commented Dec 5, 2019 at 16:27