How to prove that $ E:=ABC D $ is also positive definite?

Question

Now I think this is true:

Let $A$, $B$, $C$ and $D$ be symmetric, positive definite matrices and suppose that $E:=ABCD $ is symmetric. How might I prove that $E$ is also positive definite?

the similar question can see: How to prove that $D := ABC$ is also positive definite?

@Landscape this not true, Really? Thank you everyone

Answer 1

You cannot prove it, because there is a counterexample. Let

$$A = \begin{bmatrix} 58 & 50 & 30 \\ 50 & 66 & 38 \\ 30 & 38 & 22 \end{bmatrix}, \quad B = \begin{bmatrix} 8 & 6 & 8 \\ 6 & 5 & 4 \\ 8 & 4 & 65 \end{bmatrix}, \quad C = \begin{bmatrix} 34 & 26 & 46 \\ 26 & 56 & 52 \\ 46 & 52 & 83 \end{bmatrix}, \quad D = \begin{bmatrix} \frac{176398144863}{734764024} & \frac{2084349747}{734764024} & -\frac{401393975}{91845503} \\ \frac{2084349747}{734764024} & 123 & 74 \\ -\frac{401393975}{91845503} & 74 & 45 \end{bmatrix}.$$

Then $$ABCD = \begin{bmatrix} \frac{3747748968739763}{91845503} & \frac{4372037328164921}{91845503} & \frac{2546983749851550}{91845503} \\ \frac{4372037328164921}{91845503} & \frac{5099682487522571}{91845503} & \frac{2970895299665430}{91845503} \\ \frac{2546983749851550}{91845503} & \frac{2970895299665430}{91845503} & \frac{1730738016057502}{91845503} \end{bmatrix}.$$

$A$, $B$, $C$, $D$ are symmetric positive definite, while $ABCD$ is symmetric, but not positive semidefinite (the determinant of the principal leading submatrix of order $2$ is $-2.82211 \cdot 10^{11}$).

This was generated by few runs of the Mathematica code:

RandPDM[n_] := Module[
   {tmp},
   While[True,
    tmp = RandomInteger[7, {n, n}];
    If[Det[tmp] != 0, Break[]]
    ];
   Return[Transpose[tmp].tmp];
   ];
M1 = RandPDM[3];
M2 = RandPDM[3];
M3 = RandPDM[3];
M4 = RandPDM[3];
M4[[1, 1]] = a;
M4[[1, 2]] = b; M4[[2, 1]] = b;
M4[[1, 3]] = c; M4[[3, 1]] = c;
M4[[2, 2]] = d; M4[[2, 2]] = d;
M4[[2, 3]] = e; M4[[3, 2]] = e;
(M = M1.M2.M3.M4) // MatrixForm
sol = FindInstance[{
   Transpose[M] == M,
   a > 0, Det[M4] > 0, Det[Take[M4, 2, 2]] > 0
   }, {a, b, c, d, e}]
(M4 = (M4 /. sol[[1]])) // MatrixForm
(M = (M /. sol[[1]])) // MatrixForm
all = {M1, M2, M3, M4, M};
Map[MatrixForm, all]
Map[PositiveDefiniteMatrixQ, all]

The line Map[MatrixForm, all] outputs all five (the above are a minor edit of such output with TeXForm instead of MatrixForm), while the last line checks their positive definitess, with the output

{True, True, True, True, False}

In case you want to test these matrices for yourself, here is the Mathematica output for them:

M1 = {{58,50,30},{50,66,38},{30,38,22}};
M2 = {{8,6,8},{6,5,4},{8,4,65}},{{34,26,46},{26,56,52},{46,52,83}};
M3 = {{176398144863/734764024,2084349747/734764024,-(401393975/91845503)},{2084349747/734764024,123,74},{-(401393975/91845503),74,45}};
M4 = {{3747748968739763/91845503,4372037328164921/91845503,2546983749851550/91845503},{4372037328164921/91845503,5099682487522571/91845503,2970895299665430/91845503},{2546983749851550/91845503,2970895299665430/91845503,1730738016057502/91845503}};

Hopefully, I didn't mess up these while editing. Just in case, I'll keep Mathematica running with the above solution for few more hours, in case you have some questions.