# A signature inequality?

Given two real symmetric matrices $A$ and $B$ of common square size $n$ with no strictly negative eigenvalues, can the symmetric matrix $AB+BA$ have strictly more than $n/2$ eigenvalues which are strictly negative?

The answer to this question is yes, thanks to Junkie. My random examples did not hit a counterexample since I did them in dimension considerably greater than $3$ (typically $8$ or something similar) where it seems quite hard to find counterexamples by taking generic matrices.

This suggests however a series of new questions (which I can unfortunately no longer accept since I gave already credit to Junkie for a correct answer):

What is the maximal number of strictly negative eigenvalues of $AB+BA$ if $A$ and $B$ are definite positive symmetric matrices of common size $n\times n$?

This number is at least roughly $3n/4$ by Junkie's examples (put them along the diagonal). Can it be considerably higher?

I have for example currently no example with $4$ strictly negative eigenvalues for $n=5$. ($3$ strictly negative eigenvalues in dimension $n=5$ are easy to achieve by combining Junkie's example with an example in dimension $2$ yielding signature $(1,1)$.)

It seems that there is always at least one non-negative eigenvalue (this is obvious if $A$ and $B$ have only positive coefficients by Perron-Frobenius and it is probably not very hard in the general case).

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To my mind, tags are only useful if the keyword of the tag can't be found in the question. Any reasonable question I can think of which you might want to tag "inequality" probably already has the word "inequality" in it. – Qiaochu Yuan Jul 9 '10 at 20:02
@Qiaochu But the purpose of tags isn't simply to provide words to search on, in addition to those in the text. If I click on a tag I get a list of questions that have been tagged that way, not a list of questions containing that string. So tagging a string already in the text serves a useful purpose. – Dan Piponi Jul 12 '10 at 2:05
@Qiaochu and sigfpe: please see tea.mathoverflow.net/discussion/501/tags-what-are-they-for – Willie Wong Jul 12 '10 at 10:05
That there is at least one positive eigenvalue follows from the fact that $AB+BA$ has positive trace. To see this one can reduce to the case where $A$ is diagonal, and note that $B$ has positive diagonal entries. – Robin Chapman Jul 13 '10 at 19:02

This review seems to imply that any symmetric real matrix $C$ with positive trace is the Jordan product $(AB+BA)/2$ of two positive definite real matrices $A,B$. If so, then the maximum number of negative eigenvalues of $(AB+BA)/2$ for $n\times n$ symmetric positive definite $A,B$ is $n-1$ (it cannot be $n$ because of the positive definiteness (hence positive trace) of $A^{1/2}BA^{1/2}$, which is conjugate to $AB$).

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Ballantine's paper is dx.doi.org/10.1016/0024-3795(69)90005-6 Lemma 2 (page 43) states: for $H$ hermitian, $tr(H)>0$ is equivalent to $H=QR+RQ$ for positive definite $Q,R$. – Junkie Jul 13 '10 at 21:56
I agree that you can get n-1 negative eigenvalues, by following thru the argument. I will post some code (the numbers get rather big). – Junkie Jul 14 '10 at 1:47

Example found randomly:

EDIT: Make one with positive coefficients: $A=\pmatrix{1&2&3\cr2&5&6\cr3&6&10}$ and $B=\pmatrix{1&1&2\cr1&2&6\cr2&6&21}$.

EDIT: Here's how, with Magma I get about a 25% probability with this Magma code:

R := RealField(30);
function FindCounterExample()
S := RandomSLnZ(3,5,5); A := S*Transpose(S);
S := RandomSLnZ(3,5,5); B := S*Transpose(S);
ROOTS := Roots(CharacteristicPolynomial(A*B+B*A),R);
ROOTS := [r[1] : r in ROOTS | r[1] ge 0];
if #ROOTS eq 1 then A; B; end if;
return #ROOTS;
end function;


I get about a 25% probability of 1 positive eigenvalue, 75% of 2, and 0.15% of 3.

OUTPUT := [FindCounterExample() : i in [1..100000]];
SequenceToMultiset(OUTPUT); // {* 1^^25563, 2^^74296, 3^^141 *}


EDIT: I think this can be described as saying that there is about a 75% chance of the determinant of $AB+BA$ being negative, and when it in the 25% positive case, the chance is not too great that all the eigenvalues are positive. It can also depends on what RandomSLnZ is doing. The split might only be close to 75-25 and not exact.

EDIT: Yes when I did it with RandomSLnZ(3,3,3) I get a split of about $114+844+42$, so the 75-25 is meaningless.

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I did a few thousand random examples, without success. You were more lucky. I will accept your answer in three days after a doublecheck (I am currently without access to my favorite computer algebra system.) – Roland Bacher Jul 9 '10 at 21:15
I think that I did consider only examples with non-negative coefficients. This could explain my failure to find a counterexample. Do you have examples where $A$ and $B$ have coefficients in $/mathbb R_{/geq 0}$? – Roland Bacher Jul 10 '10 at 4:57
I understand know better what is happening and have rephrased the initial question accordingly. – Roland Bacher Jul 12 '10 at 11:55

I get about 0.1% random in 3 for 4x4. Here is a "patterned" one:

[146  37  12   0]
[ 37  10   3   0]
[ 12   3   1   0]
[  0   0   0   1]

[   1    0    0    0]
[   0 4221  202 -857]
[   0  202   10  -41]
[   0 -857  -41  174]


Here is a nonnegative one:

[139   3  47 325]
[  3   1   0   6]
[ 47   0  18 111]
[325   6 111 761]

[ 1  0  2  2]
[ 0  5 12  5]
[ 2 12 33 16]
[ 2  5 16 10]


A completely positive one:

[  757  1288    87  3416]
[ 1288  2193   148  5809]
[   87   148    10   393]
[ 3416  5809   393 15567]

[       1     4760      192     1776]
[    4760 32021426  1291596 11946513]
[     192  1291596    52097   481867]
[    1776 11946513   481867  4456990]

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I don't know whether Roland is asking his revised question (basically how does max number of strictly negative eigenvalues depend on $n$) with nonnegative entries. But I did want to mention Perron-Frobenius in this regard. I don't know whether P-F gives anything extra for symmetric matrices. Note that there is a really quick proof for strictly positive entries using Brouwer fixed-point theorem, as one can map the closed positive orthant of the unit sphere to itself by $$x \mapsto \frac{C x}{| C x |}$$ where $C$ is a square matrix, not necessarily symmetric, with strictly positive entries. – Will Jagy Jul 12 '10 at 22:22
Nice quick proof! – Roland Bacher Jul 13 '10 at 9:19
There is also the proof using the Hilbert metric on the projectivization of the positive cone (an open simplex), for which $C$ is a contraction. – BS. Jul 13 '10 at 15:47

Overnight search (about 25 million random 5x5) found:

[ 835  791 -119   -1  981]
[ 791  755 -113    0  931]
[-119 -113   17    0 -140]
[  -1    0    0    1    2]
[ 981  931 -140    2 1166]

[   5   76   -4    2  -14]
[  76 2849    0   75 -531]
[  -4    0   17    0    0]
[   2   75    0    2  -14]
[ -14 -531    0  -14   99]

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Here is Magma code that will construct a 10x10 example, with one eigenvalue about +10 and nine of -1. Note that Magma has little numerical linear algebra, so I do it myself. As can be seen in the final $P$ and $Q$, the numbers become of size about $10^{40}$ here. You can round to the nearest integer after multiplying by $10^{100}$ and get an integral answer of course. It seems much harder to require that $P$ and $Q$ have positive coefficients.

RF:=RealField(100);
function ThetaRoot(M)
h11,h12,h21,h22:=Explode(Eltseq(M));
t2:=h11^2; t1:=h22^2; t12:=(h12+h21)^2-2*h11*h22;
a:=t1+t2+t12; b:=-2*t1-t12; c:=t1;
R:=Roots(Polynomial([c,b,a]),RF); if #R eq 0 then R:=[<-b/2/a,2>]; end if;
th:=Arccos(Sqrt(R[1][1])); return th; end function;

function UnitarilySimilar(M)
t:=Trace(M); d:=Degree(Parent(M)); S:=SymmetricGroup(d);
M:=M-t/d*Parent(M)!1;
if d eq 1 or M[1][1] eq 0 then return Parent(M)!1; end if;
D:=Diagonal(M);
POS:=[i : i in [1..d] | D[i] gt 0]; NEG:=[i : i in [1..d] | D[i] lt 0];
p:=POS[1]; n:=NEG[1]; u:=S!1;
if p ne 1 then u*:=S!(p,1); end if; if n ne 2 then u*:=S!(n,2); end if;
if p eq 2 and n eq 1 then u:=S!(1,2); end if;
T1:=PermutationMatrix(BaseRing(M),u);
M1:=T1*M*Transpose(T1);
if Abs(M1[1][1]) gt Abs(M1[2][2]) then
Tt:=Parent(M)!1; Tt[1][1]:=0; Tt[1][2]:=-1; Tt[2][1]:=1; Tt[2][2]:=0;
T1:=Tt*T1; M1:=T1*M*Transpose(T1); end if;
th:=ThetaRoot(Submatrix(M1,[1,2],[1,2]));
T2:=Parent(M)!1;
T2[p][p]:=Cos(th); T2[p][n]:=-Sin(th);
T2[n][p]:=-T2[p][n]; T2[n][n]:=T2[p][p];
M2:=T2*M1*Transpose(T2);
if Abs(M2[1][1]) gt 10^(-25) then T2[p][n]:=Sin(th); T2[n][p]:=-T2[p][n];
M2:=T2*M1*Transpose(T2); end if;
U:=UnitarilySimilar(Submatrix(M2,2,2,d-1,d-1));
U:=DirectSum(<DiagonalMatrix([BaseRing(U)!1]),U>);
return U*T2*T1; end function;

M:=DiagonalMatrix([RF!10,-1,-1,-1,-1,-1,-1,-1,-1,-1]); d:=Degree(Parent(M));
U:=UnitarilySimilar(M); MT:=U*M*Transpose(U); I:=Parent(MT)!1;
for a in [1..d] do for b in [a+1..d] do // Ballantine gives "too diagonal"
I[a][b]:=Random([-2^25..2^25])/2^32; end for; end for; // perturb it
PERTURB:=I*MT*Transpose(I); B:=PERTURB; TRANS:=I*U;
Diagonal(PERTURB);
for a in [1..d] do for b in [a+1..d] do B[a][b]:=0; end for; end for;
for a in [1..d] do B[a][a]:=B[a][a]/2; end for;

function EpsilonKernel(M) d:=Degree(Parent(M));
S:=Vector([BaseRing(M)!0 : i in [1..d]]);
for e in [1..d] do T:=M;
for f in Reverse([1..e-1]) do S[f]:=T[e][f]/T[f][f];
T[e]-:=T[f]*S[f]; end for;
if Abs(T[e][e]) lt 10^(-50) then S[e]:=-1; return S/Sqrt(Norm(S)); end if;
end for; end function;

R:=[r[1] : r in Roots(CharacteristicPolynomial(M))];
return DiagonalMatrix(R),Matrix([EpsilonKernel(M-r*Parent(M)!1) : r in R]);
end function;


EDIT: In larger dimensions, the $Q=H^{-1}$ becomes time-consuming, perhaps for stability.