Find $Y\in\operatorname{GL}_n(\mathbb{Z})$ such that all eigenvalues of $YX$ are nonnegative

Question

I saw this problem some years ago and I would greatly appreciate any reference or solution.

Let $X \in \operatorname{M}_n ( \mathbb{R} )$. Prove that there is $Y \in \operatorname{M}_n ( \mathbb{Z} )$ such that $Y$ is invertible over $\mathbb Z$ and all eigenvalues of $YX$ are nonnegative.

If $X \in \operatorname{M}_n ( \mathbb{C} )$ is it possible to find $Y \in \operatorname{M}_n ( \mathbb{Z} )$ such that $Y$ is invertible over $\mathbb Z$ and all eigenvalues of $YX$ have nonnegative real part?

Answer 1

• $\DeclareMathOperator\spectrum{spectrum}\DeclareMathOperator\GL{GL}$As Nathaniel wrote, the case $X\in M_n(\mathbb{Q})$ is not difficult. Let $p>0$ be an integer s.t. $pX\in M_n(\mathbb{Z})$. The row-style Hermite normal form of $pX$ is $H=UpX$, where $U$ is integer unimodular and $H$ is upper-triangular with non-negative diagonal. Then $K=UX$ gives the result; we can also use the Smith normal form ($\spectrum(UXV)=\spectrum(VUX)$).

• Now we assume that $X\in \GL_n(\mathbb{R})$. We consider a rational approximation $Y$ of $X$; as above, $L=UY$ is upper-triangular with $>0$ diagonal $(l_i)_i$. When $X$ is non-singular, the eigenvalues $(\lambda_i)_i$ of $UX$ are close to the $(l_i)$'s.

Unfortunately, $L$ has "always" multiple eigenvalues and —in general— some $(\lambda_i)$'s are not real. We do not need all the properties of the Hermite matrix; in fact, the only properties that interest us are "$H$ is triangular with a positive diagonal".

Question. We assume that $\det(pY)$ can be written as a product of $n$ distinct positive integers. Can we modify the algorithm for constructing the Hermite form so that the diagonal of $pL$ is composed of $n$ distinct $>0$ integers?

If yes, then $\spectrum(UX)\subset (0,+\infty)$.

Answer 2

This is too long for a comment but addresses the case where matrix X entries are rational.

In detail for the case n = 2

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In the case where the matrix X has 0 for the bottom right entry and p/q for the top right entry (gcd (p,q) = 1), choose Y1 to have the entries 0, 1, 1, 0 then B1 = Y1X is lower triangular. Choose Y2 = diagonal with entries sgn($B1_{11}$) and sgn($B1_{22}$); now B2= Y2Y1X is lower triangular with non-negative diagonal entries, so has non-negative spectrum.

WLOG reduce to the case where the matrix X has 1 for the bottom right entry and p/q for the top right entry (gcd (p,q) = 1). Then choose integers u, v so that qv+pu =1. If Y1 has the entries q, -p, u, v then B1 = Y1X is lower triangular, choose Y2 = diagonal with entries sgn($B1_{11}$) and sgn($B1_{22}$); now B2= Y2Y1X is lower triangular with non-negative diagonal entries, so has non-negative spectrum.

I am fairly certain that one can extend to M2(R) by choosing a suitable approximation p/q for any real top right entry, but I have not worked out the details completely.

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This argument (i.e., reduction of an arbitrary matrix to lower triangular form via multiplication of integer matrices with determinant 1) used for M2(Q) can be extended to Mn(Q) by a technique similar to Gaussian elimination. as follows:

FOR j = n TO 2

  FOR i = 1 TO n-1

    IF Xij = 0 THEN Y = Identity

    ELSIF Xjj = 0 THEN Y = Y_rowswap(i,j)

    ELSE Y = Y_zero(i,j)

    ENDIF

    X = YX

  ENDFOR

ENDFOR

Y_rowswap (i,j) is the permutation matrix that swaps rows i and j and leaves the other rows the same.

Y_zero (i, j) is a matrix with the following entries:

q (row i,column i),

-p(row i,column j),

u(row j, column i),

v(row j,column j),

other diagonal entries = 1,

other off-diagonal entries = 0

where qXij-pXjj= 0 and qv + pu = 1

So maybe someone else (the OP?) can make the extension to Mn(R) by an argument approximating real entries by suitable rational entries...