Matrix polynomial

Question

Let $p \in \mathbb{N}, p \geq 2$ even. Prove that $$\det(X^{p}+X^{p-1}+\dots+X+I)\geq 0$$ for all invertible matrices $X$ with real entries.

The solution I found simply uses the fact that the polynomial $x^p+x^{p-1}+\dots+x+1$ doesn't have any real roots. Why is that enough?

Answer 1

• Let us assume firstly that $X$ is diagonalizable with eigenvalues $\lambda_k$.

Let $\pi(x):=x^p+x^{p-1}+\dots+x+1$; as the roots of $\pi(x)(x-1)=x^{p+1}-1$ are the $(p+1)$-th roots of unity with $(p+1)$ odd ; we know that set apart $x=1$, as $x=-1$ cannot be a root in this case, all its roots, i.e., all the roots of $\pi(x)$, are non real. Thus, the curve $y=\pi(x)$ doesn't intersect the $x$ axis ; thus, as $p(0)=1>0$, this curve remains above the $x$-axis, meaning that all values of $\pi(x)$ are $>0.$

Let $Q_X=X^p+X^{p-1}+\dots+X+I_p.$

Writing $X=PDP^{-1}$ with $D$ diagonal matrix of eigenvalues, we have:

$$Q_X=PD^pP^{-1}+PD^{p-1}P^{-1}+\dots+PDP^{-1}+PIP^{-1}=P\underbrace{(D^p+D^{p-1}+\dots+D+I)}_R P^{-1}.$$

Thus

$$\tag{1}det(Q_X)=\det(R)=\prod_{k=1}^p \pi(\lambda_k).$$

Whence $\det(Q_X)>0$ because of the positivity of polynomial $\pi(x)$ established above.

• For the general case, it suffices to use a topological argument, more precisely a continuity argument : any matrix is arbitrarily close to an invertible matrix ; thus any non-invertible matrix $X$ is the limit of a sequence of invertible matrices $X_n$; as we have a finite sum : $Q_{X_n} \to Q_X$ ; and we end up by using the fact that the determinant is a continuous function of the entries of a matrix.

Answer 2

Case 1: X is diagonal

The determinant is equal to $\prod_i f(d_i)$ where $f$ is the polynomial in the question and $d_i$ are the diagonal entries.

Case 2: X is diagonalizable

In this case, f(X) is also diagonalizable, since $PX^kP^{-1}=(PXP^{-1})^k$, so the statement follows from case 1.

Case 3: X is arbitrary

I claim the set of diagonalizable matrices is dense in the space of all matrices. From this, the statement follows from case 2. To establish the claim, let DM denote the space of diagonalizable matrices of a given size. Note that a matrix is diagonalizable if it has distinct eigenvalues, i.e., the roots of the characteristic polynomial $\phi_X(t)$ are distinct. This is the case exactly when $gcd(\phi_X, \phi'_X)=1$, which in turn is the case exactly when $Res(\phi_X, \phi'_X)\ne 0$, where $Res$ is a certain universal (i.e. doesn't depend on X) polynomial function of the coefficients of $\phi_X$ and $\phi_X'$, called the Resultant. Define the map $F$ by $F(X)=Res(\phi_X, \phi'_X)$. Summing up, we have $ F^{-1}(\mathbb{R}-\{0\})\subset DM$. Said another way, $DM^c\subset F^{-1}(0)$. But $F(X)$ is a polynomial function of the elements of $X$, which means that $F^{-1}(0)$ is Zariski-closed. Any nonempty Zariski-open set is Zariski dense, so in particular DM contains a Zariski-dense subset. Finally, $Det(X^p+...+I)$ is a polynomial function of the entries of $X$, thus continuous in the Zariski topology.