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Let $A,B$ be $d\times d$ symmetric matrices with real coefficients such that $A$ is positive semi-definite. Prove that $\det(tA+B)\in \mathbb{R}[t]$ is either $0$ or has all roots real.


Here is what I tried:

Let us at-first assume that $A$ is positive definite. Then we know that there is a non-singular matrix $S$ such that $SAS^{T}=I$ and $SBS^{T}=\Lambda$, where $\Lambda$ is some diagonal matrix. Thus the problem reduces to showing that $\det(tI+\Lambda)$ has real roots, but this is obvious because the roots are eigenvalues of $\Lambda$ and any symmetric matrix has all real eigenvalues.

However, I am stuck with the case where $A$ has a $0$ eigenvalue. My guess is that we can still find such a invertible matrix $S$ such that $SAS^{T}$ is diagonal with only $0$'s and $1$'s on the diagonal and $SBS^{T}$ is diagonal. But I am not sure how to prove this. Any help would be appreciated.

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  • $\begingroup$ Where does this come from? $\endgroup$ Commented Jul 1, 2023 at 6:20

1 Answer 1

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for some $\lambda \in\mathbb C$:
$\det\big(\lambda A + B\big)=0\implies \lambda A\mathbf x =-B\mathbf x$ for some $\mathbf x \in \mathbb C^d-\big\{\mathbf 0\big\}$

case 1: $\mathbf x \not \in \ker A\implies \mathbf x^* A \mathbf x \neq 0$ (since $A$ is PSD)
line 2 $\implies \lambda \cdot\mathbf x^* A\mathbf x =-\mathbf x^*B\mathbf x= \big(\lambda \cdot\mathbf x^* A\mathbf x\big)^*=\overline \lambda \cdot\mathbf x^* A\mathbf x$
$\implies \lambda \in \mathbb R$

case 2: $\mathbf x \in \ker A$
line 2:$\implies \mathbf 0 = \lambda A\mathbf x =-B\mathbf x \implies \mathbf x \in \ker B$

hence for any $\lambda \in \mathbb C$ we have $\big(\lambda A+B\big)\mathbf x=\mathbf 0\implies \det\big(\lambda A+B\big)=0$
$\implies \det(tA+B)\in \mathbb{R}[t]$ has an infinite number of roots, i.e. is the zero polynomial.

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