Show $x^2 A + x B +C$ is positive semi-definite

EDIT: Only $$A$$ is allowed to be positive semi-definite.

I'm interested in showing that the matrix $$f(x)=x^2 A + x B +C$$ is positive semi-definite (PSD) for any scalar $$x$$, and PSD matricex $$A$$. I use the following definition for a PSD matrix: \begin{align} u^T f(x) u &= x^2\, u^T A u + x\, u^T B u + u^T C u \\ &\geq 0 \end{align} for any vector $$u$$.

Then I use the quadratic formula to obtain the following condition: \begin{align} (u^T B u )^2 - 4(u^T A u)(u^T C u) &= u^T B u u^T B u - 4 u^T A u u^T C u \\ &= u^T (B u u^T B - A u u^T C) u \\ &\leq 0 \end{align} for any vector $$u$$.

I don't know how to get rid of $$u$$ and translate this condition to one involving $$A$$, $$B$$ and $$C$$ only.

• how can this inequality be true if you take A,B,C to be positive real numbers (1 by 1 matrices) then B^2-AC can be positive – Sandeep Silwal Jul 27 '19 at 18:30
• Of course your condition will not hold for any positive definite $A,B,C$. As a $1 \times 1$ counterexample, we can note that the "matrix" $$x^2\cdot 1 + x \cdot 5 + 6$$ is not necessarily "positive definite" (e.g. with $x = -2.5$). – Omnomnomnom Jul 27 '19 at 18:35
• Thank you for pointing that out. I have edited the question accordingly. – ToniAz Jul 27 '19 at 19:03
• For any (positive semidefinite) $A$, there will exist $B,C$ (not necessarily positive definite) such that $x^2 A + xB + C$ fails to be positive semidefinite for $x=1$. – Omnomnomnom Jul 27 '19 at 19:27

You're looking for conditions on $$A,B,C$$ that guarantee that for any $$u$$, we will have $$u^T(Buu^TB - Auu^TC)u = \operatorname{Tr}[uu^T(Buu^TB - Auu^TC)] \geq 0$$ Notably, the map $$\langle P,Q \rangle = \operatorname{Tr}(PQ^T)$$ defines an inner product over $$\Bbb R^{n \times n}$$. If we define $$\phi$$ to be the linear map given by $$\phi(X) = BXB - AXC,$$ then we're looking for conditions that guarantee $$\langle \phi(uu^T), uu^T \rangle \geq 0$$ for all choices of unit vector $$u$$. Equivalently, we're looking for conditions that guarantee $$\langle \phi(X), X \rangle \geq 0$$ whenever $$X$$ is a symmetric matrix.
Via the vectorization operator, we see that $$\phi$$ (over all of $$\Bbb R^{n \times n}$$) can be represented by the matrix $$[\phi] = B^T \otimes B - C^T \otimes A$$ where $$\otimes$$ denotes the Kronecker product. A sufficient (but not necessary, I believe) condition to guarantee that the desired condition holds is that the matrix $$[\phi] + [\phi]^T$$ is symmetric positive semidefinite.