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.

  • $\begingroup$ 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 $\endgroup$ – Sandeep Silwal Jul 27 '19 at 18:30
  • $\begingroup$ 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 $x = -2.5$). $\endgroup$ – Omnomnomnom Jul 27 '19 at 18:35
  • $\begingroup$ Thank you for pointing that out. I have edited the question accordingly. $\endgroup$ – ToniAz Jul 27 '19 at 19:03
  • $\begingroup$ 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 $x=1$. $\endgroup$ – Omnomnomnom Jul 27 '19 at 19:27

Partial answer; certainly too long for a comment.

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.

  • $\begingroup$ Why do we bring the vectorization operator into this? $\endgroup$ – ToniAz Jul 27 '19 at 20:23

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