# A direct proof of a property of symmetric 2x2-determinants

Let $f(a,b,c)=\det\begin{pmatrix}a &b\\ b& c\end{pmatrix}\in\mathbb{R}[a,b,c]$ be the determinant of a $2 \times 2$ real symmetric matrix. Let $f(x_i,y_i,z_i)\geq 0$, $x_i\geq 0$, $z_i\geq 0$ for $i\in\{1,2\}$.

(Edit: i.e. the matrix is positive semidefinite).

Then

$$f(x_1+x_2,y_1+y_2,z_1+z_2)\geq 0 \tag{\ast}$$

(something that can be formulated as convexity of the cone of symmetric $2 \times 2$ positive semidefinite (p.s.d.) matrices).

Indeed, $(*)$ can be seen by viewing $f$ as the discriminant of the quadratic polynomial $$F(a,b,c)(t)=at^2+2bt+c \tag{**}$$ with $a\geq 0$, $c\geq 0$, observing that then $f(a,b,c)\geq 0$ iff $F(a,b,c)(t)\geq 0$ for all $t$, and seeing that $$F(x_1+x_2,y_1+y_2,z_1+z_2)(t)=F(x_1,y_1,z_1)(t)+F(x_2,y_2,z_2)(t)\geq 0$$

The question is whether there is a (more) direct proof of $(*)$, and what kinds of generalisations are known. E.g. I am interested in the situation where $a,b,c$ are multivariate polynomials, and under which conditions $f(a,b,c)$ is a sum of squares (s.o.s.) of polynomials---with potential applications to efficient s.o.s. decompositions.

Edit: As well, is there an analogue of $(**)$ for higher order positive semidefinite matrices?

• One classical generalization is the convexity of the cone of positive semidefinite quadratic forms. – Ivan Izmestiev Jun 21 '17 at 10:28
• thanks, right---as I need something that "works" in $\mathbb{R}[a,b,c]$ I missed that indeed my conditions are equivalent to the eigenvalues of the 2x2 matrix in question being positive semidefinite... – Dima Pasechnik Jun 21 '17 at 11:07
• I've edited the question to work this comment in. – Dima Pasechnik Jun 21 '17 at 11:25
• Perhaps you should learn about hyperbolic polynomials, a notion due to Garding. – Denis Serre Jun 28 '17 at 11:19
• @DenisSerre I am aware of hyperbolic polynomials, but they appear to be sort of orthogonal concept---they have all roots real, as opposed to (generically) having no real roots. – Dima Pasechnik Jun 28 '17 at 12:28

Here is a direct proof: Simple computations show that

$$f\left(x_1+x_2,y_1+y_2,z_1+z_2\right) = f\left(x_1,y_1,z_1\right) + f\left(x_2,y_2,z_2\right) + \left(x_1z_2+x_2z_1-2y_1y_2\right).$$

It thus remains to check that all three addends on the right hand side are nonnegative. For $f\left(x_1,y_1,z_1\right)$ and $f\left(x_2,y_2,z_2\right)$, this follows straight from the assumptions. For $x_1z_2+x_2z_1-2y_1y_2$, we have to prove that $x_1z_2+x_2z_1 \geq 2y_1y_2$. But thanks to the nonnegativity of $x_i$ and $z_i$, we have $x_1 z_1 \geq y_1^2$ (since $x_1 z_1 - y_1^2 = f\left(x_1, y_1, z_1\right) \geq 0$) and $x_2 z_2 \geq y_2^2$ (similarly), so that the AM-GM inequality yields

$x_1z_2+x_2z_1 \geq 2 \sqrt{x_1z_2 \cdot x_2z_1} = 2 \left(\underbrace{x_1 z_1}_{\geq y_1^2} \underbrace{x_2 z_2}_{\geq y_2^2} \right)^{1/2} \geq 2 \left( y_1^2 y_2^2 \right)^{1/2} = 2 \left| y_1 y_2 \right| \geq 2 y_1 y_2$,

which is precisely what we needed to prove.

Note that this argument might not help you with your question about sum-of-squares decompositions. I am not actually sure what exactly constitutes a sum-of-squares decomposition for an inequality that only holds under assumptions...

• Thanks. AM-GM is useful. Regarding sums of squares (s.o.s.) decompositions, as Ivan noted it's a positive semidefinite matrix, and so the corresponding quadratic form over $\mathbb{R}$ is a s.o.s. Now, replace $\mathbb{R}$ with a ring of real $m$-variate polynomials; if $m=1$ then our determinant will be an s.o.s. of polynomials, as any nonnegative univariate polynomial is an s.o.s. How about $m>1$? – Dima Pasechnik Jun 21 '17 at 11:24
• Well, the statement is that if $A$ and $B$ are positive semidefinite symmetric matrices, then so is $A+B$. But how do you encode positive semidefiniteness of a matrix $C$ as a statement about s.o.s.? As "the matrix is a sum of matrices of the form $D^T D$" or as "all principal minors of the matrix are sums of squares"? The former makes the claim really easy. The latter seems more complicated -- you are asking to use s.o.s. decompositions of unknown provenance; I suspect there are no general results about this, but I don't know a good way to generate counterexamples either. – darij grinberg Jun 21 '17 at 14:20
• If $M=\begin{pmatrix}a&b\\b&c\end{pmatrix}=\sum_k\begin{pmatrix}p_k^2&p_kq_k\\p_kq_k&q_k^2\end{pmatrix}$ then $\det M$ is an s.o.s., as $\det M=\sum_i\sum_j(p_iq_j-p_jq_i)^2.$ In particular, if $M$ is a psd matrix of forms, it's not clear whether the above decomposition of $M$ as a sum of matrices of forms exists, or not. – Dima Pasechnik Jun 22 '17 at 8:39