1
$\begingroup$

I do not understand a basic concept about the convexity of a quadratic form. I read that

A quadratic form $q(h)=h^{T}\mathbf{A}h$ is convex if and only if $\mathbf{A}$ is positive semidefinite.

But if I search for the definition of the definiteness of a matrix $\mathbf{A}$, I find that:

A symmetric $n\times n$ real matrix $\mathbf{A}$ is said to be positive semidefinite if its associated quadratic form $q(h)=h^{T}\mathbf{A}h$ is non-negative for any $h\ne0$.

I am a bit confused. If I consider that "$\mathbf{A}$ positive semidefinite means $q(h) \geq 0$", the first sentence means: "$q(h)$ is convex if and only if $q(h) \geq 0$". Therefore it seems that saying that a quadratic form is convex is the same of saying it is $\ge 0$, and this seems quite strange.

$\endgroup$
6
  • 1
    $\begingroup$ There is nothing strange about. You understand it correctly. Notice there is inhomogen quadratic form which is convex but not everywhere non-negative. Like $q(h) = h^2 - 1$. $\endgroup$
    – user251257
    Apr 11, 2019 at 18:23
  • $\begingroup$ Thank you for the answer. But how can we explain the fact that q(h)=h^2 - 1 is convex but non always non negative? In conclusion is "non negative" equal to "convex"? It seems quite strange to me since the convexity of a generic function (for instance a simple function from R to R) is not due to its sign, but to its second derivative sign. So I do not understand this. $\endgroup$
    – Kinka-Byo
    Apr 11, 2019 at 19:37
  • $\begingroup$ A quadratic function essentially equals its second derivative. $\endgroup$
    – gerw
    Apr 11, 2019 at 20:25
  • 1
    $\begingroup$ The function $x^2-1$ is not of the form $x^TAx$, it’s of the form $x^TAx-c$ for a constant $c$. In 1d, a function of the form $g(x)=ax^2+bx+c$ is convex if and only if $a\geq0$, not $g(x)\geq0$ for all $x$. In 1D people often refer to functions like $g$ as “quadratics”. This is not the same as a quadratic form, which just includes the $ax^2$ part. $\endgroup$
    – David M.
    Apr 11, 2019 at 23:29
  • $\begingroup$ Thank you both. So a proof of the statment "non - negative = convex" can be simply the fact that the Hessian matrix of a quadratic form is equal to 2A (as I read on some texts)? $\endgroup$
    – Kinka-Byo
    Apr 12, 2019 at 7:09

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .