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Let $n \in \{1,2,\dots\}$, and let $A$ be an $n\times n$ matrix of real numbers. Consider the unique function $f:\mathbb{R}^n\rightarrow\mathbb{R}$ satisfying $f(x) = x^TAx$ for every $x \in \mathbb{R}^n$. Suppose $A$ is not symmetric.

  1. Is it possible for $f$ to be convex?
  2. Suppose that $f(x) \geq 0$ for every $x \in \mathbb{R}^n$. Must $f$ be convex? What if $f(x) > 0$ for every $x \in \mathbb{R}^n$?
  3. Is $A$ uniquely determined by $f$? In other words, suppose that $B$ is an $n\times n$ matrix of real numbers satisfying that $f(x) = x^TBx$ for every $x \in \mathbb{R}^n$; is it necessarily the case that $A=B$?
Evan Aad
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  • At least question 3 has been asked many times before (in different variations); https://math.stackexchange.com/questions/307381/why-do-we-assume-that-a-matrix-in-quadratic-form-is-symmetric, https://math.stackexchange.com/questions/1464994/symmetric-part-of-a-contributes-to-quadratic-form, https://math.stackexchange.com/questions/3580859/quadratic-form-if-corresponding-matrix-is-not-symmetric-matrix, https://math.stackexchange.com/questions/420917/matrix-of-quadratic-form-has-to-be-symmetric, and so on... – Hans Lundmark May 26 '21 at 17:03
  • @HansLundmark: Thanks. Actually, question 3 is the only one of the three question I asked that is not answered by any of the posts you linked to... – Evan Aad May 26 '21 at 17:27
  • Of course it's answered. The quadratic forms are equal iff the symmetric parts of $A$ and $B$ are equal, i.e. $(A+A^t)/2=(B+B^t)/2$. So $A$ is definitely not uniquely determined by $f$. – Hans Lundmark May 26 '21 at 18:52
  • Related – Rodrigo de Azevedo May 12 '23 at 05:37

1 Answers1

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$$ x^TAx = \sum_{i,j}x_i A_{ij} x_j = \frac{1}{2}\sum_{i,j}x_i (A_{ij} +A_{ji})x_j = x^TBx, $$ where $$ B = \frac{A+A^T}{2} $$ is symmetric. Thus, all the properties true for a symmetrix $A$ matrix apply.

Roger V.
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