0
$\begingroup$

Let $A \in M_n(\mathbb R)$ be a $6\times 6$ matrix given by \begin{align*} \begin{pmatrix} 1 & a_1 &b_1 & a_1^2 & a_1b_1 & b_1^2\\ 1 & a_2 &b_2 & a_2^2 & a_2b_2 & b_2^2\\ 1 & a_3 &b_3 & a_3^2 & a_3b_3 & b_3^2\\ 1 & a_4 &b_4 & a_4^2 & a_4b_4 & b_4^2\\ 1 & a_5 &b_5 & a_5^2 & a_5b_5 & b_5^2\\ 1 & a_6 &b_6 & a_6^2 & a_6b_6 & b_6^2 \end{pmatrix}. \end{align*} Suppose $a=(a_1, \dots, a_6)^T$ and $b=(b_1, \dots, b_6)^T$.

How to chose vectors $a,b$ such that the matrix $A$ is nonsingular?

$\endgroup$
1
  • 1
    $\begingroup$ Your matrix is $5 \times 6$. Perhaps you want another row? $\endgroup$ Oct 17, 2019 at 12:06

1 Answer 1

2
$\begingroup$

The necessary and sufficient condition for it to be nonsingular is of course that its determinant is nonzero. That determinant is a rather nasty polynomial in the $a_i$ and $b_i$, irreducible over the rationals, with $720$ terms, each of total degree $8$.

Of course a necessary condition is that the ordered pairs $(a_i, b_i)$ are all distinct.

One sufficient condition is that the $a_i$ and $b_i$ are algebraically independent.

$\endgroup$
3
  • $\begingroup$ Could you please explain the sufficient condition?which is $a$ and $b$ are algebraically independent imply $A$ is nonsingular $\endgroup$
    – Nate
    Oct 17, 2019 at 12:19
  • $\begingroup$ I mean the whole set $\{a_1, \ldots, a_6, b_1, \ldots, b_6\}$ algebraically independent. This says that any polynomial in the $a_i$ and $b_i$ with rational coefficients, not all $0$, has a nonzero value. The determinant is such a polynomial. $\endgroup$ Oct 17, 2019 at 12:53
  • $\begingroup$ Maybe I should mention that if $a_1, \ldots, a_6,b_1, \ldots, b_6$ are a random sample from any continuous probability distribution, this condition will be true with probability $1$. $\endgroup$ Oct 17, 2019 at 19:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .