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I have a $3\times3$ real matrix of the following form $$ A=\left( \begin{matrix} a & b & c \\ * & * & * \\ * & * & * \end{matrix} \right), $$ where $a,b,c$ are fixed. I would like to find the other entries of the matrix in such a way that $\det A = 0$ if and only if $a=b=c=0$ and no combination of $a,b,c$ appears at the denominator of the other entries.

So far I have tried with different combinations of $a,b,c$, but nothing seems to work. In particular, I always find a determinant that vanishes e.g. when $a=0$ but the others need not be zero, or something like that.

Is there a clever choice I do not see?

Gibbs
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  • So none of $a,b,c$ can appear anywhere in the $*$ entries? – Parcly Taxel Jun 07 '21 at 14:09
  • They should appear only at the numerators of the other entries. E.g. the matrix with second row $a^2, b^2, c^2$ and third row $a^3, b^3, c^3$ would be ok, but the first condition on the determinant is not satisfied as $a=0$ is enough to have vanishing determinant. – Gibbs Jun 07 '21 at 14:11
  • Not appearing in a denominator is a semantic requirement and very difficult to enforce consistently. You could require that the solution must give a valid matrix for any $a,b,c$. That's more objective. – Arthur Jun 07 '21 at 14:16
  • @Arthur yes that is a good point. Still, I am not sure whether the condition you suggest is easier to enforce. Denominators of the form $a^2+b^2+c^2$ or $|a|+|b|+|c|$ or $(a^2+b^2+c^2)^n$ would all work... – Gibbs Jun 07 '21 at 14:19
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    It's crazy how that is actually possible in dimension 4 https://math.stackexchange.com/questions/706513/calculating-the-determinant-gives-a2b2c2d22 – Laassila souhayl Jun 07 '21 at 15:14

2 Answers2

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It is impossible to replace asterisks with continuous functions of $a,b,c$, so determinant is not zero when $a,b,c\neq0$.

Let $(a,b,c)$ be a point on a unit sphere. If determinant is non-zero, you can produce an orthonormal triple out of the rows (with Gram-Schmidt process). Take the second vector. It will be tangent to the sphere. So you have created a continuous vector field on a sphere.

But we know that a sphere cannot be combed.

Vasily Mitch
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  • Thank you. I think this basically answers my question. I wonder now whether this can be done with non-continuous functions. – Gibbs Jun 08 '21 at 12:12
  • Perhaps one remark is that the Gram-Schmidt process gives you combinations of $a,b,c$ in the denominator of the other entries, but they all make sense under the assumption. Arthur pointed this out in the comments, so perhaps my problem should be phrased more precisely. – Gibbs Jun 08 '21 at 12:38
  • whether this can be done with non-continuous functions.

    Split the sphere into several pieces fitting half sphere (for example, octants). For each piece choose two non-collinear constant vectors in the non intersecting plane plane. For example octant $a,b,c\ge0$ doesn't intersect plane a+b+c=0, so we can choose two vectors $(1,0,-1)$ and $(1,-1,0)$. Repeat for every piece.

     – Vasily Mitch Jun 09 '21 at 17:44
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(Its not an answer but its too big for a comment).

Let the following matrix $A$ has constant functions of $a,b$ and $c$ in the numerator $$\det\begin{bmatrix} a & b & c \\ c & a & b \\ b^2 & c^2 & a^2 \end{bmatrix}=a^4+b^4+c^4+-abc^2-bca^2-acb^2 $$ If $a=0 \implies \det A = b^4+c^4 =0 \iff b=c=0$ (the same occours with $b,c$).

That shows us if some $a,b$ or $c$ is zero, then the determinant will necessarily vanish only if all of them are also zero. But that doesn't show us

$$\det A=0 \iff a=b=c=0$$

( we may have $a,b,c \neq 0$ such that $\det A = 0$ ).

Powder
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