3
$\begingroup$

Consider the $3 \times 3$ matrix polynomial

$$M(x)=\left(\begin{array}{ccc}Ax&Bx&Cx\\1&1&1\end{array}\right)$$

where $A, B, C$ are $2 \times 2$ real matrices. Assume that $\mbox{rank} (M(x)) \leq 2$ for all $x$ in a neighborhood of zero. Then, by a direct calculation, we have that the set $\{A,B,C\}$ is linearly dependent.

Is this a manifestation of a more general phenomenon, or pure luck?

Edit: The result fails without symmetry of the matrices, and holds for $m$ symmetric matrices $A_i\in\mathbb{R}^{n\times n}, i=1,\dots,m$. That is, $\mbox{rank}(M(x))\leq 2, \forall x\in\mathbb{R}^n$ implies $\mbox{rank}\{A_1,\dots,A_m\}\leq2$.

$\endgroup$
  • 1
    $\begingroup$ If $A$, $B$, $C$ are $2\times 2$ matrices, then there are six columns (at least in the top two rows). Do you mean that $A$, $B$, $C$ are $2\times 1$ matrices? Or are you taking $x$ itself to be a column of size two? Please explain what you intended. $\endgroup$ – David Handelman Feb 18 '17 at 15:36
  • $\begingroup$ $x$ is $2\times 1$. $\endgroup$ – Shake Baby Feb 18 '17 at 16:50
  • $\begingroup$ It would be nice if you posted your "direct calculation". $\endgroup$ – Rodrigo de Azevedo Feb 18 '17 at 17:04
  • $\begingroup$ I did it on Mathematica. $det(M(x))$ identically zero gives 3 equations on the 12 matrix entries. Solving for three variables and computing the $3\times 3$ minors of the $3\times 4$ matrix where each line correspond to one matrix entries, we get that they are all zero. $\endgroup$ – Shake Baby Feb 18 '17 at 17:11
  • $\begingroup$ And where is the "neighborhood of zero" used? $\endgroup$ – Rodrigo de Azevedo Feb 18 '17 at 17:30
2
$\begingroup$

So for all $x$ in a neighborhood of zero, by subtracting columns we find that the 2-vectors $(B-A)x$ and $(C-A)x$ are linearly dependent, i.e. scalar multiples of each other.
Suppose $(B-A)x=\lambda(C-A)x$ and $(B-A)y=\mu(C-A)y$ for an $y$ in the same neighborhood such that $x-y$ is also in the same neighborhood. Subtracting both, we find $\lambda =\mu$ and thus $(B-A)=\lambda(C-A)$, i.e. the set $\{A,B,C\}$ is linearly dependent.

I do not yet see how this argument can be extended directly to bigger matrices, like e.g. $$M(x)=\left(\begin{array}{cccc}Ax&Bx&Cx&Dx\\1&1&1&1\end{array}\right)$$ with $3\times3$ matrices $A,B,C,D$ and $x\in\mathbb R^3$, but it seems to look rather straightforward.

$\endgroup$
  • $\begingroup$ I have a prove when all matrices are symmetric and a counter-example otherwise, which I can post if anyone is interested (it is a bit lengthy though). $\endgroup$ – Shake Baby Feb 19 '17 at 21:00
  • $\begingroup$ Please post your counterexample! $\endgroup$ – Wolfgang Feb 19 '17 at 21:21
  • $\begingroup$ $M(x)=\left(\begin{array}{ccc}x&x&x\\-x+y& 0&y\\1&1&1\end{array}\right)$ $\endgroup$ – Shake Baby Feb 20 '17 at 3:31
  • $\begingroup$ This matrix has determinant identically zero (first line is a multiple of the third). $\endgroup$ – Shake Baby Feb 20 '17 at 16:29
  • $\begingroup$ Your proof, as fas as I can tell, needs non-singularity of $B-A$ and $C-A$. $\endgroup$ – Shake Baby Feb 20 '17 at 16:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.