# Rank deficient matrix polynomial implies linear dependence of coefficients

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 $\mbox{rank}\{A_1,\dots,A_m\}\leq2$.

• 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. – David Handelman Feb 18 '17 at 15:36
• $x$ is $2\times 1$. – Shake Baby Feb 18 '17 at 16:50
• It would be nice if you posted your "direct calculation". – Rodrigo de Azevedo Feb 18 '17 at 17:04
• 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. – Shake Baby Feb 18 '17 at 17:11
• And where is the "neighborhood of zero" used? – Rodrigo de Azevedo Feb 18 '17 at 17:30

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.
• $M(x)=\left(\begin{array}{ccc}x&x&x\\-x+y& 0&y\\1&1&1\end{array}\right)$ – Shake Baby Feb 20 '17 at 3:31
• Your proof, as fas as I can tell, needs non-singularity of $B-A$ and $C-A$. – Shake Baby Feb 20 '17 at 16:57