What about the rank of this matrix?

Question

Problem: Suppose $0<r<n$. Suppose $A:\mathbb{R} \to \mathcal{M}_2(\mathbb{R})$ is $\mathcal{C}^1$. Suppose that he rank of $A(t)$ il less or equal $r$ for all $t \in \mathbb{R}$. What be said about the rank of $A'(0)$?

Attempt: I tried using the fact that $rank(A)=k$ if and only $k$ is the maximum size of an invertible minor $B$. I would like to say that $rank(A'(0)) \leq r$ but I am not sure if it is true. Of course $rank(A'(0))$ could be $r$ because we can define $A(t)$ as the diagonal matrix with the first $r$ elements equal to $t$ and anywhere else equal to $0$.

Answer 1

Since $\operatorname{rank}(A(t))\le r$ for every $t$, the rank of the differential quotient $\frac{A(t)-A(0)}{t}$ is at most $2r$. It follows that the rank of $A'(0)$ is also $\le2r$. The example in the other answer has already illustrated the case where $\operatorname{rank}(A'(0))=2r$.

Answer 2

In general, the rank of $A'(0)$ can be larger than $r$. Consider for example $A(t) \colon \mathbb{R} \rightarrow M_2(\mathbb{R})$ given by $$ A(t) = \begin{bmatrix} t & t^2 \\ 1 & t \end{bmatrix}. $$ Then $A(t)$ has rank one for all $t \in \mathbb{R}$ but $$ A'(0) = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} $$ has rank two.

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In general, you can obtain an explicit relation between $A'(0)$ and $A(0)$. A matrix $A$ has rank $\leq r$ iff $\Lambda^{r+1}(A) = 0$. This is a fancy but useful way of saying that all the $(r + 1) \times (r+1)$ minors of $A$ vanish. In your case, you have $\Lambda^{r+1}(A(t)) \equiv 0$. Differentiating this condition and plugging $t = 0$ you get that we must have $$ A'(0)v_1 \wedge A(0)v_2 \wedge \dots \wedge A(0)v_{k+1} + A(0)v_1 \wedge A'(0)v_2 \wedge \dots \wedge A(0)v_{k+1} + \\ \dots + A(0)v_1 \wedge \dots \wedge A(0)v_k \wedge A'(0) v_{k+1} = 0 $$ for all $v_1,\dots,v_{k+1} \in \mathbb{R}^n$.

This is a system of linear equations on $A'(0)$ (where we assume $A(0)$ is fixed). Depending on what your initial matrix $A(0)$ is, the system might or might not put a restriction on $A'(0)$ but not necessarily a restriction on the rank of $A'(0)$.