Finding a base of matrices of rank 1 I came up with an interesting question from linear algebra that I couldn't resolve. I am interested to find out an algorithm, a characterization or perhaps a proof that only one case is possible. Of course, if some other branch of mathematics, more specialized than linear algebra, is suited better for this problem, it will useful to tell what is it.
The statement of the problem:
Let $n>1$ be a natural number. Let $\mathbb{F}$ be some algebraically closed field. Let $M_1,M_2,\dots,M_n$ be $n$ matrices over $\mathbb{F}$ of size $n\times n$ that are linearly independent, meaning that the linear closure $L=\mathcal{L}(M_1,\dots,M_n)$ (linear space span over this set of matrices) has dimension $n$. What can we say about dimension $m$ of linear subspace that is generated by matrices of rank $1$?  Namely, the dimension of $\mathcal{L}\{M\in L\mid\text{rank}M=1\}$?
I believe that in the case of $n=2$, this rank has to be always $2$. However, I couldn't prove it. Also, for other values of $n$, does it still hold that $m=n$? What are all the possible pairs $(m,n)$? Is there perhaps an algorithm or method to calculate that dimension $m$ given the matrices $M_1,M_2,\dots,M_n$?
 A: That obviously depends on the $n$ $n\times n$ matrices $M_1,\dots,M_n$ you pick and the field $\mathbb{F}$.
For example, if $n=2$ and you pick
$$M_1=\begin{pmatrix}1\\&1\end{pmatrix},M_2=\begin{pmatrix}&-1\\1&\end{pmatrix}$$
with $-1\notin(\mathbb{F}^\times)^2$ (e.g., $\mathbb{F}=\mathbb{R}$ or $\mathbb{F}=GF(p)$ with $p\equiv 3\pmod{4}$), then there are no rank 1 matrices in $L$.  But if you pick $M_1$ to be a rank-1 matrix, then it is obviously are going to be in $L$.
A: I’m turning my comment into answer. I’ll show that any pair $n \geq m > 0$ is possible. I’m unsure whether $m=0$ is possible (user10354138’s answer requires non-algebraically closed fields). The argument works in any characteristic, but it’s simpler, as a first reading, to assume that we’re in characteristic zero.
So we consider a Frobenius matrix $P$ with characteristic polynomial $\chi$, and take $M_i=P^{i-1}$.
To study the rank of matrices of $L$ (ie the $f(P)$, $f$ being a polynomial of degree less than $n$) it is easier to pretend they’re endomorphisms of some $\bigoplus_{\lambda}{E_{\lambda}}$ where the $\lambda$ are roots of $\chi$, $E_{\lambda}$ is stable under each endomorphism and has dimension $d_{\lambda}$, and $p_{\lambda}=P_{|E_{\lambda}}-\lambda I$ has nilpotency index $d_{\lambda}$.
Let $f$ be a polynomial of degree less than $n$, then the ranks $r_{\lambda}$ of $f(P)_{|E_{\lambda}}$ add up to the rank of $f(P)$. Moreover, if $\chi(\lambda)=0$, then $f_{\lambda}=f(P)_{|E_{\lambda}}=\sum_{k \geq 0}{(\frac{f^{(k)}}{k!})(\lambda)p_{\lambda}^k}$. Thus, as $p_{\lambda}$ is nilpotent, the rank of $f_{\lambda}$ is $d_{\lambda}-k$, where $k$ is the smallest integer such that $f^{(k)}/k!$ doesn’t vanish at $\lambda$, (or more precisely the minimum of such a $k$ and $d_{\lambda}$), ie the vanishing order of $f \wedge \chi$ at $\lambda$.
Thus the rank of $f(P)$ is $\sum_{\chi(\lambda)=0}{d_{\lambda}-v_{\lambda}(f)}$ where $v_{\lambda}(f)$ is the vanishing order of $f \wedge \chi$ at $\lambda$.
Thus, $f(P)$ has rank at most one iff for some root $\lambda$ of $\chi$, $\chi(x)|f(x)(x-\lambda)$, and in that case $f(P)$ is zero on $E_{\mu}$ if $\mu \neq \lambda$, but $f(P)$ acts as $(f^{(d_{\lambda}-1)}/(d_{\lambda}-1)!)(\lambda)p_{\lambda}^{n-1}$ on $E{\lambda}$.
From this, it follows that the subspace of $L$ generated by the matrices of rank one has the $\frac{\chi(x)}{x-\lambda}(P)$ as a basis so has dimension $m$.
