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 $M_1,M_2,\dots,M_n$?
Best Answer
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 $L$.