On linearly independent matrices I have been reading J.S. Milne's lecture notes on fields and Galois theory and came across the normal basis theorem (Thm 5.18 on page 66 of the notes). Trying to find my own proof, the following problem in linear algebra quickly arose:

Question: Let $F$ be a field. Given linearly independent matrices $A_1, \dots, A_n \in \operatorname{GL}_n(F)$, does there necessarily exist some $b\in F^n$ such that $A_1b, \dots, A_nb$ are linearly independent over $F$?

This is clearly not true if the matrices are not linearly independent. Also, if they are not invertible, the claim is false in general: e.g. set $n=2$ and consider
$$A_1 = \begin{pmatrix} 1 & 0 \\ 0& 0\end{pmatrix},\quad  A_2 = \begin{pmatrix} 0 & 1 \\ 0& 0\end{pmatrix}$$
Going through a case-by-case analysis, I think I can prove the claim for $n=2$, so there seems to be some hope... 
Any help would be appreciated. Thanks!
 A: The answer is no. 
Consider, for $F=\mathbb{R}$, $n=3$,
$$
A_1=\begin{bmatrix}1&0&0\\0&1&0\\ 1&2&3\end{bmatrix}, \ 
A_2=\begin{bmatrix}1&0&0\\0&1&0\\ 2&3&1\end{bmatrix}, \ 
A_3=\begin{bmatrix}1&0&0\\0&1&0\\ 3&1&2\end{bmatrix}.
$$
These three matrices are linearly independent, because the last three rows are. They are invertible, since the are triangular with nonzero diagonal. 
For any $b=\begin{bmatrix}x\\ y\\ z\end{bmatrix}\in\mathbb{R}^3$, the three vectors we obtain are
$$
A_1b=\begin{bmatrix}x\\ y\\ x+2y+3z\end{bmatrix}, \ 
A_2b=\begin{bmatrix}x\\ y\\ 2x+3y+z\end{bmatrix}, \ 
A_3b=\begin{bmatrix}x\\ y\\ 3x+y+2z\end{bmatrix}. \ 
$$
And for any choice of $x,y,z$, these vectors are linearly dependent. To see this,  we need to find coefficients, not all zero, such that $\alpha\,A_1b+\beta\,A_2b+\gamma\,A_3b=0$. The first two rows require $\alpha+\beta+\gamma=0$. And the third row requires $(x+2y+3x)\alpha+(2x+3y+z)\beta+(2x+y+2x)\gamma=0$. As it is a homogeneous system of two equations in three variables, it will always have nonzero solutions. 
