Matrices with all non-zero entries. I am reading a paper and it uses one of these facts, I would like to know if it has  a simple proof:
Let $F$ be an infinite field and $n\ge2$ and integer. Then for any non-scalar matrices $A_1,A_2,...,A_k$ in $M_{n}(F)$, there exist some invertible matrix $Q \in M_{n}(F)$ such that each matrix $QA_1Q^{-1}, QA_2Q^{-1},..., QA_{k}Q^{-1}$ have all non-zero entries.
I just don't know where to start at the first place, could have used diagonalizability but not all non-scalar matrices are diagonalizable. Maybe its too simple, please help.
Thanks in advance.
 A: For each $(i,j,\ell)$, note that the $(i,j)$ entry of $QA_\ell Q^{-1}\cdot \det(Q)$ is some polynomial $p_{ij\ell}$ in the entries in $Q$ (we multiply by $\det(Q)$ since the entries of $Q^{-1}$ are rational functions in the entries of $Q$ with denominator $\det(Q)$).  None of these polynomials are identically zero, since each $A_{\ell}$ is not scalar and so $Q$ can be chosen to make any individual one of its entries nonzero.  In more detail, the fact that $A_{\ell}$ is not scalar implies there is a vector $v\in F^n$ such that $A_{\ell}v$ is linearly independent from $v$.  If $i\neq j$, we can thus choose a basis with $v$ and $A_{\ell}v$ as the $i$th and $j$th basis vectors, and in this basis the $(i,j)$ entry of $A_{\ell}$ will be nonzero.  If $i=j$, we can instead choose a basis with $v$ as the $i$th basis vector and $A_{\ell}v-v$ as another basis vector.
The result then follows from the following well-known theorem, applied to the polynomial $p$ in the entries of $Q$ which is the product of all the $p_{ij\ell}$ and $\det(Q)$.

Theorem: Let $F$ be an infinite field and let $p\in F[x_1,\dots,x_m]$ be nonzero.  Then there exist $a_1,\dots,a_m\in F$ such that $p(a_1,\dots,a_m)\neq 0$.

Proof:  We use induction on $m$; the base case $m=0$ is trivial.  Now suppose $m>0$ and the result is known for $m-1$.  Think of $p$ as a polynomial in the variables $x_1,\dots,x_{m-1}$ with coefficients in $F[x_m]$.  At least one of its coefficients is nonzero, and that coefficient has only finitely many roots.  Since $F$ is infinite, there is some $a_m\in F$ which is not a root of that coefficient, and so the polynomial $p(x_1,\dots,x_{m-1},a_m)\in F[x_1,\dots,x_{m-1}]$ is nonzero. We can then find the required $a_1,\dots,a_{m-1}$ by the induction hypothesis.
