An $n$-by-$n$ square matrix $M$ is totally non-negative (TNN) if all its minors are non-negative. If we regard $M$ as the matrix for some linear operator $\varphi:V\rightarrow V$ under some basis $\{e_1,...,.e_n\}$, then this is the same as saying $\bigwedge^k \varphi$ is represented by a non-negative matrix under the basis $\{e_I:I\subset [n],|I|=k\}$ for all reasonable $k$. We can call this matrix $\bigwedge^k M$.
I wonder if we can get a stronger conclusion: is $\bigwedge ^k M$ totally non-negative?
None of the examples I have so far shows negative evidence. Any counter-example, proof, or reference to relevant theorems would be much appreciated.
It turns out my claim is false. Here's a counter-example:
$$M=\begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$ It's easy to check that $M$ is totally nonnegative.
Let's use the symbol $[ij]$ where $i<j$ to denote the basis element $e_i\wedge e_j$ in $\bigwedge^2 V$. Say the basis is ordered this way: $\{[12],[13],[14],[23],[24],[34]\}$. Then the matrix $\bigwedge^2 M$ is indexed by $([ij],[kl])$, whose corresponding entry is the 2-by-2 minor of $M$ obtained from choosing rows $i,j$ and columns $k,l$.
Note that $$ \wedge^2 M_{[13],[14]}=1,\quad \wedge^2 M_{[13],[23]}=1,\quad \wedge^2 M_{[14],[14]}=1,\quad \wedge^2M_{[14],[23]}=0$$
This gives us the following 2-by-2 minor of $\bigwedge^2 M$:
$$\begin{vmatrix}1 & 1 \\ 1 & 0 \end{vmatrix}$$
which is negative.