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Let $A$ be a $n$$ \times$ $(n+2)$ matrix over a field $F$ such that every set of $n$ columns (out of the ($n+2$) columns) are linearly independent over $F$. Assume that $A$ is of the form [$I_{n \times n}| D$] where $D$ is a $n \times 2$ matrix such that the first entry of both its columns is 1. I want to show that the field $F$ has at least $n+1$ elements.

My approach so far : Let the 2 columns of $D$ be $C_1= (s_1,\ldots,s_n)$ and $C_2= (t_1,\ldots,t_n)$. First of all I can conclude that the 2 columns of $D$ have all non zero entries. Further, I can conclude that $s_j \neq t_j$ where $2 \leq j\leq n$.

Please help me proceed further. Thanks !

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For $n=4$, consider the matrix

$$A=\pmatrix{ 1&0&0&0&1&2\\ 0&1&0&0&2&2\\ 0&0&1&0&3&2\\ 0&0&0&1&3&1\\ }$$

Unless I'm mistaken, I believe it is a counter example. I don't think there is a counter example for smaller $n$.

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    I think you are correct. I actually meant to prove that the field F has at least n+1 elements. And for that, showing that the matrix has atleast n 'distinct' entries would suffice. Thanks for pointing the error. I will correct my question ! – 2017-02-22
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    If it's worth anything, there is no counter-example in $\mathbb{Z}/(5)$. – 2017-02-23