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This is the last followup question to my other question. Here I have a matrix $M$ whose kernel is the field $\mathbb{F_p}$. I'm not sure what this expression is saying or why it is important. My understanding is that $\mathbb{F_p}$ is a field that contains individual non-vector elements so how can this possibly be in the kernel of matrix $M$?

For example, let us say that I have this matrix in $\mathbb{F_3}$ when is that matrix's kernel $\mathbb{F_3}$ ?

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To reiterate the previous comment, $\Bbb{F}_3$ is the 1-dimensional vector space over itself. Any matrix over a field $K$ has $K$ for its kernel whenever it has a $1$-dimensional kernel. –  Kevin Carlson Oct 27 '12 at 5:02
    
@DevenWare Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. –  Julian Kuelshammer Aug 22 '13 at 7:27

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(Comment converted to answer by request)

The key idea is that elements in the field $\mathbb{F}_p$ can be thought of as both scalars and one dimensional vectors.

So lets go through your example in detail. You have a matrix $\mathrm{M}$ in $\mathrm{Mat}_{n\times m}(\mathbb{F}_3)$. That means as map,

$M : \mathbb{F}_3^m \rightarrow \mathbb{F}_3^n$

So, $\mathrm{Ker(M)} \subseteq \mathbb{F}_3^m$. When we say that $\mathrm{Ker(M)} = \mathbb{F}_3$ what we are doing is actually abusing notation (in a very convenient way.) We actually mean that its kernel is a one dimensional subspace of $\mathbb{F}_3^m$.

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