# What does this statement mean in linear algebra :$Ker(M) = \mathbb{F_p}$ for $p$ a prime

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
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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 $\mathbb{F}_3^m$.