Dimensions with Fields and Subfields Fix a field $F$ and a subfield $K \subseteq F$. Also, consider matrix $A \in M_{m,n}(K)$. Define $V_{K} = \{x \in K^n: Ax=0\}$. This is a vector space over $K$. Define $V_{F} = \{x \in F^n: Ax=0\}$. This is a vector space over $F$. 
Prove that $dim_{K}(V_{K}) = dim_{F}(V_{F})$. 
I was originally going to approach this using rank-nullity theorem, since it seems to have been related to the kernel, but there is a hint that says how does Gaussian elimination of A differ with entries in $F$. I am sort of confused as to how is Gaussian elimination involved in this problem in the first place. Any tips on how to do this proof? Thanks. 

After performing Gaussian elimination, the dimension of the null space is the dimension of the vector space minus the number of rows which don't consist entirely of zeros.
When you perform Gaussian elimination, what arithmetic operations do you use? Can the result ever leave $K$?
