A problem on rank of a matrix over two fields This is a problem from Berkeley problems in mathematics.
If $F$ is a subfield of $K$, and $M$ has entries in $F$, how is the row rank of $M$ over $F$ related to the row rank of $M$ over $K$?
where $M$ is a n by n matrix
The solution says "If a set of rows of $M$ is linearly independent over $F$, then clearly it is also independent over K, so the rank of $M$ over $F$ is, at most, the rank of $M$ over $K$."
I have some trouble understanding this, what I thought was that if they are linearly independent over the bigger field K, they are linearly independent over F. (Because all linear combinations with scalars from F are subsumed when you are talking about linear combinations in K) However here it is the other way around
 A: The rank of a matrix is the size of its largest square submatrix with nonzero determinant. Since the determinant is determined only by the matrix's entries, it remains the same during a field extension. Thus the rank remains invariant too.
Alternatively, the rank of a matrix is just the number of nonzero diagonal entries in its Smith normal form. Since the Smith normal form computed over a subfield is still a Smith normal form over a larger field, the rank remains the same.
If one considers rational canonical form (which also is solely determined by the matrix's entries) instead, one can show not only that matrix rank is invariant on field extension, but also that similar matrices remain similar over a field extension too.
A: Doesn't necessarily need to be the case always. Consider the matrix
$$ A =
\begin{bmatrix}
    1       & 1  \\
    i       & i  \\
\end{bmatrix}
$$
$A$ has rank $2$ over $\Bbb{R}$ but $1$ over $\Bbb{C}$.
If $A$ is a matrix of rank $r$ over a field $F$ then it has rank $r$  over any extension field of $F$. But if $A$ is a matrix of rank $r$ over a field $K$ then its rank over any sub-field of $K$ is at least $r$. But it can be more also.
