Linear Algebra: Fields and Vector Spaces, Matrices I'm supposed to find $\dim_\Bbb F(K_A)$ and a basis for $K_A$ when A is the matrix over a field $\Bbb F= \Bbb F^2$ where we define $$K_A=\{\underline x=\left(
    \begin{array}{c}
      x_1 \\\vdots\\
      x_n
    \end{array}
  \right)\in \Bbb F^n;A\underline x=0\}$$ In this particular question the matrix A=    $\begin{bmatrix}1 & 1 & 1 & -1\\1 & -1 & 1 & -1 \end{bmatrix}$
Generally, If I reduce A to reduced row echelon form I get $\begin{bmatrix}1 & 0 & 1 & -1\\0 & 1 & 0 & 0 \end{bmatrix}$
a) Is this row reduction correct? 
b) Should my answers for the actual question be any different for $\Bbb F=\Bbb F^2$ then for $\Bbb F= \Bbb Q$?

If $\;\Bbb F=\Bbb F_2:=\Bbb Z/2\Bbb Z=$ the field with two elements, then $\;-1=1\;$ here, so
$$A=\begin{pmatrix}1&1&1&1\\1&1&1&1\end{pmatrix}\to\begin{pmatrix}1&1&1&1\\0&0&0&0\end{pmatrix}$$
and a basis for the null space (kernel) of the matrix can be
$$\left\{\begin{pmatrix}1\\1\\0\\0\end{pmatrix}\;,\;\;\begin{pmatrix}1\\0\\1\\0\end{pmatrix}\;,\;\;\begin{pmatrix}1\\0\\0\\1\end{pmatrix}\right\}$$
For $\;\Bbb F =\Bbb Q\;$ the answer is completely different. This time
$$A=\begin{pmatrix}1&\;\;1&1&-1\\1&-1&1&-1\end{pmatrix}\to\begin{pmatrix}1&\;\;1&1&-1\\0&-2&0&\;\;0\end{pmatrix}$$
and then the kernel has dimension $\;2\;$, with a possible basis being
$$\left\{\begin{pmatrix}1\\0\\0\\1\end{pmatrix}\;,\;\;\begin{pmatrix}0\\0\\1\\1\end{pmatrix}\right\}$$
