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 $\Bbb F= \Bbb Q$?

  • $\begingroup$ Do you mean $\mathbb{F}=\mathbb{F}_2$? What is $\mathbb{F}^2$? $\endgroup$ – user35603 Nov 20 '13 at 16:18
  • $\begingroup$ That was a general definition for $K_A$ in my case $\Bbb F=\Bbb F^2$ $\endgroup$ – Vladimir Nabokov Nov 20 '13 at 16:20
  • 2
    $\begingroup$ Ok, in general, standard usage, there is no such thing as $\;\Bbb F=\Bbb F^2\;$ : the first one can be a field, but then the second one is usually the set of all couples of elements in $\;\Bbb F\;$. Perhaps you meant $\;\Bbb F=\Bbb F_2:=$ the field with two elements? $\endgroup$ – DonAntonio Nov 20 '13 at 16:23
  • $\begingroup$ I presume so. This is whats tripping me up. I don't really understand what $\Bbb F=\Bbb F^2$ means... $\endgroup$ – Vladimir Nabokov Nov 20 '13 at 16:26
  • $\begingroup$ So, if $\mathbb{F}=\mathbb{F}_2$ then your matrix has identical rows, since $1=-1$ in $\mathbb{F}_2$. $\endgroup$ – user35603 Nov 20 '13 at 16:32

If $\;\Bbb F=\Bbb F_2:=\Bbb Z/2\Bbb Z=$ the field with two elements, then $\;-1=1\;$ here, so


and a basis for the null space (kernel) of the matrix can be


For $\;\Bbb F =\Bbb Q\;$ the answer is completely different. This time


and then the kernel has dimension $\;2\;$, with a possible basis being


  • $\begingroup$ What do you mean by $\Bbb Z/2\Bbb Z$ exactly? $\endgroup$ – Vladimir Nabokov Nov 20 '13 at 19:08
  • $\begingroup$ The field, @Vladimir: the set of all the residues modulo two. If you haven't yet studied that then, perhaps, something else was meant in your question, but nobody seems to know what, including you. $\endgroup$ – DonAntonio Nov 20 '13 at 23:59

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