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A little info regarding this field:

Addition and multiplication in $Z^n_p$ behave as usual but with the remainder taken upon division by $p$.

Ex: $Z_3$ will only consist of the three integers {$0,1,2$}.

  • $2+1=0$
  • $2+2=1$
  • $2*2=1$
  • $Z_3^3 =\begin{Bmatrix} (0,0,0),(1,0,0),...(2,2,2)\end{Bmatrix}$
  • $dim(Z_3^3)=3^3=27$

What I'm unure of is the standard basis associated with this field.

Contextually, I'm trying to find the image of a transformation: $T:Z_3^3\rightarrow Z^2_3$

Defined by $T(x)=Ax$ where $A=\begin{pmatrix}1 & 0 & 2\\ 1 & 2 & 1\end{pmatrix}$

From that I know that I can find the image by finding the largest linearly independent subset of T applied to each element of the standard basis and that will be the basis for $im(T)$.

Thanks in advance!

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There is a field with 27 elements, but you haven't described it here. Do you mean the ring $\mathbb Z_3^3$? Or do you really mean the finite field $GF(27)$? If you are not sure, then tell us how (or whether) you propose to multiply two elements of your structure. –  TonyK Feb 2 '13 at 18:27

2 Answers 2

It's a little unclear what you're asking -- a fiels does not itself (in a nontrivial way) have a "basis" or a "standard basis".

If $F$ is any field, the standard basis of the $F$-vector space $F^n$ is always the set $$ \{e_1=(1,0,0,\ldots,0),e_2=(0,1,0,\ldots,0),\ldots,e_n=(0,0,\ldots,0,1)\}$$


As for your underlying question, "find the largest linearly independent subset of T applied to each element of the standard basis" just means "consider the space spanned by the columns of $A$". Because the codomain of $T$ is two-dimensional, and $A$ has at least two columns that are linearly independent (which because there are only two just means that neither is a multiple of the other other), the image of $T$ is the entire $F^2$.

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Your claim that $\dim(\Bbb Z_3^3)=27$ is inaccurate. There are $27$ elements in $\Bbb Z_3^3$, but the dimension of $\Bbb Z_3^3$ as a vector space (not a field) over $\Bbb Z_3$ is only $3$. The standard basis is simply $$\bigl\{(1,0,0),(0,1,0),(0,0,1)\bigr\},$$ and you should be able to see that all the elements of $\Bbb Z_3^3$ can be uniquely obtained as a linear combination of these elements with $\Bbb Z_3$ coefficients.


The image of $T$ will be the subspace of $\Bbb Z_3^2$ spanned by the vectors $T(1,0,0),T(0,1,0),T(0,0,1)$. Do you know how to determine this? Since $\dim(\Bbb Z_3^2)=2$, you need to be aware that if any two of those three vectors are linearly independent, then the image will be all of $\Bbb Z_3^2$.

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