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Question goes like this:

Determine Basis of subspace U of $\mathbb{Z}_3^2$ with $$ U := \left\langle\begin{pmatrix} 2 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 2 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 17 \\ 0 \end{pmatrix}, \begin{pmatrix} 2 \\ 2 \end{pmatrix}, \begin{pmatrix} 0 \\ 2 \end{pmatrix} \right\rangle $$

I'm not 100% sure how to approach this question, but I know that $\operatorname{dim} U \leq 2$ and I tried to cancel out all the linear dependent vectors of the set.

I was then left with $\begin{pmatrix} 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 17 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 17 \\ 0 \end{pmatrix}$ is basically $\begin{pmatrix} 2 \\ 0 \end{pmatrix}$ since we are in $\mathbb{Z}_3^2$, and if I multiply that by 2 I get $\begin{pmatrix} 4 \\ 0 \end{pmatrix}$ mod 3 which is $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ which is my e1 standard basis vector.

Am I correct? Is my basis $\left\langle \begin{pmatrix} 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 17 \\ 0 \end{pmatrix}\right\rangle$ ?

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    Clearly the first and third vectors are linearly independent (they are not multiples of one another). So you get everything with those two. – 2017-02-12

1 Answers 1

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You are correct. You may solve another way, first choose a non zero element $2\choose{1}$ (say) from the set $U$. Then cancel out all scaler multiples of $2\choose{1}$. Then again choose another non zero element from the rest (if possible). Since dimension $\leq 2$. You are done.

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    Thank you. But basically there isn't just one single basis from these vectors right? There are multiple possibilities, am I correct? – 2017-02-13