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Let $\alpha$ be a linear functional in $\mathbb R^3$ such that $\alpha (e_1+e_2)=0$, $\alpha (e_1+e_3)=4$, and $\alpha (e_2+e_3)=2$. I set up a system of linear equations and was able to write $\alpha$ as $$ \alpha = e^1-e^2+3e^3 $$ I then need to determine all vectors $\mathbb v$ such that $\alpha (\mathbb v)=0$. We are given the fact that any vector of the form: $$ \begin{bmatrix} t \\ t \\ 0 \end{bmatrix} $$ will equal $0$. I was also able to use the three vectors given to find that: $$ \alpha(t\begin{bmatrix} 2 \\ -1 \\ -1 \end{bmatrix})=0 $$ How do I know when I have found all such vectors though? Are these two sufficient enough?

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    $\begingroup$ You have got a basis for $\ker \alpha$ because it has dimension two. But $v=a(1,1,0)+b(2,-1,-1)$ also satisfies $\alpha(v)=0.$ $\endgroup$ – mfl Mar 4 at 13:35
  • $\begingroup$ So any linear combination of those two vectors should span the entirety of $ker(\alpha)$ $\endgroup$ – joseph Mar 4 at 13:38
  • $\begingroup$ What is $e^1,e^2,e^3$ in contrast to $e_1,e_2,e_3$? $\endgroup$ – Dietrich Burde Mar 4 at 13:38
  • $\begingroup$ $e^i:\mathbb{R}^3\to \mathbb{R}$ is given by $e^i(e_j)=\delta^i_j.$ $\endgroup$ – mfl Mar 4 at 13:40
  • $\begingroup$ $e^1, e^2, e^3$ are the components of the dual basis to $e_1, e_2, e_3$ $\endgroup$ – joseph Mar 4 at 13:41
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All you need to do is find two linearly independent vectors that are orthogonal to $\alpha$. For example $(1,-1,0)$ and $(0,1,\frac{1}{3})$. Their span is the set of all vectors annihilated by $\alpha$.

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  • $\begingroup$ aren't the two vectors I found linearly independent also? $\endgroup$ – joseph Mar 4 at 14:53
  • $\begingroup$ you found two lines which are annihiliated by $\alpha$. That's good, because those two lines are not parallel, so that you can choose from them two linearly independent vectors, which necessarily will span the kernel of $\alpha$> $\endgroup$ – uniquesolution Mar 4 at 14:56

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