# Determine all vectors $\mathbb v$ such that $\alpha (\mathbb v)=0$

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?

• 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 $\alpha(v)=0.$ – mfl Mar 4 at 13:35
• So any linear combination of those two vectors should span the entirety of $ker(\alpha)$ – joseph Mar 4 at 13:38
• What is $e^1,e^2,e^3$ in contrast to $e_1,e_2,e_3$? – Dietrich Burde Mar 4 at 13:38
• $e^i:\mathbb{R}^3\to \mathbb{R}$ is given by $e^i(e_j)=\delta^i_j.$ – mfl Mar 4 at 13:40
• $e^1, e^2, e^3$ are the components of the dual basis to $e_1, e_2, e_3$ – joseph Mar 4 at 13:41

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 $$\alpha$$.
• 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 $\alpha$> – uniquesolution Mar 4 at 14:56