Let $S=\{(1, 2, 1), (2, \alpha, 2)\}$, where $\alpha \in \mathbb{R}$, and let $V = span(S)$. For every $\alpha \in \mathbb{R}$ I need to find a complement of $V$ in $\mathbb{R}^3$ and also provide a basis.
We have not covered orthogonal complements yet, so I am expected to find a complement based on the direct sum $V=V_1\bigoplus V_2$, where $V_1$ is a complement of $V_2$ and vice versa.
Specific to the above question, I know that I have two conditions with the direct sum to come up with a complement $V_2$:
- $V+V_2=\mathbb{R}^3$
- $V\cap V_2 = \{0\}$
I also know that I will have two different $V$ values depending on $\alpha$:
- $V = \{(1, 2, 1), (2, \alpha, 2)\}$ when $\alpha \neq 4$
- $V = \{(1, 2, 1)\}$ when $\alpha = 4$
The part where I am stuck is finding a vector $V_2$ such that the first and second conditions of the direct sum are fulfilled. I did find the vector $(1, 0, 0)$ that is linearily independent to $V$ (for any $\alpha$), but I don't know which condition this would fulfill, and what I would need to do for the other condition.
Best Answer
As you already mentioned you need to distinguish two cases.
Case $\alpha \neq 4$:
A complement in this case is given by $span((1,0,0))$ because $span((1,0,0),(1,2,1),(2,\alpha,2))$ has dimension three and is therefore equal to $\mathbb{R}^3$ and $span((1,2,1),(2,\alpha,2)) \cap span((1,0,0)) =\{0\} $ (both due to the linear independency).
Case $\alpha = 4$:
In this case $(1,2,1),(2,\alpha,2)$ are multiples of each other and a basis of the complement needs to contain $2$ vectors, for example $span((1,0,0),(0,1,0))$.