I'm missing something here. G. Strang's Linear Algebra worked example on subspaces:
$$
\mathbf{V}_{2}=\text { all vectors perpendicular to } u=(1,2,1), \text { so } u \cdot v=0
$$
Answer: S of $V_2$ is the line through (1, -1, 1). This line is perpendicular to u. The vector $x = (0,0,0) $ is in S and all its multiples cx give the smallest subspace SS = Z.
I'm not getting where it comes from that the line passes through (1, -1, 1)?
Best Answer
Most likely, this solution is actually guessed.
Here is an example of how to find a solution via calculation: We need to find a nonzero vector $v=(v_1,v_2,v_3)$ that is perpendiculat to $u=(1,2,1)$. That means that $v$ needs to satisfy the equation \begin{align*} 0 &= v \cdot u \\ &= v_1 + 2 v_2 + v_3. \end{align*} Note that this equation does not have a unique solution. For example, we can multiply $v$ by any nonzero number and this equation still stays true. Let's suppose that $v_1=1$. Then $v$ is automatically nonzero and we will be done if we find $v_2, v_3$ satisfying $$0=1 + 2 v_2 + v_3.$$ This is the equation of a line, so we can still pick one parameter freely. Let's choose $v_3=1$. Rearranging we get $v_2 = (-1-1)/2=-1$. Therefore $$v=(1,-1,1).$$
Note that we could have made different choices for $v_1$ and $v_3$. In that case we would have ended up with different vectors also perpendicular to $u$.
For example, if we had chosen $v_1=0$ then we would have ended up with the equation $$0=0+2v_1+v_3.$$ Picking $v_3=2$ we would have gotten a different perpendicular vector $\tilde v=(0,-1,2)$. You can check that $v$ and $\tilde v$ are linearly independent. Since $\mathbf V_2$ is two-dimensional, $\{v, \tilde v\}$ is a basis of $\mathbf V_2$.