Let $A_1$, $A_2$ be vectors in $R^{n}$. Show that the set of all vectors $B$ in $R^{n}$ such that $B$ is perpendicular both $A^{1}$ and $A^{2}$ is a subspace.
This was one of the tasks in S.L's linear algebra text book. I've constructed such proof:
Let $W=B$.
Notice that $O^TA_1=0=O^TA_2$ (where $O$ is a null vector), thus $O$ is in W.
Suppose $B$, $C$ are orthogonal to $A_1$ and $A_2$, then:
$$A_1^{T}(B+C)=A_1^{T}B+A_1^{T}C=0=A_2^{T}(B+C)=A_2^{T}B+A_2^{T}C$$
thus it's additive linear combination which is again in $W$.
Finally, suppose there is scalar c, then $(cB) \cdot A=c(A\cdot B)=0$.
Hence, If I'm correct, $B$ satisfies all three axioms to be a subspace.
Does this proof have any flaws? Also isn't $B$ also a kernel (null space) of $A_{1} \cup A_2$?
The main confusion to me seems to be what is $B$? is it a vector or a subspace?
Perhaps define $$W=\{ b \in \mathbb{R}^n: A_1^Tb=A_2^Tb=0 \}$$
Also, for the closure under the addition, I would write
let $b, c \in W$, then for $i\in \{1,2\}$,
$$A_i^T(b+c)=A_i^Tb+A_i^Tc=0+0=0.$$ just to highlight that $A_i^Tb=0$ and $A_i^Tc=0$.
$W$ is the nullspace of $$\begin{bmatrix} A_1^T \\ A_2^T \end{bmatrix}$$
I do not comprehend the meaning of the union of two vectors, $A_1 \cup A_2$.