In the binary operation of $\perp$ symbol - does the order matter?

Question

I have a statement written in my textbook which is, "given some matrices $A$ and $B$, then the following is true $R(A)\perp N(B^T) \iff R(A) \subseteq N(B^T)^\perp$". This is interesting to me because on the right side of the $\iff$ the order between the two arguments matters: $R(A) \subseteq N(B^T)^\perp$ is clearly different from $N(B^T)^\perp \subseteq R(A)$. This confuses me, because on the left side of the $iff$, I thought that the order of those two arguments it does not matter: $R(A)\perp N(B^T)$.

But how can the $\iff$ be true in this case if the order matters for one, but not the other?

In the binary operation of $\perp$ symbol - does the order matter? That is to say, is $A \perp B$ equivalent to $B \perp A$?

Answer 1

The order doesn't matter. The left expression $R(A)\perp N(B^T)$ is equivalent to $N(B^T)\perp R(A)$. In the right expression, note the superscript $\perp$ on one side of the $\subseteq$;

$$R(A)\subseteq N(B^T)^\perp$$

is equivalent to

$$N(B^T)\subseteq R(A)^\perp$$

but not to

$$N(B^T)^\perp\subseteq R(A).$$

Answer 2

$A\bot B$ means that $\forall (x,y)\in A\times B:\ (x,y)=0$.

Since $(x,y)=0\leftrightarrow \overline{(y,x)}=0\leftrightarrow (y,x)=0$, and since $(x,y)\in A\times B\leftrightarrow (y,x)\in B\times A$ then, the proposition above is equivalent to $\forall (y,x)\in B\times A:\ (y,x)=0$, which is $B\bot A$.