What is the difference between orthogonal subspaces and orthogonal complements? I am learning linear algebra through professor Giblert Strang's lectures on MIT OCW.
The professor says that the row space and the null space of a matrix are orthogonal subspaces. This I can follow, since any vector in the nullspace takes any linear combination of the rows to zero.
He then says that the row space and null space are more than just orthogonal subspaces, they are orthogonal complements, Because The "nullspace contains all the vectors that are perpendicular to the row space", and vice versa.
Consider:
$$A= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}=0 $$
A is rank 2, Dimension of nullspace of A=1
Null space:
$$X=c \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$$
Graphically, the row space of A is the x-y plane, and the null space is the z-axis.
It's easy to see that the nullspace does not contain all the vectors that are perpendicular to a vector in the row space. If we look at $[1 ~0~ 0]$, it has the entire y-z plane perpendicular to it, not just the z-axis.
All I can see is two orthogonal subspaces. I do not understand what additional property has earned these subspaces the term orthogonal complements. 

Refer:
Lecture Video (from 31:16 to 33:00)
Lecture Notes Page 2 paragraph 1 

There are two conceptual issues here:
(1) The orthogonal complement $X^{\perp}$ of a vector subspace $X \leq \Bbb V$ consists of the vectors in $\Bbb V$ that are perpendicular to all of the vectors in $X$, not just at least one vector.
In practice, it is convenient to use the characterization that $v \in \Bbb V$ is orthogonal to the subspace $X \leq \Bbb V$ iff for any (equivalently every) basis $(E_a)$ of $X$ we have $v \perp E_a$ for all basis elements $E_a$. Using this characterization, it is straightforward to check that the row space and null space are orthogonal complements.
(2) Given a vector space $\Bbb V$, the vector subspaces $X, Y \leq \Bbb V$ are complementary iff (a) $X$ and $Y$ are transverse, that is, $X \cap Y = \{ 0 \}$ and (b) $X$ and $Y$ together span $\Bbb V$, that is, $X + Y = \Bbb V$. Given the first condition, the second condition is equivalent to $X \oplus Y = \Bbb V$.
Given an inner product space $(\Bbb V, \,\cdot\,)$, two vector spaces $X, Y \leq \Bbb V$ of are orthogonal (we denote $X \perp Y$) iff every element of $X$ is orthogonal to every element of $Y$.
If $X$ and $Y$ are orthogonal and complementary, they are orthogonal complements
