If $x_1$ does not lie in the range space of a space, does exist an $x_2$ in null space of it? I'm working on Roger Brockett's Finite Dimesional Linear Systems book and I cannot understand some part of the proof in the book.$W(t_0,t_1)$ is Grammian Matrix
How can one say if there is no $x_1$ in the range space of W then there exist such an $x_2$ which satisfies $W(t_0,t_1)$$x_2$ and also $x_2'x_1 \neq 0$ ?
 A: For subspaces $U,V$ of $\Bbb R^n$, we have $U \subseteq V \iff V^\perp \subseteq U^\perp$.  Because $W$ is symmetric, the orthogonal complement to its range space is its null space.
We know that the span of $x_1$ is not a subset of the range space of $W$. It follows that the orthogonal complement of this range space is not a subset of the orthogonal complement of $x_1$. Thus, there exists an $x_2$ that is in the orthogonal complement of the range space of $W$ (from which it follows that $W(t_0,t_1)x_2 = 0$) but is not in the orthogonal complement of $x_1$ (from which it follows that $x_2'x_1 \neq 0$).

Another approach: given such an $x_1$, we can produce another vector $x_1^{\perp}$, which is the component of $x_1$ that is orthogonal to the range space of $W$.  If we take $x_2 = x_1^\perp$, then we find that
$$
W x_2 = (x_2' W')' = (x_2'W)' = 0.
$$
On the other hand,
$$
x_2'x_1 = (x_1^\perp)'(x_1^\perp + x_1^{||}) = \|x_1^{\perp}\|^2,
$$
where $x_1^{||} = x_1 - x_1^{\perp}$ is the projection of $x_1$ onto the range space of $W$.
