Nullspace of reflection

Question

Why is the nullspace of a reflection always an empty set? Is the zero vector included in an empty set? Also, why would the columnspace of a reflection be the entire space? I get the correct answers if I approach the questions algebraically, but I cannot seem to grasp why it is geometrically. I would really appreciate any help!

Answer 1

Any reflection is an involution ($\Phi$ is a reflection implies that $\Phi \circ \Phi = \operatorname{id}$), so any reflection must be bijective. Bijections do not have non-zero kernels, and must have full image, which means that the columnspace is the entire vector space.

Answer 2

The null space of a linear map $T \colon V \rightarrow V$ always contains the zero vector, and in particular, the null space of a reflection cannot be empty. However, it does not contain any other vector other than the zero vector. Geometrically, reflecting a vector across a (hyper)-plane cannot change a non-zero vector to a zero vector. Also, reflection must satisfy $T^2 = \operatorname{id}$. This means that $T(T(v)) = v$, or, in other words, applying a reflection twice results in the original vector. Thus, given a vector $w \in V$, if you want to find a vector $v \in V$ such that $T(v) = w$, you can take $v = -w$. This shows that $\operatorname{Im}(T) = V$.

More formally, given a direct sum decomposition $V = W_1 \oplus W_2$, the reflection of $V$ across $W_1$ is the linear map defined by $T(w_1 + w_2) = w_1 - w_2$. This means that a vector $w_1 \in W_1$ stays in place while a vector $w_2 \in W_2$ is mapped to its reflection $-w_2$. In particular, $T(w_1 + w_2) = 0$ if and only if $w_1 = w_2 = 0$ and so $\ker(T) = \{ 0_V \}$.

Answer 3

Reflection preserves length, so no non-zero vector can reflect to a vector of zero length.