Is the null space (line) of any linear transformation $T:\Bbb R^3\to\Bbb R^3$ perpendicular to the range (plane)? If the null space of a linear transformation $T:\Bbb R^3\to \Bbb R^3$ is a line and the range is a plane, would the line and plane always be perpendicular?
We know that the null space would be a line full of vectors whereas the range would be a plane full of vectors. But would the line and plane always be perpendicular? I'm not sure. 
 A: No. As an example, I can take a linear transformation which sends the $xy$-plane to the $yz$-plane, and sends any $z$ vector to zero. Then my null space is the $z$-axis yet my range is the $yz$-plane. In terms of a basis $e_1, e_2, e_3$, this linear transformation might send $e_1 \mapsto e_2, e_2 \mapsto e_3, e_3 \mapsto 0$.
A: The null space of the adjoint of $T$ is perpendicular to the image of $T$.
In general, if $T\colon V\to V$ is a linear transformation and $V$ is a finite dimensional inner product space, the adjoint of $T$ is the unique linear transformation $T^*$ such that, for every $u,v\in V$,
$$
\langle T^*u,v\rangle=\langle u,Tv\rangle
$$
If $v\in V$ and $u\in\ker T^*$, then
$$
\langle u,Tv\rangle=\langle T^*u,v\rangle=\langle 0,v\rangle=0
$$
Thus $u$ is perpendicular to every vector in the image of $T$. 
Conversely, if $\langle u,Tv\rangle=0$ for every $v\in V$, we have
$$
\langle T^*u,v\rangle=\langle u,Tv\rangle=0
$$
for every $v\in V$. In particular, $\langle T^*u,T^*u\rangle=0$, so $T^*u=0$ and therefore $u\in\ker T^*$. Thus the null space of $T^*$ contains precisely all the vectors which are perpendicular to the image of $T$.
There's no reason for $\ker T$ to be equal to $\ker T^*$.

You can easily show the existence of $T^*$ by using an orthonormal basis. If $\{e_1,\dots,e_n\}$ is such a basis of $V$, then we have, for $v\in V$,
$$
v=\sum_{i=1}^n\langle v,e_i\rangle e_i
$$
so we need, assuming $T^*$ exists,
$$
T^*u=\sum_{i=1}^n\langle T^*u,e_i\rangle e_i=
\sum_{i=1}^n\langle u,Te_i\rangle e_i
$$
and it's easy to see that this defines the required linear transformation.
In case $V=\mathbb{R}^n$ with the standard inner product, if we consider the matrix $A$ of $T$ with respect to the canonical basis, then the matrix of $T^*$ is the transpose of $A$.
