I've been thinking about this problem where I have a transformation matrix $M$ that rotates vectors in space 30 degrees clockwise about the z-axis, i.e.
$M = \begin{pmatrix} \cos{(-30)} & -\sin{(-30)} & 0 \\ \sin{(-30)} & \cos{(-30)} & 0 \\ 0 & 0 & 1 \end{pmatrix}$
I need to find vectors $\overrightarrow{a}$, $\overrightarrow{b}$, and $\overrightarrow{c}$ such that $\text{span}\{\overrightarrow{a},M\overrightarrow{a},M^2\overrightarrow{a}\}$ is a line, $\text{span}\{\overrightarrow{b},M\overrightarrow{b},M^2\overrightarrow{b}\}$ is a plane, and $\text\{\overrightarrow{c},M\overrightarrow{c},M^2\overrightarrow{c}\}$ is all of $\mathbb{R}^3$.
From my understanding of these questions it seems that $\overrightarrow{a}$ could be a vector that remains the same during z-axis rotation such as $\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$, and $\overrightarrow{b}$ could be a vector that satisfies $z=0$ during the rotation such as $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$.
However, I can't envision something that would encapsulate all of $\mathbb{R}^3$. Something like $\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ would make a sort of cone that doesn't end up covering all of $\mathbb{R}^3$, right?
What am I missing to tackle this last question, or am I approaching the question wrong? And also, is my thinking on track for the first two parts of the question?
Span of a set of vectors $\vec v_i$ is the set of all the vectors that can be obtained by all possible linear combinations of $\vec v_i$. If you have three vectors, you can get either a line, a plane, or the whole $\mathbb R^3$.
To obtain a line, all three have to be collinear. You've noticed that if the vector is along $\hat z$, the rotation will not change that vector. So for $\vec a$ you have $\vec a=M\vec a=M^2\vec a$.
You also noticed that if you have a vector $\vec b$ perpendicular to $\hat z$, $M\vec b$ and $M^2\vec b$ are in a plane. To prove that they indeed span a plane, you need to prove that two of them are independent, but the third one can be written as a linear combination of the first two. $\vec b$ and $M \vec b$ are not collinear (the angle between them is $30^circ$). However the third can be written as a linear combination of the first two. You can do this by hand, but to reduce the number of calculations, I will do a trick here. I will use $\vec b$ and $M^2\vec b$ as my independent vectors. They are not collinear, since there is an angle of $60^\circ$ between them. Since they have the same length, the vector sum of them is at $30^\circ$ to each of those, meaning that it's along $M\vec b$. So I can write $M\vec b\propto(\vec b+M^2\vec b)$, so it's a linear combination, and therefore not independent.
Now if you have your vector $\vec c$ along $(1,1,1)^T$ they are indeed on a cone. But now they are independent. To show this, calculate $x,y,z$ such as $x\vec c+yM\vec c+zM^2\vec c=0$. To show that this homogeneous system of equations has only the trivial solution, show that the determinant is not equal to $0$.