I have the following problem
a) I say that the vectors are linearly independent because they are not scalar multiples of each other.
b) This one I'm struggling with. How do I find the dimension of the vector space? Is it 3 dimension due to the fact that the vectors only have 3 coordinates? Is it so simple?
c) I'm not sure what to say here. I don't think it belongs belongs to the span. If this is true, can it be explained by the same reasoning as in a)?
On
Answering each of your questions:
a) True, they are linearly independent (it's easily seen that they're not schalar multiples of each other). You can see this from noticing that in the second component, $a_1$ has $-1$ and $a_2$ has $1$, but the other components aren't the same with opposite symbols.
b) Since they are linearly independent, then the vector space generated by both vectors (or how it's written: $\langle a_a,a_2\rangle$) is of dimension $2$ (it would have been dimension $1$ if they were linearly dependent).
c) Since $a_1$ is not a schalar multiple of $a_2$ (because they are linearly independent), you conclude that $a\notin \langle a_2\rangle$.
As you see, all of your questions have to do with the same important question: are those vectors linealy independent? The answer to that is yes, and the following statements are concluded from there.
On
B) Let $W=span(a_1,a_2)$. $\dim(W)=2$. In fact the dimension of a subaspace is the cardinality of a basis. In this case, since $a_1$ and $a_2$ are generators(by definition) of the subspace, and are linearly indpendent they are a basis
C)If $\; a_1 \in span(a_2) \Rightarrow a_1=ka_2$, which is not true.
On
$a)$ For the vectors to be independent, the relationship $\lambda_{1}a_1+\lambda_{2}a_2=0 $
must yield $\lambda_1=\lambda_2=0$.
You can verify that this is the case here.
$b)$ You have two linearly independent vectors. This means that any $v \in V$, where $V$ is the vector space created by $a_1, a_2$ can be written as their linear combination. That is, $v=\kappa_1a_1+\kappa_2a_2$.
Now all you have to do is (re)check the definitions found on your course notes/book to see if it holds true.. Same for $c)$.
These questions are related to the concept of a subspace. The ambient space here is $\mathbb{R}^3$, but there are many subspaces of $\mathbb{R}^3$.
What are the subspaces of $\mathbb{R}^3$? There are two trivial spaces. The space consisting only of the vector $\langle 0, 0, 0\rangle$ is a zero-dimensional subspace and the whole of $\mathbb{R}^3$ is a three dimensional subspace of $\mathbb{R}^3$. The other subspaces are one and two dimensional.
The usual way to define a subspace is to specify a spanning set, that is, a set of vectors. If $S$ is a set of vectors, then the spanning set of $S$, usually denoted $\textrm{span}(S)$, is the set of all vectors that can be written as linear combinations of elements of $S$.
In this case, you want to know the dimension of $\textrm{span}\left(\left\{\vec{a_1}, \vec{a_2}\right\}\right)$.
It can't be zero dimensional because it contains non-zero vectors. And it can't be three dimensional (or higher) because there is a spanning set with two elements.
Could this subspace be one dimensional? It it were there would have to be a single vector $\vec{x}$ that spanned the whole subspace. But that would mean that there would be (non-zero) scalars $\lambda_1$ and $\lambda_2$ such that:
But then we could write $$\frac{1}{\lambda_1}\vec{a_1} - \frac{1}{\lambda_2}\vec{a_2}=\mathbf{0}$$ which would imply that $\vec{a_1}$ and $\vec{a_2}$ are not linearly independent, contradicting your answer to question 1.
All of this is a roundabout way to get to the conclusion that if $S$ is a finite set of linearly independent vectors, then $$\dim(\textrm{span}(S))=\vert S\vert.$$
For part 3, you already have the right idea. To show that $\vec{a_1}$ is not in $\textrm{span}(\{\vec{a_2}\})$, you just need to show that there is no $\lambda$ such that $\lambda\vec{a_2}=\vec{a_1}$, which will follow directly from the fact that $\vec{a_1}$ and $\vec{a_2}$ are linearly independent.