According to my solution of below question is as $v_1$ and $v_2$ are not combination of each other that is why they are linear independent and they are span of $\mathbb{R}^3$. Is it correct? can anyone give me proper and detail solution? I am just self-learning linear algebra.
Thanks in advance.
On
Let's settle some of the confusion.
We say a set of vectors $v_1 \dots v_k$ are linearly independent whenever
$$a_1v_1+\dots+a_kv_k = 0 \quad \implies \quad a_1,\dots,a_k = 0$$
Another way of saying the same thing, for your case, is that if
$$a_1v_1 + a_2v_2 = 0 \quad \Leftrightarrow \quad a_1v_1 = -a_2v_2$$
then the only solution is $a_1 = a_2 = 0$, this is probably what you mean when you write that they are not a combination of each other. A formal proof would require proving the defining implication.
Every set of vectors span (as a verb) their span (as a noun). A basis for a space $V$ is a linearly independent set of vectors that span $V$. So prove that your vectors are linearly independent to conclude that they are a basis for their span. But not for the whole of $\mathbb{R}^3$. See if you can find a vector $v \in \mathbb{R}^3$ which is not in $\textbf{span}\{v_1,v_2\}$.
We say $u \in \textbf{span}\{v_1,\dots, v_k\} $ precisely when we can find $a_1,\dots,a_k$ (in your chosen field, this case $\mathbb{R}$) such that $$u = a_1v_1+\dots + a_kv_k $$ So $v_1 \in \textbf{span}\{v_1,v_2\}$ because I can find $a_1 = 1, a_2 = 0$ so that $$ v_1 = 1\cdot v_1 + 0\cdot v_2$$ For a general vector $v \in \mathbb{R}^3$, $v \in \textbf{span}\{v_1,v_2\}$ precisely when the equation $$v = a_1v_1 + a_2v_2 $$ has solutions $a_1, a_2$. Try this! Take vector, write up the above equation, and solve it $a_1,a_2$. You will fail precisely when $v \not\in \textbf{span}\{v_1,v_2\}$.
Given a set $S$ of vectors, they form a basis of $\operatorname{span}S$ when and only when they are linearly independent. So, yes, since your vectors are linearly independent, they do form a basis of the subspace of $\mathbb{R}^3$ that they span.