I am reading now linear algebra book where it is said that for three vectors $\vec{u}, \vec{v}, \vec{w}$, $c\vec{u}+d\vec{v}+e\vec{w}$ is the linear combination that fills three-dimensional space. But if $\vec{w}$ happens to be $c\vec{u}+d\vec{v}$, then the vector $\vec{w}$ is in the plane of first two. So we do not get the full three-dimensional space.
This confuses me. Actually in my thoughts then we can't even get full two-dimensional plane. Because with $c\vec{u}$ I can get any vector just putting any value to scalar c. Then any sum $c\vec{u}+d\vec{v}$ will be actually the value of some $c\vec{u}$. I feel that I am missing something. Please clarify this and if it is possible show an example of full three-dimensional space of linear combinations.
How about the canonical base? $$ \mathbb{R}^3 = \{ x_1 (1,0,0)^T + x_2 (0,1,0)^T + x_3 (0,0,1)^T \mid x_i \in \mathbb{R} \} $$ Then $$ x = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} $$ is the vector $x$ as linear combination of the canonical base vectors.
The linear independence of a set of vectors hinges on the question, if the null vector can be combined in more than one way ($x_i = 0$ works always) out of them, thus if there is a set of $x_i$ not all of them zero. This relates to the number of solutions of the above linear system for $x=0$. It needs to have exactly one solution $(x_1, x_2, x_3)^T = (0,0,0)^T = 0$.
For the canonical basis vectors, which form the identity matrix, this is the case.