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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.

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    I would expect $u,v,w$ to be taken linearly independent. – 2017-02-10

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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.

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    Great! Thank you! So in my case when $x_1\vec{u}+x_2\vec{v}$ is $x_1\vec{u}$ for some $x_1$ then they are just not linearly independent. Is it right? – 2017-02-10
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    Yes, as you can combine $w$ from $u$ and $v$ you have $x_1 u + x_2 v - 1 \cdot w = 0$ with $(x_1, x_2, -1) \ne 0$. You would also have $\det(u,v,w) = 0$ either from algebraic properties of $\det$ or geometrically that all three vectors could not span more than a plane which has zero volume. – 2017-02-10
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    I am very interested, then which space, I mean which dimensions are defined by vectors $(5, 1, 0), (1, \frac{1}{5}, 0), (0, 0, 1)$? Because first and second vectors are dependent. But adding them together gives us a vector with all three cordinates defined. So can we say that this basis fully defines 3d even with two dependent vectors in the basis? Or is it imortant to have a basis of N independent vectors in order to define N-dimensional space? – 2017-03-21
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    Those vectors will span the plane $(5,1,0) c_1 + (0,0,1) c_2$ with $c_1, c_2 \in \mathbb{R}$, a two dimensional subspace of $\mathbb{R}^3$. Indeed to span a space of dimension $N$ you need $N$ linear independent vectors. – 2017-03-22