Spanning and lineary dependency for finding out if W forms a basis of R^n

Question

I've come across a point of confusion on the concept of 'span' and 'lineary independency' in finding out whether a set of vectors form the basis of $R^n$.

So for $R^n$ and some $W$ where $\begin{bmatrix}x\\y\\z\end{bmatrix}$ = $c_1\begin{bmatrix}1\\1\\0\end{bmatrix}$ + $c_2\begin{bmatrix}0\\1\\1\end{bmatrix}$ + $c_3\begin{bmatrix}1\\0\\1\end{bmatrix}$, you would check if there exists a non-trivial or trivial solution for the system, and if there is a trivial solution, then $W$ would not span $R^n$, thus failing to meet the requirement of forming the basis of $R^n$. But then, if the set of vectors are linearly independent, then it forms a basis of $R^n$. How is this possible? A system with a trivial solution would mean the vectors are linearly independent and that W does not span $R^n$ (as far as I understood). How is it that two contradictory statements can be true when considering whether a system of vectors forms the basis of $R^n$?

Answer 1

I think you're mixing up a couple of different claims. If you have non-trivial solutions to a homogeneous system, then that shows lack of linear independence. If a set of $3$ vectors in $\Bbb R^3$ fail to be linearly independent, then they don't span $R^3$.

When you have $3$ vectors in $\Bbb R^3$ (or generally $n$ vectors in $\Bbb R^n$), the following is true: if a linear system written using the vectors has multiple solutions for a given target vector, then there is some dependence relation, which makes the set not independent, so it doesn't span its space.

We're depending here on the fact that the number of vectors in the set is the same as the dimension of the space. You might find it easier to understand the various claims thinking about $2$ vectors in $\Bbb R^2$, at least for some claims.

Answer 2

A system of vectors $\vec v_i$ are defined linearly indepedent if

$$\sum c_i \vec v_i=0 \iff c_i=0$$

then if the linear system with $A=[v_1\, v_2\,...\,v_n]$

$$A\vec c=0$$

has only the trivial solution*, that is $\vec c=\vec 0$, then the vectors $\vec v_i\in \mathbb{R}^n$ are linearly independent and if they are $n$ they are a basis for $\mathbb{R}^n$.

Answer 3

Two things can go wrong:

• for some $(x,y,z)$, you get no solutions (i.e. the vectors do not span),
• for some $(x,y,z)$, you get more than one solution (i.e. the vectors are not linearly independent).

These problems can coexist: you can have either or both of them happening simultaneously. If neither happens, then you are in this situation:

• you get exactly one solution for every $(x,y,z)$ (i.e. the vectors form a basis).

The special case $(x,y,z) = (0,0,0)$ is enough to check linear independence, but it won't tell you anything about spanning.