I have a question here i can not explain my answer if it is right i do not sure
Let S = {A = (a1, a2) , B = (a2, b2)} be a spanning set for R2 and some element x = (a, b) ∈ R2 We have x = c1 A + c2 B & x = c3 A + c4 B "x has more than one representation" such that ( c1 ≠ c3 & c2 ≠ c4 )
Does S form a Basis for R2 ??
My Solution is no because the given of c1 ≠ c3 and c2 ≠ c4 make the S linearly dependent such that x must be equal to zero and the scalers are not equal so it has one condition of basis that it has A, B span but not linear independent where set to be basis it must have span and linearly independent when I search for equivalent solutions I found that any span equal to the dimension of vector space is linearly independent SO ... I don't know the right answer Any help?
You can try to prove this much more general result(well a theorem rather) that given a set of $n$ vectors $\{v_{1},v_{2},...v_{n}\}$ . They form a basis for $V$ if and only if every vector $v$ in $V$ can be uniquely expressed as a linear combination of $v_{1},v_{2},...v_{n}$.
That is for any $v\in V$. If $v=\sum_{i=1}^{n}c_{i}v_{i}=\sum_{i=1}^{n}d_{i}v_{i}$.
Then $c_{i}=d_{i}\,,\forall i=1,2,..n$.
Now it is easy to see that $\{v_{1},v_{2},...v_{n}\}$ generate the space, i.e $\text{span}(\{v_{1},v_{2},...v_{n}\})=V$.
and $0=\sum_{i=1}^{n}0\cdot v_{i}$. Thus if $\sum_{i=1}^{n}c_{i}v_{i}=0$ then $c_{i}=0\,,\forall i=1,2,...n$ by uniqueness assumption. Hence $\{v_{1},v_{2},...v_{n}\}$ is a linearly independent spanning set and hence a basis.
Conversely if $\{v_{1},v_{2},...v_{n}\}$ is a basis then the representation is unique by Linear Independence . That is if $\sum_{i=1}^{n}c_{i}v_{i}=\sum_{i=1}^{n}d_{i}v_{i}$ then $c_{i}-d_{i}=0$.