If we let $v_1, …, v_n$ be vectors in a vector space, and let $w_1, …, w_m$ be linear combinations of $v_1, …, v_n$. If we suppose that $v$ is a linear combination of $w_1, …, w_m$ can I prove that $v$ is a linear combination of $v_1, …, v_n$?
As all of $w_1, …, w_m$ are linear combinations, they can each be written as $(\alpha_{1_1}v_1+…+\alpha_{n_1}v_n),…,(\alpha_{1_n}v_1+…+\alpha_{n_n}v_n)$ I think, and from this, supposing $v$ is a linear combination of them, we get that it is equal to:
$$\beta_1(\alpha_{1_1}v_1+…+\alpha_{n_1}v_n),…,\beta_n(\alpha_{1_n}v_1+…+\alpha_{n_n}v_n)=(\beta_1+\alpha_{1_1}+…+\alpha_{1_n})v_1+…+(\beta_n+\alpha_{1_n}+…+\alpha_{n_n})v_n = v$$
where all of the betas and alphas are scalar coefficients for each vector in $v_1, …, v_n$, is this sufficient proof for it being a linear combination?
The notation gets quite messy so I am not sure if there's a simpler way. Any help would be appreciated.
Best Answer
If you keep track of all the lists it gets quite messy.
Proof: $v$ is a linear combination of $w_j$, which are linear combinations of $v_i$ ($i \in \{1,\ldots, n\}$, $j \in \{1,\ldots, m\}$):
$$w_j = \sum_{i=1}^n a_{ij} v_i$$
$$v = \sum_{j=1}^m b_j w_j$$
Thus,
$$v = \sum_{j=1}^m \sum_{i=1}^n a_{ij} b_j v_i \\ = \sum_{i=1}^n \left( \sum_{j=1}^m a_{ij} b_j \right) v_i\text{,}$$
and $v$ is a linear combination of $v_i$ with coefficients $c_i = \sum_{j=1}^m a_{ij} b_j$ (same answer as Ofir's). $\square$
Observe that if $\mathbf{A} = [a_{ij}]$ and $\mathbf{b} = (b_1,\ldots, b_m)$, we can regard $c_i$ as elements of the vector $\mathbf{c} = \mathbf{A} \mathbf{b}$.