Solution sets for elements different elements in the span of vectors

Question

I'm trying to see if I have a grasp of the understanding for simple linear algebra concepts and terminology.

Suppose we have vectors $v_1, ... , v_n$ and vectors $b_1$ and $b_2$ are elements of the span of $v_1, ... , v_n$. Is this the same as saying that there are some weights $c_1, ... , c_n$ where $c_1v_1 + ... + c_nv_n = b_1$ as well as some weights $d_1, ... , d_n$ where $d_1v_1 + ... + d_nv_n = b_2$?

Furthermore is it safe to say that all the possible solutions (all the possible weights $c_1, ..., c_n$) to make $c_1v_1 + ... + c_nv_n = b_1$ are never equal to any possible solution to make $d_1v_1 + ... + d_nv_n = b_2$, where $b_1 \neq b_2$ (ie. no list of numbers in the solution set for $b_1$ is the same as any list of numbers in the solution set for $b_2$)?

Thanks everyone for the detailed responses!

Answer 1

If vectors $\{v_1,\cdots, v_n\}$ form a basis for $V,$ then, for all $b \in V$, there is a unique set of coefficients $\{c_1,\cdots, c_n\}$ such that $c_1v_1 + \cdots + c_nv_n = b.$

Now what you have above has not said that $\{v_1,\cdots, v_n\}$ form a basis, or that they are linearly independent.

If they are not linearly independent there could be multiple sets of coefficients such that $c_1v_1 + \cdots + c_nv_n = b.$

However, it could never be that $c_1v_1 + \cdots + c_nv_n = b_1$ and $c_1v_1 + \cdots + c_nv_n = b_2$ unless $b_1 = b_2.$

Answer 2

For your first questions (second paragraph) the answer is yes. To be in the span of a collection of vectors means, by definition, that it is a linear combination of those spanning vectors. So $b_1\in\operatorname{Span}(v_1,\ldots,v_n)$ if and only if $b_1=c_1v_1+\cdots+c_nv_n$ for scalars $c_1,\ldots,c_n$.

For your second question (third paragraph) the answer is also yes. It is clear that if $c_1=d_1,\ldots, c_n=d_n$ then we would have to have $b_1=b_2$, since the linear combinations are identical. The converse also clearly holds, so $b_1=b_2$ if and only if the possible values for each $c_i$ and $d_i$ coincide. However, if $v_1,\ldots v_n$ do not form a basis, then the $c_i$'s and $d_i$'s are not necessarily unique.

Answer 3

Let us assume we live in a $\mathbb{R}$-vector space, for the sake of simplicity.

Obvious but important remark

Please notice that there can be some $i \in \{ 1, \ldots, n \}$ where $c_{i_0} = 0$, for example, consider the span in $\mathbb{R}^2$ of $v_1 = (1, 2), v_2 = (1, 1), v_3 = (1, 3)$, then let us write $b_1 = (2, 3) = 1 \times v_1 + 1 \times v_2 + 0 \times v_3$.

Basis

Another thing, if $(v_1, \ldots, v_n)$ form a basis of your space, then it is safe to say, when you have $b_1 = c_1 v_1 + \ldots + c_n v_n$, that, if you can write $b_1 = d_1 v_1 + \ldots + d_n v_n$, then $c_i = d_i$ for all $i \in \{ 1, 2, \ldots, n \}$ (a basis enables you to write your vector in a unique way).

Can I have $b_2 = b_1$ even if $\forall i \in \{ 1, 2 \ldots, n \}, d_i \neq c_i$ ?

If you have $b_2 = d_1 v_1 + \ldots + d_n v_n$ and $b_1 = c_1 v_1 + \ldots + c_n v_n$, consider $b_2 - b_1 = (d_1 - c_1) v_1 + \ldots + (d_n - c_n) v_n$.

You can see that $b_2 - b_1 = 0$ only if all these coefficients cancel out, e.g, for the $n = 1$ case, $(d_1 - c_1) v_1 = 0$, either $d_1 = c_1$, either $v_1 = 0_E$.

In general, it's complex to say what will be null if $b_2 = b_1$, because some vectors might be null or hide other vectors which will cancel others, here is the role of a basis, to help you to say that: all $d_i = c_i$ if $b_2 = b_1$.

Span

But, if you consider a span of too many vectors, then some vectors might be redundant in the space (look at the first example in $\mathbb{R}^2$) and there might be more than one solution in the solution set of weights.

Pay attention to special cases, like if you consider the span of no vectors, you will get the famous $\{ 0_E \}$, the null space of your vector space $E$.

In this case, you have an infinite amount of solutions for weights, because: $\forall x \in \{ 0_E \}, \forall y \in \mathbb{R}, x = y x$.