System of vectors, linear dependence

[nuuskur]

Homework Statement

Prove that if in a system of vectors: $S_a =\{a_1, a_2, ..., a_n\}$ every vector $a_i$ is a linear combination of a system of vectors: $S_b = \{b_1, b_2, ..., b_m\}$, then $\mathrm{span}(S_a)\subseteq \mathrm{span}(S_b)$

Homework Equations

The Attempt at a Solution

We know due to $a_j$ being a linear combination, that every $a_j\in S_a = \sum\limits_{j=1}^m c_j\cdot b_j$ where $b_j\in S_b, c_j\in\mathbb{R}\setminus\{0\}$
But where should I go from here? Suggestions?

 

[Orodruin]

I suggest taking a vector in $span(S_a)$ and show that it is necessarily in $span(S_b)$.

 

[Fredrik]

The Attempt at a Solution

We know due to $a_j$ being a linear combination, that every $a_j\in S_a = \sum\limits_{j=1}^m c_j\cdot b_j$ where $b_j\in S_b, c_j\in\mathbb{R}\setminus\{0\}$

The $c_j$ can be zero. The span of a set S is the set of all linear combinations of elements of S, including linear combinations where one or more (maybe all) of the coefficients are zero.

I would do what Orodruin said, and avoid notations like

every $a_j\in S_a = \sum\limits_{j=1}^m c_j\cdot b_j$ where...​

It's $a_j$ that's equal to a linear combination, not $S_a$. Oddly enough, the phrase

every $a_j\in S_a$ is equal to $\sum\limits_{j=1}^m c_j\cdot b_j$ where...​

would be considered acceptable.

 

[nuuskur]

Alright. Let's denote the systems:
$A = \{a_1, a_2, ..., a_n\}\\B = \{b_1, b_2, ..., b_m\}$
Let's denote the linear span of a system $L(A), L(B)$. Then the respective linear spans would be:
$L(A) = \left\{a\ |\ a = \sum\limits_{k=1}^n \lambda _k\cdot a_k, \lambda _k\in\mathbb{R}, a_k\in A \right\}\\<br /> L(B) = \left\{b\ |\ b = \sum\limits_{k=1}^m \lambda _k\cdot b_k, \lambda _k\in\mathbb{R}, b_k\in B \right\}$
We know that every vector $a\in A$ is a linear combination of the vectors in system $B$, that is:
$a = \sum\limits_{k=1}^m\lambda _k\cdot b_k$ where $\lambda _k\in\mathbb{R}, b_k\in B$.
Considering that a linear span is a vector space, then it is closed under multiplication with a scalar. Therefore, every $a\in L(A)$ implies $a\in L(B)\Leftrightarrow L(A)\subseteq L(B)_{\square}$

 

[Fredrik]

Alright. Let's denote the systems:
$A = \{a_1, a_2, ..., a_n\}\\B = \{b_1, b_2, ..., b_m\}$
Let's denote the linear span of a system $L(A), L(B)$. Then the respective linear spans would be:
$L(A) = \left\{a\ |\ a = \sum\limits_{k=1}^n \lambda _k\cdot a_k, \lambda _k\in\mathbb{R}, a_k\in A \right\}\\<br /> L(B) = \left\{b\ |\ b = \sum\limits_{k=1}^m \lambda _k\cdot b_k, \lambda _k\in\mathbb{R}, b_k\in B \right\}$
We know that every vector $a\in A$ is a linear combination of the vectors in system $B$, that is:
$a = \sum\limits_{k=1}^m\lambda _k\cdot b_k$ where $\lambda _k\in\mathbb{R}, b_k\in B$.

Writing this down is a good start, but I don't follow your argument here:

Considering that a linear span is a vector space, then it is closed under multiplication with a scalar. Therefore, every $a\in L(A)$ implies $a\in L(B)\Leftrightarrow L(A)\subseteq L(B)_{\square}$

I would just start with a simple statement like "Let $x\in L(A)$." Then you can use the definition of $L(A)$ to say something about $x$. This statement will involve the $a_k$. Then you can use what you know about the $a_k$ to say something else. And so on. At some point you should be able to conclude that $x\in L(B)$. Then you will have proved that $L(A)\subseteq L(B)$.

Be careful with your statements. The quoted statement above is saying that every vector in the subspace $L(A)$ implies some statement. Statements are implied by other statements, not by vectors.