1
$\begingroup$

I'm more frequently coming across phrases such as "vector $b$ spanned by $\{b_1, \dots , b_n\}$" and "$A$ spans $B$" while studying linear algebra. What do the terms "spanned by" and "spans" mean in this context?

For example: Does "$A$ spans $B$" mean that span$(A)$ = $B$ (where span() is the function whose output is the span of set $A$)?

$\endgroup$
  • $\begingroup$ It means we can find linear combinations of the vectors in the spanning set to get any vector in the spanned vector space. $\endgroup$ – mathreadler Aug 25 '17 at 15:48
  • 1
    $\begingroup$ Yes, "A spans B" means span(A)=B. I don't think I've ever called an individual vector spanned by a set though - instead I'd say the vector is in the span. $\endgroup$ – anon Aug 25 '17 at 15:49
  • 2
    $\begingroup$ I guess your textbook explains such a basic concept. $\endgroup$ – Francesco Polizzi Aug 25 '17 at 15:49
  • $\begingroup$ @Francesco Polizzi You guess wrong. I'm not using a textbook at the moment. $\endgroup$ – Aluthren Aug 25 '17 at 15:54
  • 2
    $\begingroup$ @Aluthren: so you are studying from some random sources? Bad move. Anyway, that's not my business. $\endgroup$ – Francesco Polizzi Aug 25 '17 at 16:00
8
$\begingroup$

If $V$ is a vector space, and $A$ is a subset of $V$, and $W$ is a vector subspace of $V$, then the phrase "$A$ spans $W$" means that each vector in $W$ can be written as a linear combination of vectors from $A$. Stated succinctly, $A$ spans $W$ if $\operatorname{span}(A) = W$, where $$ \operatorname{span}(A) = \big\{\sum_{\text{finite}}\alpha_iv_i\bigm| \text{$\alpha_i$ is a scalar, and $v_i\in A$}\big\}. $$

You will also hear "$W$ is spanned by $A$" if $A$ spans $W$. You will not hear phrases like "The vector $b$ is spanned by vectors $b_1,\dots,b_n$," since it is vector spaces that are spanned, not individual vectors. Instead, you may hear something like "The vector $b$ lies in the span of the vectors $b_1,\dots,b_n$."

$\endgroup$
  • $\begingroup$ Thank you for the clarification. $\endgroup$ – Aluthren Aug 25 '17 at 16:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.