I am in a University Linear Algebra course and am confused by the term span and its relation to both matrices and vectors. Can someone help clarify what they mean?


  1. Can it only be made of vectors (1 row matrix)? (or can you "solve a span" of large matrices? $3\times 5, 4\times 9$, etc.)

  2. In reduced echelon form will each vector/matrix act the same inside that span? (as an example, my professor says suppose matrix $A$ in $\mathbb{R}^4$ has the property that $Ax = b$ gives one unique solution, but how can each $B$-value act the same? Esp. if you can create inconsistent solutions.

  3. With $\mathbb{R}^3$ or $\mathbb{R}^n$ does that mean this is just a concept of a span containing ALL vectors with $N$ columns(or is it rows?).

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    $\begingroup$ It's hard to understand what you're asking about. You'll get better results if you ask about a particular well-stated problem that you don't understand or don't know how to solve. $\endgroup$ Jan 30 '15 at 17:40
  • $\begingroup$ Instead of concentrating on the word "span", try writing down the definition of the span of a set of vectors, with some examples that have been shown you. Do you understand what the definition says? Do you understand why the examples turn out the way they do? $\endgroup$ Jan 30 '15 at 17:45
  • $\begingroup$ Just to make sure I am correct. the Definition i have is that the span is a set of all linear combinations of a list of vectors. that makes sense, but can there be cases where vectors must be solved? and then in turn, that would give the span some property even though there are infinite combos of that vector? or is that totally unrelated to Ax=b form? $\endgroup$ Jan 30 '15 at 17:51
  • $\begingroup$ What do you mean by "cases where vectors must be solved"? $\endgroup$ Jan 30 '15 at 18:25
  • $\begingroup$ @Omnomnomnom: I think that the OP is referring to echelon row reduction, or something of that sort. $\endgroup$
    – Tom Au
    Jan 30 '15 at 21:44

If you have an n-space, you need at least n vectors to "fill" that n-space. And that's true only if the vectors are linearly independent.

(If any two vectors are linearly dependent, they "collapse" into one, and you have only n-1 vectors that don't quite fill the n-space.) In row reduced echelon form, that means you have one (or more) rows will all zeroes.

If you have the n independent (e.g. column) vectors, they can be put into a matrix with dimension n (by definition).

  • $\begingroup$ so to be n independent they must all have a pivot? (each row). otherwise its missing a value and makes unable to fill the space? $\endgroup$ Jan 31 '15 at 20:47
  • $\begingroup$ @Maximus12793: Yes. $\endgroup$
    – Tom Au
    Jan 31 '15 at 21:34

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