0
$\begingroup$

I was given a set of vectors containing $\{ [111], [-3 -3 -3],[555]\}$ I'm not sure if memory serves me right, but aren't the $2$ other vectors scalar multiples of $[111]$? Wouldn't that mean the dimension of the basis is $1$?

The correction states the dimension is $3$, and I am currently confused.

Thanks

$\endgroup$
  • 1
    $\begingroup$ Your language doesn't make much sense: we don't speak of the dimension of a basis but of the dimension of a vector space. If the vector space is spanned by a finite number of vectors, the dimension of the vector space is defined to be the number of vectors in any basis of the space. In other words, the dimension of a (finite-dimensional) space is the number (cardinality) of vectors in a basis for the space. $\endgroup$ – symplectomorphic Feb 29 '16 at 22:32
  • $\begingroup$ Basically I'm trying to imply that: I find the basis of the set, and the dimension is simply the number of vectors in that basis. $\endgroup$ – AAS.N Feb 29 '16 at 23:30
  • $\begingroup$ Yes, you have the right idea. But you didn't tell us what your "set" (presumably you mean vector space) is. You've only told us it contains certain vectors, all of which lie in the span of $(1,1,1)$. But this leaves open the possibility that your set contains vectors not on that line. So it very well could be the case that your set has dimension 3. You need to clarify exactly what space you're dealing with. The dimension will then be the number of vectors in the basis of that space. $\endgroup$ – symplectomorphic Mar 1 '16 at 0:37
0
$\begingroup$

Yes, the dimension of the basis is 1. Any vector of these 3 forms a basis.

$\endgroup$
  • $\begingroup$ This is not necessarily true. The OP says his space contains the three vectors. This leaves open the possibility that there may be other vectors in the space that are not in the span of $(1,1,1)$. $\endgroup$ – symplectomorphic Feb 29 '16 at 23:51

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.