Re: Maximum number of linearly independent vectors

In article
<62dff4d4-4cbb-4618-8967-ef3bcd4fd52a@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Tonico <Tonicopm@xxxxxxxxx> wrote:

On Sep 3, 10:15 pm, Mc Lauren Series <mclaurenser...@xxxxxxxxx> wrote:
On Sep 4, 12:06 am, Robert Israel





<isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
Mc Lauren Series <mclaurenser...@xxxxxxxxx> writes:

In a set of vectors, there can be a maximum of two 2-D vectors which
are linearly independent. Any set of three or more 2-D vectors are
linearly dependent.

Similalry, there can be a maximum of three 3-D vectors which are
linearly independent. Can this be generalized for N-D vectors that
there can be a maximum of N N-D vectors which are linearly
independent?

Look up the definition of "dimension".
--
Robert Israel              isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics        http://www.math.ubc.ca/~israel
University of British Columbia            Vancouver, BC, Canada

Here, I am using it in the sense of components. By 2-D vectors, I mean
vectors having two components. Sorry for the confusion. What is the
answer to my question?-


The answer is yes, as long as you're talking of vectors with
components from a field, like the reals of complex.
The appropiate context though is what Robert I. already told you: look
up "dimension of vector space", in linear algebra books (or in google,
of course).

Tonio

With all due respect, it's not quite as easy as looking up
the definition of dimension. A basis for R^n (where R^n is
is what OP would refer to as the set of all n-D vectors) is
a linearly independent set of vectors that spans R^n, and
there's a theorem that says that every basis of R^n has
exactly n elements, and the dimension of a vector space is
defined to be the number of elements in a basis. So
every basis for, say, R^7 has 7 elements; how does that imply
that any set of 8 vectors in R^7 is linearly dependent? Well,
Tonio, you know, and Robert knows, and I know (I just taught it
to a class last week, as it happens), but it does take a little work
to prove it.

--
Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)
.

Relevant Pages

  • Re: example of module...
    ... I know an example with the ring being Z x Z (where Z is the ring of ... First is a basis, so all other bases will have dimension 1 by ... is a linear combination equal to 0 with nonzero coefficients; ...
    (sci.math)
  • Re: row operations leave column rank untouched?
    ... they remain so after the row operations. ... possibility that some previously linearly independent columns become ... Instead we prove that any linearly dependent columns after some row ... the same linear relationships among columns always hold before and ...
    (sci.math)
  • Re: Some help vector spaces
    ... Now the vectors that span U are linearly dependent since ... where denotes span. ... Similarly for V is a linear combination of the other ... I'm trying to find a basis for U + V. ...
    (sci.math)
  • Re: basis vector problem
    ... >hence it is linearly independent. ... >vectors and give an argument as to why my choice of basis is correct? ... set can be linearly DEpendent would be if r is a linear ... , so r is a linear combination ...
    (sci.math)
  • Re: linear independence of identity matrix vectors
    ... >>>gotten using linear combination of the rest. ... span do not have to be linearly independent. ... This explicitly shows a linear dependence of the vector ... on the standard basis vectors. ...
    (sci.math)