Are #x#, #x^2#, #x^3# linearly dependent?

Question

5200 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-04T23:27:01

No. At least not normally.

In order to be linearly dependent, there would have to exist constants #a#, #b#, #c# not all zero such that:

#ax+bx^2+cx^3=0#

for all values of #x#.

If #x# assumes Real values, or values in any field of characteristic #0# then there are no such constants #a, b, c#.

To see this, consider putting #x=1#, #x=2# and #x=3#, to get the set of simultaneous equations:

#{ (a+b+c=0), (2a+4b+8c=0), (3a+9b+27c=0) :}#

This has unique solution #a=b=c=0#

Hence #ax+bx^2+cx^3=0# can only hold identically if #a=b=c=0#

#color(white)()#
Exceptions

If #x# takes values in a field of non-zero characteristic then #x#, #x^2# and #x^3# can be linearly dependent.

For example, in #GF(2)# (the field with two elements) we can put #a=b=1# and #c=0#, to find that:

#ax+bx^2+cx^3=0#

for all (two possible) values of #x#.

In addition, note that if we put #a=-1#, #b=0# and #c=1# then we have:

#ax+bx^2+cx^3 = x^3-x = x(x-1)(x+1)#

which is #0# for all elements of #GF(3)#