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Couple of Linear Algebra Questions

  1. Oct 9, 2007 #1

    First off, a stupid question: Why does linear indipendance not imply orthogonality? i mean we define the latter such that the inner product is zero i sort of see it as a case of the chicken and the egg, are we using two things to define each other in a cyclical way? Also what extra constraints are placed on orthogonal vectors besides them being linearly independant? I mean if i have two vectors in R2 surely if they are LI they are at right angles? Or am i missing something here...

    Also could someone please explain group action with a really simply example as i totally don't get it, and also why you canonly have vector spaces over fields and not rings ( i understand the basic differences between them).

  2. jcsd
  3. Oct 9, 2007 #2

    matt grime

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    Orthogonal only makes sense if you have already defined your inner product. Thus some set of vectors can be orthogonal in some inner product, and not in others. There is no cyclicity here at all.

    As for the other kettle of fish, pick any set S, and let G be the set of permutations of S. This is a group action on S. Pick any regular n-gon, and let G be the set of symmetries of it. G acts on the n-gon.

    A vector space is _by defininition_ over a field. If you alter the definition to be over a ring the resulting object is called a module. They have many different properties.
    Last edited: Oct 9, 2007
  4. Oct 9, 2007 #3


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    In R2, for example, the vectors <1, 0> and <1, 1> are certainly independent but not orthogonal. In this very simple case, "independent" just means "not parallel". That certainly does not imply "at right angles".

    You cannot have a vector space over a ring because such things are not called "vector spaces"- they are called "modules". The reason they are given different names (which is really your question) is that they have very different properties. In particular, the lack of a multiplicative inverse cause problems with finding bases.
    Last edited by a moderator: Oct 10, 2007
  5. Oct 9, 2007 #4


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    orthogonal does NOT imply independent, non zero and orthogonal does so.
  6. Oct 9, 2007 #5
    Ah thanks guys yeh i figured out my problem with LIness, i had it wrong in my head, when i wrote it down on paper i could see i was confusing it with orthogonality but under a different name.

    Cheers for the reponses!
  7. Oct 20, 2007 #6
    could you please elaborate i dont see the connection (sorry im sure its obvious)
  8. Oct 21, 2007 #7


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    A "vector space" is defined over a field- in particular evey member of the field, except 0, has a multiplicative inverse so we can do the following:
    Suppose [itex]{v_1, v_2, \cdot\cdot\cdot, v_n}[/itex] is a DEPENDENT set of vectors in vector space V over field F. Then, by definition of "dependent", there exist a set [itex]a_1, a_2, \cdot\cdot\cdot, a_n}[/itex] of members of F, not all 0, such that [itex]a_1v_1+ \cdot\cdot\cdot+ a_nv_n= 0[/itex]. In particular, if [itex]a_k[/itex] is not 0, we can write that as [itex]a_kv_k= -a_1v_1- \cdot\cdot\cdot- a_nv_n[/itex] and so, since [itex]a_k\ne 0[/itex], [itex] v_k= -(a_1/a_k)v_1- \cdot\cdot\cdot- (a_n/a_k)v_n[/itex]. That is, [itex]v_k[/itex] can be written as a linear combination of the other vectors in the set. We can continue that until we reach a linearly independent subset such that all the other vectors in the set can be written as linear combinations of the vectors in the subset: a basis for the span of the original set of vectors.

    If, instead, we have a module (a "vector space" over a ring) we cannot, in general, "divide by" that [itex]a_k[/itex] and so we cannnot guarentee that every set of dependent vectors contains a basis for its span- or even that a module has a "basis"!
  9. Oct 21, 2007 #8
    Ah ok gotcha, thanks!
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