1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Orthogonality and Orthonormality: Take 2

  1. Oct 11, 2015 #1

    RJLiberator

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data

    Which of the following sets of vectors in ℂ^3 is an orthogonal set? Which is an orthonormal set? Which is an orthonormal basis?

    [tex] \begin{pmatrix}
    1/\sqrt{2}\\
    0\\
    1/\sqrt{2}
    \end{pmatrix},
    \begin{pmatrix}
    -1/\sqrt{2}\\
    0\\
    1/\sqrt{2}
    \end{pmatrix},
    \begin{pmatrix}
    0\\
    (1+i)/\sqrt{2}\\
    0
    \end{pmatrix}[/tex]

    2. Relevant equations


    3. The attempt at a solution

    Question 1:
    If I can conclude that the set is not orthogonal, then I can automatically conclude that the set is not orthonormal.
    My answer: YES.

    Question 2:
    If I can conclude that the set is orthogonal, but not orthonormal, then I can conclude that the set is not an orthonormal basis.
    My Answer: YES

    I want to get these two understandings out of the way first. These are by definition, is this correct?


    Next, My work for this particular set:

    If we let the first vector = v, second vecotr = w, third vector = z. By simple dot product algebra
    v*w=0
    v*z=0
    w*z = 0

    So we conclude that this set is orthogonal.

    Now, we check for orthonormality by taking each vector and inner product it with itself. The interesting note here is that when you take the inner produce of z and itself you must use the complex conjugate of z and z itself. so <z*|z> which equals 1, instead of <z|z> which equals -i.

    Here, in using that special element of the inner product we conclude that all 3 vectors inner producted with themselves is indeed 1 and so we see orthonormality.

    Now, for orthonormal basis, we conclude that is IS an orthonormal basis considering there is a theorem that states that any orthonormal set of N vectors in V is an orthonormal basis for V. Since this set has 3 vectors and it is in ℂ ^ 3 we conclude that this is an orthonormal basis as well.



    Is my analysis spot on? I feel like I am starting to finally get the hang of it.
     
  2. jcsd
  3. Oct 11, 2015 #2
    I'd say this is spot on.

    You nicely illustrated the philosophy behind your reasoning, invoked the right theorems.
    One thing I would add regarding using the complex conjugate is that you need this to define a well-behaved norm. (i.e. recall the definitions you used and argue how to satisfy this for a complex vector)
    And perhaps how this norm relates to cartesian norms as opposed to for example ##||z||_1 = \sum_{i=1}^3 |z_i|## (Taxicab or Manhattan norm).
     
  4. Oct 11, 2015 #3

    RJLiberator

    User Avatar
    Gold Member

    Thank you kindly for the look over and the words of further advice.
    :D
     
  5. Oct 11, 2015 #4

    ogg

    User Avatar

    Since you only provide ONE set of vectors, I assume your question is only about it. ie is it orthogonal, orthonormal and would it form a basis for C3. Since it spans the space (that is, every vector in C3 can be expressed in terms of it) it is all 3. (I actually am trusting you to have done the math right, looks right to me by inspection).
     
  6. Oct 11, 2015 #5

    RJLiberator

    User Avatar
    Gold Member

    Indeed, I do not want to bore physics forums with more sets, I wanted to analyze this one as an example of my understanding.

    Thank you ogg for your check over as well.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Orthogonality and Orthonormality: Take 2
Loading...