Results 1 to 10 of 10

Math Help - Vector orthogonality, orthonormal basis and subspace

  1. #1
    Senior Member x3bnm's Avatar
    Joined
    Nov 2009
    Posts
    300
    Thanks
    16

    Vector orthogonality, orthonormal basis and subspace

    If a vector v is orthogonal to orthonormal basis of a vector space W then why is that v is orthogonal to every vector in W?

    Is there any theory for this? Is this true? I can't remember any proof of this in my Linear Algebra book.(maybe I forgot)

    Can anyone kindly help me figure out the proof of this?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member x3bnm's Avatar
    Joined
    Nov 2009
    Posts
    300
    Thanks
    16

    Re: Vector orthogonality, orthonormal basis and subspace

    Don't worry I found the solution. For those who are interested please follow on.

    We have to show that any vector that is orthogonal to orthonormal basis of a subspace is also orthogonal to any vector in that subspace.

    Suppose orthonormal basis of W is
    r_a =& \{r_{a1}, r_{a2},...,r_{an}\}, r_b=\{r_{b1},r_{b2},...,r_{bn}\},...,r_n    = \{r_{n1}, r_{n2},..., r_{nn}\}




    Also Suppose m is a vector and m = \{m_1,m_2, ...,m_n\} and is orthogonal to each orthonormal basis of subspace W so

    m.r_a = m_1.r_{a1} + m_2.r_{a2} + ...+m_n.r_{an} = 0

    m.r_b = m_1.r_{b1} + m_2.r_{b2} + ...+m_n.r_{bn} = 0

     m.r_n = m_1.r_{n1} + m_2.r_{n2} + ...+m_n.r_{nn} = 0



    Let v be a vector in W so v \in W so

    \begin{align*}v =& c_a.r_a + c_b.r_b + ... + c_n.r_n   \\=& c_a(r_{a1}, r_{a2},...,r_{an}) + c_b(r_{b1},r_{b2},...,r_{bn})+...+ c_n(r_{n1}, r_{n2},..., r_{nn}) \end{align*}

    as a vector in a space can be written as linear combination of it's basis.




    It is necessary to show that m.v = 0



    \begin{align*}m.v =& \Big[m_1.(c_a.r_{a1}) + m_2.(c_a.r_{a2}) + ... + m_n.(c_a.r_{an})\Big] + \Big[m_1.(c_b.r_{b1}) + \\& m_2.(c_b.r_{b2}) + ... + m_n.(c_b.r_{bn})\Big] + ...+ \\ &\Big [m_1.(c_n.r_{n1}) + m_2.(c_n.r_{n2}) + ... + m_n.(c_n.r_{nn})\Big] \\ =& c_a(m_1.r_{a1} + m_2.r_{a2} + ... + m_n.r_{an}) +  c_b(m_1.r_{b1} + m_2.r_{b2} + ... + m_n.r_{bn}) + ... + \\& c_n(m_1.r_{n1} + m2.r_{n2} + ... + m_n.r_{nn}) \\ =& 0 + 0 + ... + 0 \\ =& 0 \end{align*}

    As was to be shown.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,317
    Thanks
    697

    Re: Vector orthogonality, orthonormal basis and subspace

    a (real) inner product is linear in each of its arguments....the complex version of this theorem is only slightly harder to prove (because the conjugate of 0 is still 0).
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member x3bnm's Avatar
    Joined
    Nov 2009
    Posts
    300
    Thanks
    16

    Re: Vector orthogonality, orthonormal basis and subspace

    Quote Originally Posted by Deveno View Post
    a (real) inner product is linear in each of its arguments....the complex version of this theorem is only slightly harder to prove (because the conjugate of 0 is still 0).
    It is interesting to know that there are two different methods for same theorem.

    Is it possible to give the proof or point me to a direction where i can get the proof for complex version? I really like to understand this theorem.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member x3bnm's Avatar
    Joined
    Nov 2009
    Posts
    300
    Thanks
    16

    Re: Vector orthogonality, orthonormal basis and subspace

    Quote Originally Posted by Deveno View Post
    a (real) inner product is linear in each of its arguments....the complex version of this theorem is only slightly harder to prove (because the conjugate of 0 is still 0).

    Deveno did you mean this?


    Suppose orthonormal basis of W is
    r_a = \{a_{a1} + b_{a1} i, a_{a2} + b_{a2} i, ...., a_{an} + b_{an} i\}

    r_b=\{a_{b1} + b_{b1} i, a_{b2} + b_{b2} i, ...., a_{bn} + b_{bn} i\}

    \vdots

    r_n=\{a_{n1} + b_{n1} i, a_{n2} + b_{n2} i, ...., a_{nn} + b_{nn} i\}


    Let v is a vector and v \in W so

    \begin{align*}v =& c_a.r_a + c_b.r_b + ... + c_n.r_n\\ =& c_a(a_{a1} + b_{a1} i, a_{a2} + b_{a2} i, ...., a_{an} + b_{an} i) + \\ & c_b(a_{b1} + b_{b1} i, a_{b2} + b_{b2} i, ...., a_{bn} + b_{bn} i)+...+\\ & c_n(a_{n1} + b_{n1} i, a_{n2} + b_{n2} i , ...., a_{nn} + b_{nn} i) \end{align}.


    Since m is vector and m = \{m_1 + n_1.i, m_2 + n_2.i, ...,m_n + n_n.i\} and orthogonal to each orthonormal basis:

    \begin{align*}m.r_a =& (m_1 + n_1.i).(a_{a1} - b_{a1} i) + (m_2 + n_2.i).(a_{a2} - b_{a2} i) + ...+ (m_n + n_n.i).(a_{an} - b_{an} i) = 0\\ m.r_b =& (m_1 + n_1.i).(a_{b1} - b_{b1} i) + (m_2 + n_2.i).(a_{b2} - b_{b2} i) + ...+ (m_n + n_n.i).(a_{bn} - b_{bn} i) = 0\\ \vdots \\  m.r_n =& (m_1 + n_1.i).(a_{n1} - b_{n1} i) + (m_2 + n_2.i).(a_{n2} - b_{n2} i) + ...+ (m_n + n_n.i).(a_{nn} - b_{nn} i) = 0\end{align*}


    \begin{align*}m.v =& m.(\overline{c_{a}.r_a}) + m.(\overline{c_{b}.r_b}) +...+ m.(\overline{c_{n}.r_n})\\ =& \Big[c_a.(m_1 + n_1.i).(a_{a1} - b_{a1} i) + c_a.(m_2 + n_2.i).(a_{a2} - b_{a2} i) + ... + c_a.(m_n + n_n.i).(a_{an} - b_{an} i)\Big] \\ &+ \Big[c_b.(m_1 + n_1.i).(a_{b1} - b_{b1} i) + c_b.(m_2 + n_2.i).(a_{b2} - b_{b2} i) + ...  + c_b.(m_n + n_n.i).(a_{bn} - b_{bn} i))\Big] \\ &+ ...+ \Big [c_n.(m_1 + n_1.i).(a_{n1} - b_{n1} i) + c_n.(m_2 + n_2.i).(a_{n2} - b_{n2} i) + ... + c_n.(m_n + n_n.i).(a_{nn} - b_{nn} i)\Big] \\ =& c_a.\Big[ (m_1 + n_1.i).(a_{a1} - b_{a1} i) + (m_2 + n_2.i).(a_{a2} - b_{a2} i) + ... + (m_n + n_n.i).(a_{an} - b_{an} i) \Big] \\ & +  c_b \Big[ (m_1 + n_1.i)(a_{b1} - b_{b1} i) + (m_2 + n_2.i).(a_{b2} - b_{b2} i) + ...+ (m_n + n_n.i).(a_{bn} - b_{bn} i)\Big] \\ &+ ... +  c_n\Big[ (m_1 + n_1.i).(a_{n1} - b_{n1} i) +  (m_2 + n_2.i).(a_{n2} - b_{n2} i) + ..+ (m_n + n_n.i).(a_{nn} - b_{nn} i) \Big] \\  =& 0 + 0 + ... + 0 \\  =& 0 \end{align*}

    As was to be shown.
    Last edited by x3bnm; September 25th 2011 at 05:40 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,317
    Thanks
    697

    Re: Vector orthogonality, orthonormal basis and subspace

    that's the general idea, but (and i could be wrong because physicists and mathematicians have slightly differing conventions on complex inner products which is confusing) i believe that the conjugate sign should be over your c's not the basis vectors. that is for a complex inner product:

    <x,y+z> = <x,y> + <x,z> but <x,ay> = \overline{a}<x,y>
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member x3bnm's Avatar
    Joined
    Nov 2009
    Posts
    300
    Thanks
    16

    Re: Vector orthogonality, orthonormal basis and subspace

    Quote Originally Posted by Deveno View Post
    that's the general idea, but (and i could be wrong because physicists and mathematicians have slightly differing conventions on complex inner products which is confusing) i believe that the conjugate sign should be over your c's not the basis vectors. that is for a complex inner product:

    <x,y+z> = <x,y> + <x,z> but <x,ay> = \overline{a}<x,y>
    Ok. I'll remember that. I can now close the thread. Thanks Deveno.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member x3bnm's Avatar
    Joined
    Nov 2009
    Posts
    300
    Thanks
    16

    Re: Vector orthogonality, orthonormal basis and subspace

    Quote Originally Posted by Deveno View Post
    that's the general idea, but (and i could be wrong because physicists and mathematicians have slightly differing conventions on complex inner products which is confusing) i believe that the conjugate sign should be over your c's not the basis vectors. that is for a complex inner product:

    <x,y+z> = <x,y> + <x,z> but <x,ay> = \overline{a}<x,y>
    I had to reopen this thread because I've a question.

    I'm having confusion over complex inner product used in this theorem. Sorry I didn't understand your explanation.

    \text{if } x = \{a + b.i \} \text{ and } y = \{ c + d.i \}

    Am I correct with this one for common inner product:
    \langle x, y \rangle = (a + b.i)(c + d.i) = a.c + a.d.i + b.c.i - b.d

    and this one for complex inner product (according to your last post....maybe I misunderstood can you kindly clarify..sorry):

    \langle x, y \rangle = [ \,\, (a + b.i).[(-1)(c + d.i)]\,\,  ] = ((-1).a.c + (-1).a.d.i + (-1).b.c.i - (-1).b.d)
    Last edited by x3bnm; September 26th 2011 at 07:39 AM.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member x3bnm's Avatar
    Joined
    Nov 2009
    Posts
    300
    Thanks
    16

    Re: Vector orthogonality, orthonormal basis and subspace

    Quote Originally Posted by Deveno View Post
    that's the general idea, but (and i could be wrong because physicists and mathematicians have slightly differing conventions on complex inner products which is confusing) i believe that the conjugate sign should be over your c's not the basis vectors. that is for a complex inner product:

    <x,y+z> = <x,y> + <x,z> but <x,ay> = \overline{a}<x,y>
    So you are sayin' to use:

    \langle x,cy \rangle =  -c\langle x,y \rangle

    for conjugate of complex number.

    Okay I understand that. Thanks.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2

    Re: Vector orthogonality, orthonormal basis and subspace

    Quote Originally Posted by Deveno View Post
    that's the general idea, but (and i could be wrong because physicists and mathematicians have slightly differing conventions on complex inner products which is confusing) i believe that the conjugate sign should be over your c's not the basis vectors. that is for a complex inner product:

    <x,y+z> = <x,y> + <x,z> but <x,ay> = \overline{a}<x,y>
    The physicists have it all over the mathematicians on this one (although even some mathematicians have started to do things the same way as the physicists). In my opinion, the inner product should definitely be conjugate linear in the second argument. The reason for that is the Dirac bra-ket notation: \langle x|y\rangle. You can "break it up" to get your bras: \langle x|, and your regular vectors, which are kets: |y\rangle. It's quite a powerful notation. The inner product being conjugate linear in the second argument is part of what makes the notation work so well.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Orthonormal basis
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: April 3rd 2011, 06:30 AM
  2. Orthonormal Basis
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: January 13th 2010, 04:54 AM
  3. orthonormal basis help
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: December 9th 2009, 03:17 PM
  4. Orthonormal basis
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: June 12th 2009, 08:13 PM
  5. linear algebra- vector-basis-subspace
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: October 10th 2008, 07:23 PM

/mathhelpforum @mathhelpforum