# Vector orthogonality, orthonormal basis and subspace

• Sep 25th 2011, 09:47 AM
x3bnm
Vector orthogonality, orthonormal basis and subspace
If a vector $\displaystyle v$ is orthogonal to orthonormal basis of a vector space $\displaystyle W$ then why is that $\displaystyle v$ is orthogonal to every vector in $\displaystyle W$?

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?
• Sep 25th 2011, 11:14 AM
x3bnm
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 $\displaystyle W$ is
$\displaystyle r_a =& \{r_{a1}, r_{a2},...,r_{an}\}, r_b=\{r_{b1},r_{b2},...,r_{bn}\},...,r_n = \{r_{n1}, r_{n2},..., r_{nn}\}$

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

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

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

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

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

\displaystyle \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 $\displaystyle m.v = 0$

\displaystyle \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.
• Sep 25th 2011, 12:51 PM
Deveno
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).
• Sep 25th 2011, 01:10 PM
x3bnm
Re: Vector orthogonality, orthonormal basis and subspace
Quote:

Originally Posted by Deveno
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.(Happy)

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.
• Sep 25th 2011, 05:23 PM
x3bnm
Re: Vector orthogonality, orthonormal basis and subspace
Quote:

Originally Posted by Deveno
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 $\displaystyle W$ is
$\displaystyle r_a = \{a_{a1} + b_{a1} i, a_{a2} + b_{a2} i, ...., a_{an} + b_{an} i\}$

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

$\displaystyle \vdots$

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

Let $\displaystyle v$ is a vector and $\displaystyle v \in W$ so

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

\displaystyle \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*}

\displaystyle \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.
• Sep 25th 2011, 05:49 PM
Deveno
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:

$\displaystyle <x,y+z> = <x,y> + <x,z>$ but $\displaystyle <x,ay> = \overline{a}<x,y>$
• Sep 25th 2011, 05:54 PM
x3bnm
Re: Vector orthogonality, orthonormal basis and subspace
Quote:

Originally Posted by Deveno
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:

$\displaystyle <x,y+z> = <x,y> + <x,z>$ but $\displaystyle <x,ay> = \overline{a}<x,y>$

Ok. I'll remember that. I can now close the thread. Thanks Deveno.
• Sep 25th 2011, 06:54 PM
x3bnm
Re: Vector orthogonality, orthonormal basis and subspace
Quote:

Originally Posted by Deveno
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:

$\displaystyle <x,y+z> = <x,y> + <x,z>$ but $\displaystyle <x,ay> = \overline{a}<x,y>$

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

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

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

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

$\displaystyle \langle x, y \rangle = [ \,\, (a + b.i).[(-1)(c + d.i)]\,\, ] = ((-1).a.c + (-1).a.d.i + (-1).b.c.i - (-1).b.d)$
• Sep 29th 2011, 03:36 AM
x3bnm
Re: Vector orthogonality, orthonormal basis and subspace
Quote:

Originally Posted by Deveno
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:

$\displaystyle <x,y+z> = <x,y> + <x,z>$ but $\displaystyle <x,ay> = \overline{a}<x,y>$

So you are sayin' to use:

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

for conjugate of complex number.

Okay I understand that. Thanks.
• Sep 29th 2011, 07:04 AM
Ackbeet
Re: Vector orthogonality, orthonormal basis and subspace
Quote:

Originally Posted by Deveno
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:

$\displaystyle <x,y+z> = <x,y> + <x,z>$ but $\displaystyle <x,ay> = \overline{a}<x,y>$

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: $\displaystyle \langle x|y\rangle.$ You can "break it up" to get your bras: $\displaystyle \langle x|,$ and your regular vectors, which are kets: $\displaystyle |y\rangle.$ 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.