2
$\begingroup$

Lets say $(V, \left < \right> )$ is an inner product space. How does one obtain an orthonormal basis $(e_i)$ from just the inner product by directly calculating?

By that I mean if $x \in V$, then to obtain say an orthogonal basis, we find $y$ such that $(x,y) =0$. To get norm $1$, we reinforce that $(y,y) = 1$.

Let's actually put this into context with an example. I will adapt this from this link from physics with some slight changes.

Consider the skew-Hermitian matrix group from Linear Algebra with $0$ trace $B = \{ A \in M_{n\times n}(\mathbb{C}) : A = -A^* \}$ with the inner product $(A_i, A_j) = -\frac{1}{2}tr(A_iA_j)$ (I added $1/2$ that we obtain orthonormal matrices). It can be shown that an orthonormal basis contains the Gell-Mann matrices.

Now one can obtain those Gall-Mann by just deducing the skew-Hermitian property without writing down all the diffuclt procedures I wrote in the beginning - writing a generic element $x \in V$ and figuring out what the formulas are for $y$ so that $(x,y) = 0$ and $(y,y) =1$. But this seems like it's only possible if we know what space we are working on. How does one do this if we have no idea on the underlying space? I am imagine this gets more difficult if the group is more complicated.

$\endgroup$
5
  • $\begingroup$ You can start from an arbitrary basis of your vector space, and from it compute an orthnormal one using the Gram-Schmidt process : en.m.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process $\endgroup$
    – Joel Cohen
    Aug 15, 2016 at 5:48
  • $\begingroup$ @JoelCohen, what if you don't even have that information? Put it into context, what is a basis for the set $B$ I described? $\endgroup$ Aug 15, 2016 at 5:49
  • $\begingroup$ denote $E_{i, j} $ the matrix that has a single one at position $(i, j) $, and zeros everywhere else. A basis of your spaces is given by the matrices $E_{i, j}-E_{j, i}$, $i(E_{i, j}+E_{j, i}) $ and $iE_{i, i}$ for $j < i$. You can find using gauss elimination on the system of equations defining your space. $\endgroup$
    – Joel Cohen
    Aug 15, 2016 at 6:22
  • $\begingroup$ @jacobsmith, did you mean to have a comma in your formula $-\frac12 tr(A_i,A_j)$? I would think this should be the trace of the product of the matrices $A_i$ and $A_j$. $\endgroup$
    – Spencer
    Aug 15, 2016 at 6:26
  • $\begingroup$ @Spencer, yes, sorry. $\endgroup$ Aug 15, 2016 at 6:29

2 Answers 2

2
$\begingroup$

You seem to have some confusion on how to get a starting basis to even orthogonalize. I'll show the method I use in the case $n=2$.

The space $B$ will be a subpace of $M_{2\times2}(\mathbb{C})$. Take some generic matrix $A$ and set up the equation for a skew Hermitian matrix.

$$ A^\dagger + A = 0 $$

$$ \left(\begin{array} \ a & b \\ c & d \end{array}\right)^\dagger+ \left(\begin{array} \ a & b \\ c & d \end{array}\right)=0 $$

$$ \left(\begin{array} \ \bar{a} & \bar{c} \\ \bar{b} & \bar{d} \end{array}\right)+ \left(\begin{array} \ a & b \\ c & d \end{array}\right)=0 $$

$$ \left(\begin{array} \ \bar{a} + a & \bar{c} + b \\ \bar{b}+c & \bar{d}+d \end{array}\right)=0 $$

From this we get three independent equations.

$$\Re(a)=0$$ $$\Re(d)=0$$ $$b=-\bar{c}$$

add to this the equation which comes from the requirement that the trace be zero.

$$ a+d=0$$

So the matrix has the general form,

$$ \left( \begin{array} \ ix & y+iz \\ -(y-iz) & -ix \end{array}\right)$$

where now $x,y,z,w$ are all real.

$$ \left( \begin{array} \ xi & y+iz \\ -(y-iz) & -ix \end{array}\right) = x \left( \begin{array} \ i & 0 \\ 0 & -i \end{array}\right) + y \left( \begin{array} \ 0 & 1 \\ -1 & 0 \end{array}\right) + z \left( \begin{array} \ 0 & i \\ i & 0 \end{array}\right) $$

Clearly the four matrices multiply $x,y,$ and $z$ respectively form a basis for the space $B$.

This method can be used to get a basis for $B$ for any given $n$. Once you have this basis you can then orthogonalize it using Graham Schmidt as indicated by the other answer.

If you try my method for general $n$ using index notation you may be able to derive the more general result.


Notice that the result ended up just being $i$ times the Pauli matrices. These are in fact orthogonal with respect to your inner product. If you look at the 3x3 Gell-Mann matrices you will see the same structure of the Pauli matrices in them. This may lead to a clever way of deducing the orthogonal basis.

$\endgroup$
2
  • $\begingroup$ Huh I can't believe I actually forgot about something this simple. Just factor... $\endgroup$ Aug 15, 2016 at 7:04
  • $\begingroup$ So it appears that the inner product is only for normalization. $\endgroup$ Aug 15, 2016 at 7:04
1
$\begingroup$

There is no canonical choice of orthonormal basis for an inner product space, just as there is no canonical choice of basis for a vector space. One doesn't just spring forth from the situation or arise naturally in any way.

Of course you can take a basis and create an orthonormal basis from it using the Gram-Schmidt process, but I assume you would consider that brute force.

$\endgroup$
2
  • $\begingroup$ Yes resorting to Gram-Schdmit is brute force, it's actually no different from the basic procedure I described. Well all I want is to find one orthonormal basis. I want to work with the information that I don't even know how to get an arbitrary basis on my set. $\endgroup$ Aug 15, 2016 at 5:51
  • $\begingroup$ My point is: how do you find even just one basis for a vector space, if all you're told is "$V$ is a vector space"? You can't get it for free. My point is that the same thing is true about finding an orthonormal basis of an inner product space. $\endgroup$ Aug 15, 2016 at 5:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .