# How does one figure out the orthonormal basis from just the inner product without brute force for an inner product space?

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.

• 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 Aug 15, 2016 at 5:48
• @JoelCohen, what if you don't even have that information? Put it into context, what is a basis for the set $B$ I described? Aug 15, 2016 at 5:49
• 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. Aug 15, 2016 at 6:22
• @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 $A_j$. Aug 15, 2016 at 6:26
• @Spencer, yes, sorry. Aug 15, 2016 at 6:29

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.

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

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.

• 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. Aug 15, 2016 at 5:51
• 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. Aug 15, 2016 at 5:53