Orthogonal Matrix in Index notation

Question

The following is an excerpt from Prof David Tong's notes on Classical Dynamics. This is in the section on Kinematics:

I am unable to understand how the equation circled in red, is the same as the condition for an orthogonal matrix given after 'in other words.' I may be missing something obvious, but I'm not sure what it is. What I have tried is the following:

$$R_{ac}R_{bc} = \delta_{ab} \implies R_{ac}R_{cb}^T = \delta_{ab} \implies RR^T = I$$

But how does this guarantee that $R^TR = I$ ?

Further, for two matrices $A_{m\times n}$ and $B_{n\times l}$ the product in index notation is given by:

$$(AB)_{ml} = \sum_n A_{mn}B_{nl}$$

For a general square matrix M, the condition on it being orthogonal is that:

$$M^TM = MM^T = I$$ If I try to write the first condition in index notation I seem to get:

$$(M^TM)_{jj} = \sum_iM_{ji}M_{ij}$$

But how do I get the non diagonal elements of this product at all? Do I change the index on one of the matrices?

Throughout the rest of this section wherever I expect a certain index ordering, I find that he has written down the opposite (for example here, I think it should be $R_{ac}R_{cb} = \delta_{ab}$) so I think there is something I don't understand with the notation, or something more fundamental that I am missing.

It would be better to attach the picture in light-mode. – VIVID Dec 23 '20 at 15:01 — VIVID, Dec 23 '20 at 15:01
This is the first occurrence of $R_{xy}$ in the section – newtothis Dec 23 '20 at 15:04 — newtothis, Dec 23 '20 at 15:04
Oh yes, it seems to be mentioned at the end. – VIVID Dec 23 '20 at 15:07 — VIVID, Dec 23 '20 at 15:07
I've attached the picture in light mode as well. – newtothis Dec 23 '20 at 15:09 — newtothis, Dec 23 '20 at 15:09

score 1 · Answer 1 · answered Dec 23 '20 at 16:31

1

You essentially have the right idea in the first and second equations you wrote down. I think you're just confused about how to apply index notation.

For what Tong wrote down (exclusively in index notation) we have

$$\delta_{ab} = R_{ac} R_{bc} $$

It is easier to understand this formula if we write it in expanded summation notation:

$$\delta_{ab} = \sum_{c} R_{ac} R_{bc}$$

Then we see that we have

$$\delta_{ab} = \sum_{c} R_{ac} R_{bc} = \sum_{c} R_{ac} R_{cb}^{T},$$

which by your third equation implies $\delta_{ab} = (RR^T)_{ab}$. Since $\delta_{ab}$ defines the components of the identity matrix, we have $I = RR^{T}$.

answered Dec 23 '20 at 16:31

motherboard

652

Yes, I was able to get this as well. My question is, shouldn't we also prove $R^TR = I$ ? If this implies that, I fail to see how. – newtothis Dec 23 '20 at 17:32
It is a general fact that if $A$ and $B$ are square matrices and $AB=I$, then $BA=I$. Hence the fact that $RR^T=I$ implies that $R^T R=I$. See for instance https://math.stackexchange.com/q/3852/323432. – projectilemotion Dec 23 '20 at 17:47

Orthogonal Matrix in Index notation

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