Summary of what's gone on before. Got through the index notation for gradients and whatnot. I was left a little bit baffled by the notation for the orthogonal transpose identity. Consequently, I'm digging back into it.

In this set of notes, the transpose, orthogonal identity, $M_{ki}M_{kj} = \delta_{ij}$, is first hammered out. It then becomes obvious what's going on. The dummy summing of the two row indices gave us the equivalent of a matrix multiply where it's row times row instead of row times column. The rows of a matrix that should have been transposed however are the same as the columns of one that wasn't. In other words by forcing a different type of matrix multiply, they teased out the transpose for free.

Here's the hammering through bit.

Let's take as an example, the simple rotation matrix about the z axis. Keep in mind that it has already been explained above why this will work for any orthogonal matrices and that this is just a concrete version that I already happened to work out before figuring out the general case above.

$M_z = \begin{pmatrix}

cos \theta & sin \theta & 0\\

-sin \theta & cos \theta & 0\\

0 & 0 & 1

\end{pmatrix}$

Now, with the multiply defined as above, we will get three resulting matrices for $k=1$, $k=2$, and $k=3$, and then sum them all together. The k's are fixed, so the only combinations of terms that need to be multiplied are the i's and the j's . For each k, we'll get the 9 terms indexed by i and j. So, for the k's, 1 through 3, respectively, we get:

$\begin{pmatrix}

cos^2 \theta & cos \theta sin \theta & 0\\

sin \theta cos \theta & sin^2 \theta & 0\\

0 & 0 & 0

\end{pmatrix}+\begin{pmatrix}

sin^2 \theta & -sin \theta cos \theta & 0\\

-cos \theta sin \theta & cos^2 \theta & 0\\

0 & 0 & 0

\end{pmatrix}+\begin{pmatrix}

0 & 0 & 0\\

0 & 0 & 0\\

0 & 0 & 1

\end{pmatrix}$

which adds term by term to give $\delta_{ij}$.

NOTE: Look at this a lot! The multiply method of swizzling through all the combinations of i and j tends to slip my mind!

In this set of notes, the transpose, orthogonal identity, $M_{ki}M_{kj} = \delta_{ij}$, is first hammered out. It then becomes obvious what's going on. The dummy summing of the two row indices gave us the equivalent of a matrix multiply where it's row times row instead of row times column. The rows of a matrix that should have been transposed however are the same as the columns of one that wasn't. In other words by forcing a different type of matrix multiply, they teased out the transpose for free.

Here's the hammering through bit.

Let's take as an example, the simple rotation matrix about the z axis. Keep in mind that it has already been explained above why this will work for any orthogonal matrices and that this is just a concrete version that I already happened to work out before figuring out the general case above.

$M_z = \begin{pmatrix}

cos \theta & sin \theta & 0\\

-sin \theta & cos \theta & 0\\

0 & 0 & 1

\end{pmatrix}$

Now, with the multiply defined as above, we will get three resulting matrices for $k=1$, $k=2$, and $k=3$, and then sum them all together. The k's are fixed, so the only combinations of terms that need to be multiplied are the i's and the j's . For each k, we'll get the 9 terms indexed by i and j. So, for the k's, 1 through 3, respectively, we get:

$\begin{pmatrix}

cos^2 \theta & cos \theta sin \theta & 0\\

sin \theta cos \theta & sin^2 \theta & 0\\

0 & 0 & 0

\end{pmatrix}+\begin{pmatrix}

sin^2 \theta & -sin \theta cos \theta & 0\\

-cos \theta sin \theta & cos^2 \theta & 0\\

0 & 0 & 0

\end{pmatrix}+\begin{pmatrix}

0 & 0 & 0\\

0 & 0 & 0\\

0 & 0 & 1

\end{pmatrix}$

which adds term by term to give $\delta_{ij}$.

NOTE: Look at this a lot! The multiply method of swizzling through all the combinations of i and j tends to slip my mind!

## Comments

It's not terribly enlightening, but I often found it helpful to prove index identities to myself using "brute force" by programming the summation into a computer and confirming the left-hand side is the same as the right-hand side. The reason I would do this is sort of two-fold.

Sometimes it was helpful just to prove to myself the identity worked. I didn't really believe it until I saw it in action.

And sometimes I misunderstood the index notation and got the identity wrong! The brute force would make that clear.

I like your idea about programming the identities. I guess I'll get better with practice, but these are pretty difficult for me to see now, and a computer program, as you pointed out would force me to confront the inconsistencies in my thinking about indices more quickly.