Prove $\ell_{ki}\ell_{kj}=\delta_{ij}$

Question

Prove $\ell_{ki}\ell_{kj}=\delta_{ij}$

where $\{\hat{\mathbf{e}}_i\}$ and $\{\hat{\mathbf{e}}_i'\}$ are sets of orthonormal basis vectors for $i\in\{1,2,3\}$, $\ell$'s are the direction cosines such that $\ell_{ij}=\cos{(\hat{\mathbf{e}}_i',\hat{\mathbf{e}}_j)}=\hat{\mathbf{e}}_i'\cdot\hat{\mathbf{e}}_j$, and $\delta$ is the Kronecker delta function such that $$\delta_{ij}=\begin{cases} 1, i=j\\ 0, i\ne j\end{cases}$$ This uses indicial notation (free indices and dummy indices and associated summation conventions). I'm not sure how universal the notation is but I'm using the notation from "The Linearized Theory of Elasticity" by William S. Slaughter. I think I've provided as much as I need, but please ask if you feel I'm missing information.

I know in the end I'm going to have to reduce the problem to $\ell_{ki}\ell_{kj}=\hat{\mathbf{e}}_i\cdot\hat{\mathbf{e}}_j=\delta_{ij}$ or $\ell_{ki}\ell_{kj}=\hat{\mathbf{e}}_i'\cdot\hat{\mathbf{e}}_j'=\delta_{ij}$. And I understand how to prove $\ell_{ik}\ell_{jk}=\delta_{ij}$: $$\begin{align*} \ell_{ik}\ell_{jk}&=(\hat{\mathbf{e}}_i'\cdot\hat{\mathbf{e}}_k)\ell_{jk}\\ &=\hat{\mathbf{e}}_i'\cdot(\ell_{jk}\hat{\mathbf{e}}_k)\\ &=\hat{\mathbf{e}}_i'\cdot\hat{\mathbf{e}}_j'\quad(\text{because in general }\hat{\mathbf{e}}_i'=\ell_{ij}\hat{\mathbf{e}}_j)\\ &=\delta_{ij} \end{align*}$$

However, when I try to follow the same type of substitutions I seem to hit a dead end. Can someone provide me a hint, not a full solution? I would like to solve it on my own.

EDIT: I think I have to use $\hat{\mathbf{e}}_i=\ell_{ji}\hat{\mathbf{e}}_j'$, still working on that route, though.

Answer 1

Yeah, I needed to rewrite everything and start on a new piece of paper before I realized the EDITed reverse transformation was relevant.

The proof is as follows: $$\begin{align*} \ell_{ki}\ell_{kj}&=(\hat{\mathbf{e}}_k'\cdot\hat{\mathbf{e}}_i)\ell_{kj}\\ &=(\ell_{kj}\hat{\mathbf{e}}_k')\cdot\hat{\mathbf{e}}_i\\ &=\hat{\mathbf{e}}_j\cdot\hat{\mathbf{e}}_i\quad(\text{because of my edited relationship in the question})\\ &=\delta_{ji}\\ &=\delta_{ij} \end{align*}$$

Answer 2

There is a simpler proof. Since you are transforming from one set of orthonormal Cartesian coordinates to another, your change of basis matrix $[l]$ is orthogonal (its transpose is also its inverse).

Thus, $[l][l]^T = [l]^T[l]=[I]$.

Clearly then, $l_{kj}l_{ki}$ = $l^T_{ik}l_{kj} = \delta_{ij}$, since this is just the matrix product $[l]^T [l]$ in components.