Why does the following expression equal the Kronecker delta?

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I am currently pursuing a self study in tensor analysis and tensor algebra, utilizing the book Tensors - The mathematics of Relativity Theory and Continuum Mechanics - by Anadijiban Das, and came across the following Theorem 1.2.15:

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In looking at equation (i) (or statement (1.9)) why does the left hand side equal $\delta^j_i$? Is there some kind of index simplification law that I am missing? I am very new to tensors so please give a gentle yet precise explanation on why this is.

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Conceptually, $\lambda$ and $\mu$ are just the coordinate transforms back and forth between the two bases. This means they are inverses of each other as coordinate transforms, and so their composition must be the identity transform, which is the Kronecker delta.

To see it algebraically, plug one equation in 1.8 into the other. You get $$ \mathbf{e}_j = \mu_j^k\lambda_k^i \mathbf{e}_i. $$ Now, since the $\mathbf{e}_i$ are linearly independent, the only way to write $\mathbf{e}_j$ as a linear combination of the basis vectors is $\mathbf{e}_j$. Thus, $\mu_j^k\lambda_k^i = 0$ if $i\ne j$ and $\mu_j^k\lambda_k^i = 1$ if $i = j$, which is the definition of $\delta_j^i$. So $\mu_j^k\lambda_k^i = \delta_j^i$.