Substitution Rule for Matrices

Theorem
Let $\mathbf A$ be a square matrix of order $n$.

Then:
 * $(1): \quad \ds \sum_{j \mathop = 1}^n \delta_{i j} a_{j k} = a_{i k}$
 * $(2): \quad \ds \sum_{j \mathop = 1}^n \delta_{i j} a_{k j} = a_{k i}$

where:
 * $\delta_{i j}$ is the Kronecker delta
 * $a_{j k}$ is element $\tuple {j, k}$ of $\mathbf A$.

Proof
By definition of Kronecker delta:
 * $\delta_{i j} = \begin {cases}

1 & : i = j \\ 0 & : i \ne j \end {cases}$

Thus:
 * $\delta_{i j} a_{j k} = \begin {cases}

a_{i k} & : i = j \\ 0 & : i \ne j \end {cases}$ and:
 * $\delta_{i j} a_{k j} = \begin {cases}

a_{k i} & : i = j \\ 0 & : i \ne j \end {cases}$

from which the result follows.