Proof

Let A and B be matrices. Then
$(A B)^{t} = B^{t} A^{t}$

Suppose that A is an $n \times m$ matrix and B is a $m \times ℓ$ matrix as follows

$A = [\begin{matrix} a_{11} & a_{12} & \dots & a_{1 m} \\ a_{21} & a_{22} & \dots & a_{2 m} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{n 1} & a_{n 2} & \dots & a_{n m} \end{matrix}], B = [\begin{matrix} b_{11} & b_{12} & \dots & b_{1 ℓ} \\ b_{21} & b_{22} & \dots & b_{2 ℓ} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ b_{m 1} & b_{m 2} & \dots & b_{m ℓ} \end{matrix}],$

Let us denote the $(i, j)$ -entry in A and B by $[A]_{i j}$ and $[B]_{i j}$ .

Then we have

$\begin{array}{rcl} [A]_{i j} = a_{i j} and [A^{t}]_{i j} = a_{j i} =: A_{i j}^{'} \\ [B]_{i j} = b_{i j} and [B^{t}]_{i j} = b_{j i} =: B_{i j}^{'} \\ [A B]_{i j} = \sum_{k = 1}^{n} a_{i k} b_{k j} and [(A B)^{t}]_{i j} = [A B]_{j i} = \sum_{k = 1}^{n} a_{j k} b_{k i} \end{array}$

Therefore, we obtain

$[A B^{t}]_{i j} = [A B]_{j i} = \sum_{k = 1}^{n} a_{j k} b_{k i} = \sum_{k = 1}^{n} b_{k i} a_{j k} = \sum_{k = 1}^{n} B_{i k}^{'} A_{k j}^{'} = [B^{t} A^{t}]_{i j}$

Linear Algebra

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Proof