linear algebra ----- (AB)T = BTAT 证明_证明(ab)t=btat-优快云博客

本文详细解析了矩阵乘法转置的性质，即(AB)T=BTAT，通过定义矩阵A和B的元素，解释了如何通过矩阵A和B的乘积C的转置CT来验证这一性质。文章深入探讨了矩阵转置的概念及其在数学证明中的应用。

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Given an $m×nm\times n$ -matrix $A$ and an $n×pn\times p$ -matrix $B$ . Prove that $AB)^T = B^TA^T$ .

Here is my attempt:

Write the matrices $A$ and $B$ as $A = [a_{ij}]$ and $B = [b_{ij}]$ , meaning that their $(i,j)\left(i,j\right)$ -th entries are $a_{ij}$ and $b_{ij}$ , respectively.

Let $C=AB=[c_{ij}]$ , where $cij=∑k=1naikbkjc_{ij} = \sum_{k=1}^n a_{ik}b_{kj}$ , the standard multiplication definition.

We want $AB)^T = C^T = [c_{ji}]$ . That is the element in position $j, i$ is $∑k=1naikbkj\sum_{k=1}^n a_{ik}b_{kj}$ . For instance, if $i = 2, j = 3$ , then the element in $2, 3$ of $C$ is that sum, but the element in position $3, 2$ of the transpose is that sum.

I need to get the same value for the element in position $3, 2$ of the right side.

The transpose matrices are $B^T=[b_{ji}], A^T=[a_{ji}]$ . They are size $\times n$ and $\times m$ . That is, they switch rows and columns.

Let $D = B^T A^T = [d_{ji}]$ . I write the indices backwards because if I want the element in position $3, 2$ , that is, $i = 2, j = 3$ just like on the other side.