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I know the identity $(AB)^T = B^TA^T$. I tried to research the exact reason why this is the case and could not find a good description. My professor said the reason is because "you ought to put on your socks before your shoes" which did not really resonate with me...

Ty Jensen
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2 Answers2

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One way of defining the transpose is the matrix $M$ that satisfies $\langle Mx, y \rangle = \langle x, A y \rangle $ for all $x,y$.

It is not hard to show that $M=A^T$ where $A^T$ is defined in the usual way.

Consequently, $ \langle (AB)^Tx, y \rangle = \langle x, ABy \rangle = \langle A^Tx, By \rangle = \langle B^T A^Tx, y \rangle $ and hence $(AB)^T = B^T A^T$.

copper.hat
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I don't have the reputation to comment, but to add to the answer above, the adjoint of a linear operator $T$ on a vector space $V$, denoted $T^*$, is the unique function such that $$\langle Tv, w \rangle = \langle v, T^*w \rangle$$ for all $v, w \in V$.

It is straightforward to show that $T^* $ is a linear map, that $(TS)^* = S^*T^\ast$ for any other linear operator $S$ on $V$, and that the matrix representation of $T^\ast$ with respect to any orthonormal basis of $V$ is the conjugate transpose of the matrix representation of $T$ with respect to that basis. In the case of a real $n \times n$ matrices, using the dot product on $\mathbb{R}^n$, these results imply $(AB)^T = B^T A^T$.

The book "Linear Algebra Done Right" by Sheldon Axler covers these results, for example.

Ameer Q
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